last updated: May 1, 2025

Stochastic Processes and Financial Mathematics
(part one)

Chapter A Solutions to exercises (part one)

Chapter 1
  • 1.1 At time 1 we would have 10(1+r) in cash and 5 units of stock. This gives the value of our assets at time 1 as 10(1+r)+5S1.

  • 1.2

    • (a) Assuming we don’t buy or sell anything at time 0, the value of our portfolio at time 1 will be x(1+r)+yS1, where S1 is as in the solution to 1.1. Since we need to be certain of paying off our debt, we should assume a worst case scenario for S1, that is S1=sd. So, we are certain to pay off our debt if and only if

      x(1+r)+ysdK.

    • (b) Since certainty requires that we cover the worst case scenario of our stock performance, and d<1+r, our best strategy is to exchange all our assets for cash at time t=0. This gives us x+ys in cash at time 0. At time 1 we would then have (1+r)(x+ys) in cash at time 0 and could pay our debt providing that

      (1+r)(x+ys)K.

  • 1.3

    • (a) We borrow s cash at time t=0 and use it to buy one stock. Then wait until time t=1, at which point we will a stock worth S1, where S1 is as in 1.1. We sell our stock, which gives us S1 in cash. Since S1sd>s(1+r), we repay or debt (plus interest) which costs us s(1+r). We then have

      S1s(1+r)>0

      in cash. This is an arbitrage.

    • (b) We perform the same strategy as in (a), but with the roles of cash an stock swapped: We borrow 1 stock at time t=0 and then sell it for s in cash. Then wait until time t=1, at which point we will have s(1+r) in cash. Since the price of a stock is now S1, where S1 is as in 1.1, and S1us<(1+r)s, we now buy one stock costing S1 and repay the stockbroker, leaving us with

      s(1+r)S1>0

      in cash. This is an arbitrage.

  • 1.4 We can calculate, using integration by parts, that

    E[X]=xfX(x)dx=0xλeλxdx=1λ. Similarly, we can calculate that

    E[X2]=x2fX(x)dx=0x2λeλxdx=2λ2

    This gives that

    var(X)=E[X2]E[X]2=1λ2.

  • 1.5 We can calculate

    E[Xn]=0P[Xn=0]+n2P[Xn=n2]=n2n=n

    as n. Also,

    P[|Xn|>0]=P[Xn=n2]=1n0

    as n.

  • 1.6 We have

    P[Yy]=P[Xμσy]=P[Xμ+yσ]=μ+yσ12πσ2e(xμ)22σ2dx. We want to turn the above integral into the distribution function of the standard normal. To do so, substitute z=xμσ, and we obtain

    P[Yy]=y12πσ2ez22σdz=y12πez22dz. Hence, Y has the distribution function of N(0,1), which means YN(0,1).

  • 1.7 For p1,

    1xpdx=[xp+1p+1]x=1

    and we obtain a finite answer only if p+1<0. When p=1, we have

    1x1dx=[logx]x=1=,

    so we conclude that 1xpdx is finite if and only if p>1.

  • 1.8 As n,

    • en0 because e>1.

    • sin(nπ2) oscillates through 0,1,0,1 and does not converge.

    • cos(nπ)n converges to zero by the sandwich rule since |cos(nπ)n|1n0.

    • i=1n2i=12n is a geometric series, and 12n10=1.

    • i=1n1i tends to infinity, because

      12+13+14+15+16+17+18+19+110++111+112+113+114+115+116+.

      each term contained in a is greater than or equal to 12.

  • 1.9 Recall that n=1n2<. Define n1=1 and then

    nr+1=inf{kN;k>nr and |xk|r2}.

    Then for all r we have |xnr|r2. Hence,

    r=1|xnr|r=1r2<.

Chapter 2
  • 2.1 The result of rolling a pair of dice can be written ω=(ω1,ω2) where ω1,ω2{1,2,3,4,5,6}. So a suitable Ω would be

    Ω={(ω1,ω2);ω1,ω2{1,2,3,4,5,6}}.

    Of course other choices are possible too. Since our choice of Ω is finite, a suitable σ-field is F=P(Ω).

  • 2.2

    • (a) Consider the case of F. We need to check the three properties in the definition of a σ-field.

      • 1. We have F so the first property holds automatically.

      • 2. For the second we check compliments: ΩΩ=, Ω{1}={2,3}, Ω{2,3}={1} and ΩΩ=; in all cases we obtain an element of F.

      • 3. For the third, we check unions. We have {1}{2,3}=Ω. Including into a union doesn’t change anything; including Ω into a union results in Ω. This covers all possible cases, in each case resulting in an element of F.

      So, F is a σ-field. Since F is just F with the roles of 1 and 2 swapped, by symmetry F is also a σ-field.

    • (b) We have {1}FF and {2}FF, but {1}{2}={1,2} and {1,2}FF. Hence FF is not closed under unions; it fails property 3 of the definition of a σ-field.

      However, FF is the intersection of two σ-fields, so is automatically itself a σ-field. (Alternatively, note that FF={,Ω}, and check the definition.)

  • 2.3

    • (a) Let Am be the event that the sequence (Xn) contains precisely m heads. Let Am,k be the event that we see precisely m heads during the first k tosses and, from then on, only tails. Then,

      P[Am,k]=P[m heads in X1,,Xk, no heads in Xk+1,Xk+2,]=P[m heads in X1,,Xk,]P[ no heads in Xk+1,Xk+2,]=P[m heads in X1,,Xk,]×0=0. If Am occurs then at least one of Am,1,Am,2, occurs. Hence,

      P[Am]k=0P[Am,k]=0

      .

    • (b) Let A be the event that the sequence (Xn) contains finitely many heads. Then, if A occurs, precisely one of A1,A2, occurs. Hence,

      P[A]=m=0P[Am]=0.

      That is, the probability of having only finite many heads is zero. Hence, almost surely, the sequence (Xn) contains infinitely many heads.

      By symmetry (exchange the roles of heads and tails) almost surely the sequence (Xn) also contains infinitely many tails.

  • 2.4

    • (a) F1 contains the information of whether a 6 was rolled.

    • (b) F2 gives the information of which value was rolled if a 1, 2 or 3 was rolled, but if the roll was in {4,5,6} then F1 cannot tell the difference between these values.

    • (c) F3 contains the information of which of the sets {1,2}, {3,4} and {5,6} contains the value rolled, but gives no more information than that.

  • 2.5

    • (a) The values taken by X1 are

      X1(ω)={0 if ω=3,1 if ω=2,4,4 if ω=1,5.

      Hence, the pre-images are X11(0)={3}, X11(1)={2,4} and X11(4)={1,5}. These are all elements of G1, so X1mG1. However, they are not all elements of G2 (in fact, none of them are!) so X1mG2.

    • (b) Define

      X2(ω)={0 if ω=1,2,1 otherwise.

      Then the pre-images are X21(0)={1,2} and X21(1)={3,4,5}. Hence X2mG2.

      It isn’t possible to use unions/intersections/complements to make the set {3,4,5} out of {1,5},{2,4},{3} (one way to see this is note that as soon as we tried to include 5 we would also have to include 1, meaning that we can’t make {3,4,5}). Hence X2mG1.

  • 2.6 We have X(TT)=0, X(HT)=X(TH)=1 and X(HH)=2. This gives

    σ(X)={,{TT},{TH,HT},{HH},{TT,TH,HT},{TT,HH},{TH,HT,HH},Ω}.

    To work this out, you could note that X1(i) must be in σ(X) for i=0,1,2, then also include and Ω, then keep adding unions and complements of the events until find you have added them all. (Of course, this only works because σ(X) is finite!)

  • 2.7 Since sums, divisions and products of random variables are random variables, X2+1 is a random variable. Since X2+1 is non-zero, we have that 1X2+1 is a random variable, and using products again, XX2+1 is a random variable.

    For sin(X), recall that for any xR we have

    sin(x)=n=0(1)n(2n+1)!x2n+1.

    Since X is a random variable so is (1)n(2n+1)!X2n+1. Since limits of random variables are also random variables, so is

    sinX=limNn=0N(1)n(2n+1)!X2n+1.

  • 2.8 Since X0 we have E[Xp]=E[|X|p] for all p.

    E[X]=1x12x3dx=21x2dx=2[x11]1=2<

    so XL1, but

    E[X2]=1x22x3dx=21x1dx=2[logx]1=

    so XL2.

  • 2.9 Let 1pq< and XLq. Our plan is to generalize the argument that we used to prove (2.7).

    First, we condition to see that

    |X|p=|X|p𝟙{|X|1}+|X|p𝟙{|X|>1}.

    Hence,

    E[|X|p]=E[|X|p𝟙{|X|1}+|X|p𝟙{|X|>1}]E[1+|X|q]=1+E|X|q]< Here we use that, if |X|1 then |X|p1, and if |X|>1 then since pq we have |X|p<|X|q. So XLp.

  • 2.10 First, we condition to see that

    (A.1)X=X𝟙{X<a}+X𝟙{Xa}.

    From (A.1) since X0 we have

    XX𝟙{Xa}a𝟙{Xa}. This second inequality follows by looking at two cases: if X<a then both sides are zero, but if Xa then we can use that Xa. Using monotonicity of E, we then have

    E[X]E[a𝟙{Xa}]=aE[1{Xa}]=aP[Xa].

    Dividing through by a finishes the proof.

    This result is known as Markov’s inequality.

  • 2.11 Since X0, Markov’s inequality (from 2.10) gives us that

    P[Xa]1aE[X]=0

    for any a>0. Hence,

    P[X>0]=P[X>1]+P[1X>0]=P[X>1]+P[n=1{1nX>1n+1}]=P[X>1]+n=1P[1nX>1n+1]P[X1]+n=1P[X>1n+1]=0.

Chapter 3
  • 3.1 We have S2=X1+X2 so

    E[S2|σ(X1)]=E[X1|σ(X1)]+E[X2|σ(X1)]=X1+E[X2]=X1. Here, we use the linearity of conditional expectation, followed by the fact that X1 is σ(X1) measurable and X2 is independent of σ(X1) with E[X2]=0.

    Secondly, S22=X12+2X1X2+X22 so

    E[S22]=E[X12|σ(X1)]+2E[X1X2|σ(X1)]+E[X22|σ(X1)]=X12+2X1E[X2|σ(X1)]+E[X22]=X12+1 Here, we again use linearity of conditional expectation to deduce the first line. To deduce the second line, we use that X12 and X1 are σ(X1) measurable (using the taking out what is known rule for the middle term), whereas X2 is independent of σ(X1). The final line comes from E[X2|σ(X1)]=0 (which we already knew from above) and that E[X22]=1.

  • 3.2 For in, we have XimFn, so Sn=X1+X2++Xn is also Fn measurable. For each i we have |Xi|2 so |Sn|2n, hence SnL1. Lastly,

    E[Sn+1|Fn]=E[X1++Xn|Fn]+E[Xn+1|Fn]=(X1++Xn)+E[Xn+1]=Sn. Here, we use linearity of conditional expectation in the first line. To deduce the second, we use that XimFn for in and that Xn+1 is independent of Fn in the second. For the final line we note that E[Xn+1]=132+23(1)=0.

  • 3.3

    • (a) Since XimFn for all in, we have Mn=X1X2XnmFn. Since |Xi|c for all i, we have |Mn|cn<, so Mn is bounded and hence MnL1 for all n. Lastly,

      E[Mn+1|Fn]=E[X1X2XnXn+1|Fn]=X1XnE[Xn+1|Fn]=X1XnE[Xn+1]=X1Xn=Mn. Here, to deduce the second line we use that XimFn for in. To deduce the third line we use that Xn+1 is independent of Fn and then to deduce the forth line we use that E[Xn+1]=1.

    • (b) By definition of conditional expectation (Theorem 3.1.1), we have MnL1 and MnmGn for all n. It remains only to check that

      E[Mn+1|Gn]=E[E[Z|Gn+1]|Gn]=E[Z|Gn]=Mn. Here, to get from the first to the second line we use the tower property.

  • 3.4 Note that the first and second conditions in Definition 3.3.3 are the same for super/sub-martingales as for martingales. Hence, M satisfies these conditions. For the third condition, we have that E[Mn+1|Fn]Mn (because M is a supermartingale) and E[Mn+1|Fn]Mn (because M is a submartingale. Hence E[Mn+1|Fn]=Mn, so M is a martingale.

  • 3.5

    • (a) If m=n then E[Mn|Fn]=Mn because Mn is Fn measurable. For m>n, we have

      E[Mm|Fn]=E[E[Mm|Fm1]|Fn]=E[Mm1|Fn].

      Here we use the tower property to deduce the first equality (note that m1n so Fm1Fn) and the fact that (Mn) is a martingale to deduce the second inequality (i.e. E[Mm|Fm1]=Mm1). Iterating, from m,m1,n+1 we obtain that

      E[Mm|Fn]=E[Mn+1|Fn]

      and the martingale property then gives that this is equal to Mn, as required.

    • (b) If (Mn) is a submartingale then E[Mm|Fn]Mn, whereas if (Mn) is a supermartingale then E[Mm|Fn]Mn.

  • 3.6 We have (Mn+1Mn)2=Mn+122Mn+1Mn+Mn2 so

    E[(Mn+1Mn)2|Fn]=E[Mn+12|Fn]2E[Mn+1Mn|Fn]+E[Mn2|Fn]=E[Mn+12|Fn]2MnE[Mn+1|Fn]+Mn2=E[Mn+12|Fn]2Mn2+Mn2=E[Mn+12|Fn]Mn2 as required. Here, we use the taking out what is known rule (since MnmFn) and the martingale property of (Mn).

    It follows that E[Mn+12|Fn]Mn2=E[(Mn+1Mn)2|Fn]0. We have Mn2mFn and since MnL2 we have Mn2L1. Hence (Mn2) is a submartingale.

  • 3.7 Using that E[Xn+1|Fn]=aXn+bXn1 we can calculate

    E[Sn+1|Fn]=E[αXn+1+Xn|Fn]=α(aXn+bXn1)+Xn=(αa+1)Xn+αbXn1.

    We want this to be equal to Sn, and Sn=αXn+Xn1. So we need

    αa+1=α and αb=1.

    Hence, α=1b is our choice. It is then easy to check that

    E[Sn+1|Fn]=1b(aXn+bXn1)+Xn=(ab+1)Xn+Xn1=1bXn+Xn1=Sn

    and thus Sn is a martingale, as required.

  • 3.8 Note that

    E[X1|σ(Sn)]=E[X2|σ(Sn)]==E[Xn|σ(Sn)]

    is constant, by symmetry (i.e. permuting the X1,X2,,Xn does not change the distribution of Sn). Hence, if we set E[Xi|σ(Sn)]=α then

    nα=i=1nE[Xi|σ(Sn)]=E[i=1nXi|σ(Sn)]=E[Sn|σ(Sn)]=Sn. Here we use the linearity of conditional expectation and the fact that Sn is σ(Sn) measurable. Hence,

    α=E[X1|σ(Sn)]=Snn.

Chapter 4
  • 4.1 Since Zn=eSn and SnmFn, also ZnmFn. Since |Sn|n, we have |eSn|en<, so Zn is bounded and hence ZnL1. Hence also MnmFn and MnL1.

    Lastly,

    E[eSn+1|Fn]=E[eXn+1eSn|Fn]=eSnE[eXn+1|Fn]=eSnE[eXn+1]=eSn(12e+12e1)=eSne+1e2. Here, we use the taking out what is known rule and the fact that Xn+1 (and hence also eXn+1) is independent of Fn. We also calculate the expectation of the discrete random variable Xn+1, using that P[Xn+1=1]=P[Xn+1=1]=12. This gives us that

    E[Mn+1|Fn]=MnE[Zn+1|Fn]Zn where to deduce the last line we use that e+1e2>1. Thus Mn is a martingale and Zn is a submartingale.

  • 4.2 Since XimFn for all in, we have MnmFn. We have |Xi|1 so |Sn|n and |Mn|max{(q/p)n,(q/p)n}=(q/p)n (note p>q), which implies that SnL1 and MnL1 for all n.

    We have

    E[Sn+1|Fn]=E[Sn+Xn+1|Fn]=Sn+E[Xn+1|Fn]=Sn+E[Xn+1]>Sn. Here we use the taking out what is known rule, followed by the fact that Xn+1 is independent of Fn and the relationship between conditional expectation and independence. To deduce the final step we use that E[Xn+1]=p(1)+q(1)=pq>0. Therefore, we have E[Sn+1|Fn]Sn, which means that Sn is a submartingale.

    We now look at Mn. We have

    E[Mn+1|Fn]=(q/p)SnE[(q/p)Xn+1|Fn]=(q/p)SnE[(q/p)Xn+1]=(q/p)Sn=Mn. Here we use the taking out what is known rule, followed by the fact that Xn+1 is independent of Fn and the relationship between conditional expectation and independence. To deduce the final step we use that

    E[(q/p)Xn+1]=p(q/p)1+q(q/p)1=p+q=1.

  • 4.3 Since XiFn for all in we have SnmFn where Fn=σ(X1,,Xn). Further, if we set m=max(|a|,|b|) then |Sn|nm so SnL1.

    We have

    E[Sn+1|Fn]=E[Sn+Xn+1|Fn]=Sn+E[Xn+1|Fn]=Sn+E[Xn+1]=Sn+apabpb where pb=1pa. Therefore, Sn is a martingale if and only if apa=bpb.

  • 4.4 Clearly SnmFn (recall that, by default, we take (Fn) to be the generated filtration of (Sn)) hence Sn2mFn and Sn2nFn. We have |Sn|leqn so E[|Sn2n]E[Sn2]+E[n]=2n by the triangle law and monotonicity of expectation, hence Sn2n L1. Similarly Sn2L2,

    We have

    E[Sn+12(n+1)|Fn]=E[(Xn+1+Sn)2(n+1)|Fn]=E[Xn+12|Fn]+2E[SnXn+1|Fn]+E[Sn2|Fn](n+1)=E[Xn+12]+2SnE[Xn+1]+Sn2(n+1)=1+2Sn(0)+Sn2(n+1)=Sn2n Here, the second line follows by linearity. The third line follows by the taking out what is known rule and the independence rule, using that Xn+1 is independent of Fn, whilst SnmFn. The fourth line follows because Xn+12=1 and E[Xn+1]=12(1)+12(1)=0. Hence Mn=Sn2n is a martingale.

    Subtracting n from both sides, we obtain E[Sn+12|Fn]=Sn2+1. Hence (Sn2) is a submartingale.

  • 4.5 The natural filtration is Fn=σ(X1,,Xn). We can write

    Sn+1=𝟙{Sn=0}+𝟙{Sn0}(Sn+Xn+1).

    Hence, if SnmFn then Sn+1mFn+1. Since S1=X1mFn, a trivial induction shows that SnmFn for all n. hence, LnmFn and also SnLnmFn.

    We have nSnLnn+1 (because the walk is reflected at zero and can increase by at most 1 in each time step) so SnLn is bounded and hence SnLnL1.

    Lastly,

    E[Sn+1|Fn]=E[𝟙{Sn=0}+𝟙{Sn0}(Sn+Xn+1)|Fn]=𝟙{Sn=0}+𝟙{Sn0}(Sn+E[Xn+1|Fn])=𝟙{Sn=0}+𝟙{Sn0}Sn(A.2)=𝟙{Sn=0}+Sn. Here we use linearity and the taking out what is known (SnmFn) rule, as well as that Xn+1 is independent of Fn. Hence,

    E[Sn+1Ln+1|Fn]=E[Sn+1i=0n𝟙{Si=0}|Fn]=Sn+𝟙{Sn=0}i=0n𝟙{Si=0}=Sni=0n1𝟙{Si=0}=SnLn Here we use (A.2) to deduce the second line, as well as taking out what is known. Hence, SnLn is a martingale.

  • 4.6 Let Bn be number of red balls in the urn at time n (i.e. after the nth draw is complete). Then, the proportion of red balls in the urn just after the nth step is

    Mn=Bnn+3.

    Essentially the same argument as for the two colour urn of Section 4.2 shows that Mn is adapted to the natural filtration and also in L1. We have

    E[Mn+1|Fn]=E[Mn+1𝟙{(n+1)th draw is red}+Mn+1𝟙{(n+1)th draw is not red}|Fn]=E[Bn+1n+4𝟙{(n+1)th draw is red}+Bnn+4𝟙{(n+1)th draw is not red}|Fn]=Bn+1n+4E[𝟙{(n+1)th draw is red}|Fn]+Bnn+4E[𝟙{(n+1)th draw is not red}|Fn]=Bn+1n+4Bnn+3+Bnn+4(1Bnn+3)=Bn(n+4)(n+3)(n+4)=Mn. Hence (Mn) is a martingale.

  • 4.7 By the result of 3.6, Sn2 is a submartingale.

    • (i) Mn=Sn2+n is Fn measurable (because SnmFn) and since |Sn|n we have |Mn|n2+n<, so MnL1. Further, using the submartingale property of Sn2,

      E[Sn+12+n+1|Fn]=E[Sn+12|Fn]+n+1Sn2+n

      so Mn is a submartingale. But E[M1]=20=E[M0], so (by Lemma 3.3.6) we have that (Mn) is not a martingale.

    • (ii) We have shown in Section 4.1 that (Sn) is a martingale, and in Exercise 4.4 that Mn=Sn2n is a martingale. Hence Sn2n+SnFn and E[|Sn2+Snn|]E[|Sn2n|]+E[|Sn|]< by the triangle law, so also Sn2+SnnL1. Also,

      E[Sn+12+Sn+1(n+1)|Fn]=E[Sn+12(n+1)|Fn]+E[Sn+1|Fn]=Sn2n+Sn as required. To deduce the second line we use the martingale properties of (Sn) and (Mn). As all martingales are submartingales, this case is also a submartingale.

      More generally, the sum of two martingales is always a martingale. The same is true for submartingales and also for supermartingales.

    • (iii) Mn=Snn is Fn measurable (because SnmFn) and since |Sn|n we have |Sn|1, so MnL1. Since Sn is a martingale, we have

      E[Sn+1n+1|Fn]=1n+1E[Sn+1|Fn]=Snn+1Snn.

      Hence Mn=Snn is not a martingale. Also, since Sn may be both positive or negative, we do not have Snn+1Snn, so Sn is also not a submartingale.

  • 4.8 We showed that Sn2n was a martingale in 4.7. We try to mimic that calculation and find out what goes wrong in the cubic case. So, we look at

    E[(Sn+1Sn)3|Fn]=E[Sn+13|Fn]3SnE[Sn+12|Fn]+3Sn2E[Sn+1|Fn]E[Sn3|Fn]=E[Sn+13|Fn]3Sn(Sn2+1)+3Sn2SnSn3=E[Sn+13|Fn]Sn33Sn. Here we use linearity, taking out what is known and that fact that E[Sn+12|Fn]=Sn2+1 (from 4.7). However, also Sn+1Sn=Xn+1, so

    E[(Sn+1Sn)3|Fn]=E[Xn+13|Fn]=E[Xn+13]=0. because Xn+1 is independent of Fn and Xn+13=Xn (it takes values only 1 or 1) so E[Xn+13]=0. Putting these two calculations together, we have

    (A.3)E[Sn+13|Fn]=Sn3+3Sn.

    Suppose (aiming for a contradiction) that there is a deterministic function f:NR such that Sn3f(n) is a martingale. Then

    E[Sn+13f(n+1)|Fn]=Sn3f(n).

    Combining the above equation with (A.3) gives

    f(n+1)f(n)=3Sn

    but this is impossible because Sn is not deterministic. Thus we reach a contradiction; no such function f exists.

Chapter 5
  • 5.1 The value of the portfolio (1,3) at time 1 will be

    1(1+r)+3S1=(1+r)+3S1

    where S1 is a random variable with P[S1=su]=pu and P[S1=sd]=pd.

  • 5.2

    • (a) We need a portfolio h=(x,y) such that V1h=Φ(S1)=1, so we need that

      x(1+r)+ysu=1x(1+r)+ysd=1. Solving this pair of linear equations gives x=11+r and y=0. So our replicating portfolio consists simply of 11+r cash and no stock.

      Hence, the value of this contingent claim at time 0 is x+sy=11+r.

    • (b) Now we need that

      x(1+r)+ysu=3x(1+r)+ysd=1. Which has solution x=11+r(32sususd) and y=2s(ud).

      Hence, the value of this contingent claim at time 0 is x+sy=11+r(32sususd)+2ud.

  • 5.3

    • (a) If we buy two units of stock, at time 1, for a price K then our contingent claim is Φ(S1)=2S12K.

    • (b) In a European put option, with strike price K(sd,su), we have the option to sell a single unit of stock for strike price K. It is advantageous to us to do so only if K>S1, so

      (A.4)Φ(S1)={0 if S1=su,KS1 if S1=sd.=max(KS1,0)

    • (c) We S1=su we sell a unit of stock for strike price K and otherwise do nothing. So our contingent claim is

      Φ(S1)={KS1 if S1=su,0 if S1=sd.

    • (d) Holding both the contracts in (b) and (c) at once, means that in both S1=su and S1=sd, we end up selling a single unit of stock for a fixed price K. Our contingent claim for doing so is Φ(S1)=KS1.

  • 5.4

    • (a) The contingent claims for the call and put options respectively are max(S1K,0) from (5.6) and max(KS1,0) from (A.4). Using the risk neutral valuation formula from Proposition 5.2.6 we have

      Π0call=11+rEQ[max(S1K,0)],Π0put=11+rEQ[max(KS1,0)].

    • (b) Hence,

      Π0callΠ0put=11+rEQ[max(S1K,0)max(KS1,0)]=11+rEQ[max(S1K,0)+min(S1K,0)]=11+rEQ[S1K]=11+r(EQ[S1]K) By (5.4) (or you can use Proposition 5.2.6 again) we have 11+rEQ[S1]=S0=s, so we obtain

      Π0callΠ0put=sK1+r.

  • 5.5 In the binomial model, the contingent claim of a European call option with strike price K is Φ(ST)=max(STK).

  • 5.6 A tree-like diagram of the possible changes in stock price, with the value of the contingent claim Φ(ST)=max(KS1,0) written on for the final time step looks like

    (A recombining tree of the stock prices with the contingent claim marked)

    By (5.3), the risk free probabilities in this case are

    qu=(1+0.25)0.52.00.5=0.5,qd=1qu=0.5.

    Using Proposition 5.2.6 recursively, as in Section 5.6, we can put in the value of the European put option at all nodes back to time 0, giving

    (The previous picture with the value of the contract marked)

    Hence, the value of our European put option at time 0 is 12.

    Changing the values of pu and pd do not alter value of our put option, since pu and pd do not appear in the pricing formulae.

  • 5.7 A tree-like diagram of the possible changes in stock price, with the value of the option at all nodes of the tree shown in (green) square boxes, and hedging portfolios h=(x,y) written (in red) for each node of the tree:

    (The previous picture with the hedging portfolios marked)

    The hedging portfolios are found by solving linear equations of the form x+dSy=Φ~(dS), x+uSy=Φ~(uD), where Φ~ is the contingent claim and S the initial stock price of the one period model associated to the given node. The value of the contingent claim at each node is then inferred from the hedging portfolio as V=x+Sy.

  • 5.8 A tree-like diagram of the possible changes in stock price, with the value of the option at all nodes of the tree shown in (green) square boxes, and hedging portfolios h=(x,y) written (in red) for each node of the tree:

    (A recombining tree of the stock price with the contingent claim and hedging portfolios marked)

    The hedging portfolios are found by solving linear equations of the form x+dSy=Φ~(dS), x+uSy=Φ~(uD), where Φ~ is the contingent claim and S the initial stock price of the one period model associated to the given node. The value of the contingent claim at each node is then inferred from the hedging portfolio as V=x+Sy.

    The hedging portfolio is the same at each node. This is because the contingent claim of our call option always exceeds zero i.e. we are certain that at time T=2 we would want to exercise our call option and buy a single unit of stock for a price K=60. Since there is no interest payable on cash, this means that we can replicate our option (at all times) by holding a single unit of stock, and 60 in cash.

  • 5.9 We have that (St) is adapted (see Section 5.4) and since St(dt,ut) we have also that StL1. Hence, St is a martingale under P if and only if EP[St+1|Ft]=St. That is if

    EP[St+1|Ft]=E[Zt+1St|Ft]=StEP[Z|Ft]=StEP[Z] is equal to St. Hence, St is a martingale under P if and only if EP[Zt+1]=upu+dpd=1.

    Now consider Mt=logSt, and assume that 0<d<u and 0<s, which implies St>0. Since St is adapted, so is Mt. Since StSt(dt,ut) we have Mt(tlogd,tlogu) so also MtL1. We have

    Mt=log(S0i=1tZi)=log(S0)+i=1tZi.

    Hence,

    EP[Mt+1|Ft]=logS0+i=1tlog(Zi)+E[log(Zt+1)|Ft]=Mt+EP[logZt+1]=Mt+pulogu+pdlogd. Here we use taking out what is known, since S0 and ZimFt for it, and also that Zt+1 is independent of Ft. Hence, Mt is a martingale under P if and only if pulogu+pdlogd=0.

Chapter 6
  • 6.1 Since Xn0 we have

    E[|Xn0|]=E[Xn]=122n=2(n+1)

    which tends to 0 as n. Hence Xn0 in L1. Lemma 6.1.2 then implies that also Xn0 in probability and in distribution. Also, 0Xn2n and 2n0, so by the sandwich rule we have Xn0 as n, and hence P[Xn0]=1, which means that Xn0 almost surely.

  • 6.2

    • (a) We have XnL1X and hence

      |E[Xn]E[X]|=|E[XnX]|E[|XnX|]0.

      Here we use the linearity and absolute value properties of expectation. Hence, E[Xn]E[X].

    • (b) Let X1 be a random variable such that P[X1=1]=P[X1=1]=12 and note that E[X1]=0. Set Xn=X1 for all n2, which implies that E[Xn]=00 as n. But E[|Xn0|]=E[|Xn|]=1 so Xn does not converge to 0 in L1.

  • 6.3 We have Xn=X when U=2, and |XnX|=1n when U=0,1. Hence, |XnX|1n and by monotonicity of E we have

    E[|XnX|]1n0 as n,

    which means that XnL1X.

    We can write

    P[XnX]=P[XnX,U=0]+P[XnX,U=1]+P[XnX,U=2]=P[1+1n1,U=0]+P[11n1,U=1]+P[00,U=2]=P[U=0]+P[U=1]+P[U=2]=1. Hence, Xna.s.X.

    By Lemma 6.1.2, it follows that also XnX in probability and in distribution.

  • 6.4 We have Xn=X1=1Y for all n, so Y=0Xn=1 and Y=1Xn=0. Hence |YXn|=1 for all n. So, for example, with a=12 we have P[|XnY|a]=1, which does not tend to zero as n. Hence Xn does not converge to Y in probability.

    However, Xn and y have the same distribution, given by

    P[Yx]=P[Xnx]={0 if x<012 if x[0,1)1 if x1

    so we do have P[Xnx]P[Yx] as n. Hence XndY.

  • 6.5 Let (Xn) be the sequence of random variables from 6.1. Define Yn=X1+X2++Xn.

    • (a) For each ωΩ, we have Yn+1(ω)=Yn(ω)+Xn+1(ω). Hence, (Yn(ω) is an increasing sequence. Since Xn(ω)2n for all n we have

      |Yn(ω)|21+22++2ni=12i=1,

      meaning that the sequence Yn(ω) is bounded above by 1.

    • (b) Hence, since bounded increasing sequences converge, for any ωΩ the sequence Yn(ω) converges.

    • (c) The value of Y1 is either 0 or 12, both with probability 12. The value of Y2 is either 0,14,12,34, all with probability 14. The value of Y3 is either 0,18,14,38,12,58,34,78 all with probability 18.

    • (d) We can guess from (c) that the Yn are becoming more and more evenly spread across [0,1], so we expect them to approach a uniform distribution as n.

    • (e) We will use induction on n. Our inductive hypothesis is

      • (IH)n: For all k=0,1,,2n1, it holds that P[Yn=k2n]=2n.

      In words, this says that Yn is uniformly distributed on {k2n;k=0,1,,2n1}.

      For the case n=1, we have Y1=X1 so P[Y1=0]=P[Y1=12]=12, hence (IH)1 holds.

      Now, assume that (IH)n holds. We want to calculate P[Yn+1=k2(n+1) for k=0,1,,2n+11. We consider two cases, dependent on whether k is even or odd.

      • If k is even then we can write k=2j for some j=0,,2n1. Hence,

        P[Yn+1=k2(n+1)]=P[Yn+1=j2n]=P[Yn=j2n and Xn+1=0]=P[Yn=j2n]P[Xn+1=0]=2n12=2(n+1). Here we use that Yn and Xn+1 are independent, and use (IH)n to calculate P[Yn=j2n].

      • Alternatively, if k is odd then we can write k=2j+1 for some j=0,,2n1. Hence,

        P[Yn+1=k2(n+1)]=P[Yn+1=j2n+2(n+1)]=P[Yn=j2n and Xn+1=2(n+1)]=P[Yn=j2n]P[Xn+1=2(n+1)]=2n12=2(n+1). Here, again, we use that Yn and Xn+1 are independent, and use (IH)n to calculate P[Yn=j2n].

      In both cases we have shown that P[Yn+1=k2(n+1)]=2(n+1), hence (IH)n+1 holds.

    • (f) We know that YnY almost surely, hence by Lemma 6.1.2 the convergence also holds in distribution. Hence P[Yna]P[Ya].

      From part (e), for any a[0,1] we have

      (k1)2na<k2nP[Yna]=k2n.

      Hence aP[Yna]a+2n, and the sandwich rule tells us that P[Yna]a. Hence P[Ya]=a, for all a[0,1], which means that Y has a uniform distribution on [0,1].

  • 6.7 Since 𝟙{Xn+1}1{Xn} we have Xn+1Xn, and since X>0 we have Xn0. Further, for all ωΩ, whenever X(w)<n we have Xn(ω)=X(ω), and since P[X(ω)<]=1 this means that P[Xn(ω)X(ω)]=1. Hence Xna.s.X.

    Therefore, the monotone convergence theorem applies and E[Xn]E[X].

  • 6.6 Define Xn=Yn. Then Xn0 and Xn+1Xn, almost surely. Hence, (Xn) satisfies the conditions for the dominated convergence theorem and there exists a random variable X such that Xna.s.X and E[Xn]E[X]. Define Y=X and then we have Yna.s.Y and E[Yn]E[Y].

  • 6.8

    • (a) The equation

      (A.5)E[i=1nXi]=i=1nE[Xi]

      follows by iterating the linearity property of E n times (or by a trivial induction). However, in E[i=1Xi]=i=1E[Xi] both sides are defined by infinite sums, which are limits, and the linearity property does not tell us anything about limits.

    • (b) We’ll take a limit as n on each side of (A.5). We’ll have to justify ourselves in each case.

      First, consider the right hand side. Since we have Xn0 almost surely, E[Xn]0 by monotonicity of E. Therefore, i=1nE[Xi] is an increasing sequence of real numbers. Hence it is convergent (either to a real number or to +) and, by definition of an infinite series, the limit is written i=1E[Xi].

      Now, the left hand side, and here we’ll need the monotone convergence theorem. Define Yn=i=1nXi. Then Yn+1Yn=Xn+10 almost surely, so Yn+1Yn. Since Xi0 for all i, also Yn0, almost surely. Hence the monotone convergence theorem applies and E[Yn]E[Y] where Y=i=1Xi.

      Putting both sides together (and using the uniqueness of limits to tell us that the limits on both sides must be equal as n) we obtain that E[i=1Xi]=i=1E[Xi].

    • (c) Since P[Xi=1]=P[Xi=1], we have E[Xi]=0 and hence also i=1E[Xi]=0. However, the limit i=1Xi is not well defined, because Yn=i=1nXi oscillates (it jumps up/down by ±1 on each step) and does not converge. Hence, in this case E[i=1Xi] is not well defined.

  • 6.6

    • (a) We have both P[Xnx]P[Xx] and P[Xnx]P[Yx], so by uniqueness of limits for real sequences, we have P[Xx]=P[Yx] for all xR. Hence, X and Y have the same distribution (i.e. they have the same distribution functions FX(x)=FY(x)).

    • (b) By definition of convergence in probability, for any a>0, for any ϵ>0 there exists NN such that, for all nN,

      P[|XnX|>a]<ϵ and P[|XnY|>a]<ϵ.

      By the triangle inequality we have

      (A.6)P[|XY|>2a]=P[|XXn+XnY|>2a]P[|XXn|+|XnY|>2a].

      If |XXn|+|XnY|>2a then |XXn|>a or |XnY|>a (or possibly both). Hence, continuing (A.6),

      P[|XY|>2a]P[|XnX|>a]+P[|XnY|>a]2ϵ.

      Since this is true for any ϵ>0 and any a>0, we have P[X=Y]=1.

  • 6.7 Suppose (aiming for a contradiction) that there exists and random variable X such that XnX in probability. By the triangle inequality we have

    |XnXn+1||XnX|+|XXn+1|

    Hence, if |XnXn+1|>1 then |XnX|>12 or |Xn+1X|>12 (or both). Therefore,

    P[|XnXn+1|>1]P[|XnX|>12]+P[|Xn+1X|>12].

    Since XnX in probability, the right hand side of the above tends to zero as n. This implies that

    P[|XnXn+1|>1]0

    as n. But, Xn and Xn+1 are independent and P[Xn=1,Xn+1=0]=14 so P[|XnXn+1|>1]14 for all n. Therefore we have a contradiction, so there is no X such that XnX in probability.

    Hence, by Lemma 6.1.2, we can’t have any X such that XnX almost surely or in L1 (since it would imply convergence in probability).

Chapter 7
  • 7.1

    • (a) We have (CM)n=i=1n0(SiSi1)=0.

    • (b) We have (CM)n=i=1n1(SiSi1)=SnS0=Sn.

    • (c) We have

      (CM)n=i=1nSi1(SiSi1)=i=1n(X1++Xi1)Xi=12(i=1nXi)212i=1nXi2=Sn22n2 In the last line we use that Xi2=1.

  • 7.2 We have

    (XM)n=i=1n(αCn1+βDn1)(MnMn1)=αi=1nCn1(MnMn1)+βi=1nDn1(MnMn1)=α(CM)n+β(CM)n as required.

  • 7.3

    • (a) We note that

      |Sn|i=1n|Xi|i=1n1=n<.

      Hence SnL1. Since XiFn for all in, we have also that SnmFn. Lastly,

      E[Sn+1|Fn]=E[Sn+Xn+1|Fn]=Sn+E[Xn+1]=Sn. Therefore, Sn is a martingale. In the above, in the second line we use that SnmFn and that Xn+1 is independent of Fn. To deduce the last line we use that E[Xi]=12n2(1)=12n2(1)+(11n2)(0)=0.

      We next check that Sn is uniformly bounded in L1. We have E[|Xi|]=12i2|1|+12i2|1|+(11i2)|0|=1i2. Using that |Sn||X1|++|Xn|, monotonicity of expectations gives that

      E[|Sn|]i=1nE[|Xi|]i=1n1i2i=11i2<.

      Hence supnE[|Sn|]<, so (Sn) is uniformly bounded in L1.

      Therefore, the martingale convergence theorem applies, which gives that there exists some real-valued random variable S such that Sna.s.S.

    • (b) Note that (Sn) takes values in Z and that, from part (b), almost surely we have that Sn converges to a real value. It follows immediately by Lemma 7.4.1 that (almost surely) there exists NN such that Sn=S for all nN.

  • 7.4 Here’s some R code to plot 1000 time-steps of the symmetric random walk. Changing the value given to set.seed will generate a new sample.

    > T=1000
    > set.seed(1)
    > x=c(0,2*rbinom(T-1,1,0.5)-1)
    > y=cumsum(x)
    > par(mar=rep(2,4))
    > plot(y,type="l")
    
    
    

    Adapting it to plot the asymmetric case and the random walk from exercise 7.3 is left for you.

    In case you prefer Python, here’s some Python code that does the same job. (I prefer Python.)

    import numpy as np
    import matplotlib.pyplot as plt
    
    p = 0.5
    rr = np.random.random(1000)
    step = 2*(rr < p) - 1
    
    start = 0
    positions = [start]
    for x in step:
        positions.append(positions[-1] + x)
    
    plt.plot(positions)
    plt.show()
    
    
    

    Extending this code as suggested is left for you. For the last part, the difference in behaviour that you should notice from the graph is that the walk from Exercise 7.3 eventually stops moving (at some random time), whereas the random walk from Q2 of Assignment 3 never keeps moving but makes smaller and smaller steps.

  • 7.5 We will argue by contradiction. Suppose that (Mn) was uniformly bounded in L1. Then the martingale convergence theorem would give that Mna.s.M for some real valued random variable M. However, Mn takes values in Z and we have Mn+1Mn for all n, so Lemma 7.4.1 gives that (Mn) cannot converge almost surely. This is a contradiction.

  • 7.6

    • (a) Take the offspring distribution to be P[G=0]=1. Then Z1=0, and hence Zn=0 for all n1, so Zn dies out.

    • (b) Take the offspring distribution to be P[G=2]=1. Then Zn+1=2Zn, so Zn=2n, and hence Zn.

  • 7.7

    • (a) We have

      E[Mn+1|Fn]=E[Mn+1𝟙{nth draw is red}|Fn]+E[Mn+1𝟙{nth draw is black}|Fn]=E[Bnn+3𝟙{nth draw is red}|Fn]+E[Bn+1n+3𝟙{nth draw is black}|Fn]=Bnn+3E[𝟙{nth draw is red}|Fn]+Bn+1n+3E[𝟙{nth draw is black}|Fn]=Bnn+3Bnn+2+Bn+1n+3(1Bnn+2)=Bn2(n+2)(n+3)+Bn(n+2)+(n+2)Bn2Bn(n+2)(n+3)=(n+1)Bn+(n+2)(n+2)(n+3) This is clearly not equal to Mn=Bnn+2, so Mn is not a martingale.

      This result might be rather surprising, but we should note that the urn process here is not ‘fair’, when considered in terms of the proportion of (say) red balls. If we started with more red balls than black balls, then over time we would expect to see the proportion of red balls reduce towards 12, as we see in part (b):

    • (b) Your conjecture should be that, regardless of the initial state of the urn, P[Mn12 as n]=1. (It is true.)

  • 7.8

    • (a) Let Bn be the number of red balls in the urn at time n, and then Mn=BnK. Let us abuse notation slightly and write B=r for the event that the ball X is red, and X=b for the event that the ball X is black. Let X1 denote the first ball drawn on the (n+1)th draw, and X2 denote the second ball drawn.

      Let Fn=σ(B1,,Bn). It is clear that Mn[0,1], so MnL1, and that Mn is measurable with respect to Fn. We have

      E[Mn+1|Fn]=E[Mn+1𝟙{X1=r,X2=r}|Fn]+E[Mn+1𝟙{X1=r,X2=b}|Fn]+E[Mn+1𝟙{X1=b,X2=r}|Fn]+E[Mn+1𝟙{X1=b,X2=b}|Fn]=E[BnK𝟙{X1=r,X2=r}|Fn]+E[Bn+1K𝟙{X1=r,X2=b}|Fn]+E[Bn1K𝟙{X1=b,X2=r}|Fn]+E[BnK𝟙{X1=b,X2=b}|Fn]=BnKE[𝟙{X1=r,X2=r}|Fn]+Bn+1KE[𝟙{X1=r,X2=b}|Fn]+Bn1KE[𝟙{X1=b,X2=r}|Fn]+BnKE[𝟙{X1=b,X2=b}|Fn]=BnKBnKBnK+Bn+1KBnKKBnK+Bn1KKBnKBnK+BnKKBnKKBnK=1K3(Bn3+2Bn2(KBn)+Bn(KB)nBn(KBn)+Bn(KBn)2)=1K3(K2Bn)=Mn

    • (b) Since Mn[0,1] we have E[|Mn|]1, hence Mn is uniformly bounded in L1. Hence, by the martingale convergence theorem there exists a random variable M such that Mna.s.M.

      Since Mn=BnK, this means that also Bna.s.KM. However, Bn is integer valued so, by Lemma 7.4.1, if Bn is to converge then it must eventually become constant. This can only occur if, eventually, the urn contains either (i) only red balls or (ii) only black balls.

      In case (i) we have Mn=0 eventually, which means that M=0. In case (ii) we have Mn=1 eventually, which means that M=1. Hence, M{0,1} almost surely.

  • 7.9

    • (a) Consider the usual Pólya urn, from Section 4.2, started with just one red ball and one black ball. After the first draw is complete we have two possibilities:

      • 1. The urn contains one red ball and two black balls.

      • 2. The urn contains two red balls and one black ball.

      In the first case we reach precisely the initial state of the urn described in 7.9. In the second case we also reach the initial state of the urn described in 7.9, but with the colours red and black swapped.

      If Mna.s.0, then in the first case the fraction of red balls would tend to zero, and (by symmetry of colours) in the second case the limiting fraction of black balls would tend to zero; thus the limiting fraction of black balls would always be either 1 (the first case) or 0 (the second case). However, we know from Proposition 7.4.3 that the limiting fraction of black balls is actually uniform on (0,1). Hence, P[Mn0]=0.

    • (b) You should discover that having a higher proportion of initial red balls makes it more likely for M (the limiting proportion of red balls) to be closer to 1. However, since M is a random variable in this case, you may need to take several samples of the process (for each initial condition that you try) to see the effect.

      Follow-up challenge question: In fact, the limiting value M has a Beta distribution, Be(α,β). You may need to look up what this means, if you haven’t seen the Beta distribution before. Consider starting the urn with n1 red and n2 black balls, and see if you can use your simulations to guess a formula for α and β in terms of n1 and n2 (the answer is a simple formula!).

  • 7.10

    • (a) We have that (Mn) is a non-negative martingale, hence supnE[|Mn|]=supnE[Mn]=E[M0]=1. Thus (Mn) is uniformly bounded in L1 and the (first) martingale convergence theorem applies, giving that Mna.s.M for a real valued M. Since Mn0 almost surely we must have M0 almost surely.

    • (b) The possible values of Mn are W={(q/p)z;zZ}. Note that Mn+1Mn for all nN.

      On the event that M>0 there exists ϵ>0 such that Mnϵ for all sufficiently large n. Choose mN such that (q/p)mϵ and then for all sufficiently large n we have MnWm={(q/p)z;zm,zZ}. For such n, since MnMn+1 we have |MnMn+1||(q/p)m(q/p)m1|>0, which implies that (Mn) could not converge. Hence, the only possible almost sure limit is M=0.

      If (Mn) was uniformly bounded in L2 then by the (second) martingale convergence theorem we would have E[Mn]E[M]. However E[Mn]=E[M0]=1 and E[M]=0, so this is a contradiction. Hence (Mn) is not uniformly bounded in L2.

    • (c) We have (q/p)nSa.s.0. Noting that q/p>1 and hence log(q/p)>0, taking logs gives that Snlog(q/p)a.s., which gives Sna.s..

      For the final part, if (Sn) is a simple asymmetric random walk with upwards drift, then (Sn) is a simple asymmetric random walk with downwards drift. Applying what we’ve just discovered thus gives Sna.s., and multiplying by 1 gives the required result.

  • 7.11

    • (a) This is a trivial induction: if Sn is even then Sn+1 is odd, and vice versa.

    • (b) By part (a), p2n+1=0 since 0 is an even number and S2n+1 is an odd number.

      If Sn is to start at zero and reach zero at time 2n, then during time 1,2,,2n it must have made precisely n steps upwards and precisely n steps downwards. The number of different ways in which the walk can do this is (2nn) (we are choosing exactly n steps on which to go upwards, out of 2n total steps). Each one of these ways has probability (12)2n of occurring, so we obtain p2n=(2nn)22n.

    • (c) From (b) we note that

      p2(n+1)=(2n+2)(2n+1)(n+1)(n+1)22p2n=2(2n+1n+1)22pn=(2n+12n+2)p2n=(112(n+1))12p2n.

      Hence, using the hint, we have

      p2(n+1)exp(12(n+1))p2n.

      Iterating this formula obtains that

      p2nexp(12n)exp(12(n1))exp(12(1))p0(A.7)=exp(12i=1n1i). Recall that i=1n1i as n, and that ex0 as x. Hence, the right hand side of the expression above tends to zero as n. Hence, since pn0, by the sandwich we have p2n0 as n.

  • 7.12

    • (a) We note that

      E[Sn+1f(n+1)|Fn]=Snf(n+1)+1f(n+1)E[Xn+1|Fn]=Snf(n+1)+1f(n+1)E[Xn+1]=Snf(n+1) Here we use that SnmFn and that Xn+1 is independent of Fn.

      Since Snf(n) is a martingale we must have

      (A.8)Snf(n+1)=Snf(n).

      Since P[Sn0]>0 this implies that 1f(n+1)=1f(n), which implies that f(n+1)=f(n). Since this holds for any n, we have f(1)=f(n) for all n; so f is constant.

    • (b) If Snf(n) is to be a supermartingale then, in place of (A.8), we would need

      (A.9)Snf(n+1)Snf(n).

      Note that we would need this equation to hold true with probability one. To handle the we now need to care about the sign of Sn. The key idea is that Sn can be both positive or negative, so the only way that (A.9) can hold is if 1f(n+1)=1f(n).

      We will use some of the results from the solution of exercise 7.11. In particular, from for odd nN from 7.11(b) we have P[Sn=0]=0, and for even nN from (A.7) we have P[Sn=0]<1. In both cases we have P[Sn0]>0. Since Sn and Sn have the same distribution we have P[Sn<0]=P[Sn<0]=P[Sn>0], and hence

      P[Sn0]=P[Sn<0]+P[Sn>0]=2P[Sn<0]=2P[Sn>0].

      Hence, for all nN we have P[Sn>0]>0 and P[Sn<0]>0.

      Now, consider n1. We have that here is positive probability that Sn>0. Hence, by (A.9), we must have 1f(n+1)1f(n), which means that f(n+1)f(n). But, there is also positive probability that Sn<0, hence by (A.9) we must have 1f(n+1)1f(n), which means that f(n+1)f(n). Hence f(n+1)=f(n), for all n, and by a trivial induction we have f(1)=f(n) for all n.

  • 7.13

    • (a) Since Mn=Znμn we note that

      (A.10)1μ2(n+1)(Zn+1μZn)2=(Mn+1Mn)2. Further,

      Zn+1μZn=i=1Zn(Xin+1μ)

      so it makes sense to define Yi=Xin+1μ. Then E[Yi]=0,

      E[Yi2]=E[(Xin+1μ)2]=var(Xin+1)=σ2,

      and Y1,Y2,,Yn are independent. Moreover, the Yi are identically distributed and independent of Fn. Hence, by taking out what is known (ZnmFn) we have

      E[(Zn+1μZn)2|Fn]=E[(i=1ZnYi)2|Fn]=i=1ZnE[Yi2|Fn]+iji,j=1ZnE[YiYj|Fn]=i=1ZnE[Yi2]+iji,j=1ZnE[Yi]E[Yj]=ZnE[Y1]2+0=Znσ2. So, from (A.10) we obtain

      E[(Mn+1Mn)2|Fn]=Znσ2μ2(n+1).

    • (b) By part (a) and (3.4), we have

      E[Mn+12|Fn]=Mn2+Znσ2μ2(n+1).

      Taking expectations, and using that E[Zn]=μn, we obtain

      E[Mn+12]=E[Mn2]+σ2μn+2.

  • 7.14

    • (a) Note that Xn+1Xn0 for all n, because taking an inf over k>n+1 is an infimum over a smaller set than over k>n. Hence, (Xn) is monotone increasing and hence the limit X(ω)=limnXn(ω) exists (almost surely).

    • (b) Since |Mn|infk>n|Mk|, we have Xn|Mn|. By definition of inf, for each ϵ>0 and nN there exists some nn such that Xn|Xn|ϵ. Combining these two properties, we obtain

      |Mn|ϵXn|Mn|.

    • (c) Letting n, in the above equation, which means that also n because nn, we take an almost sure limit to obtain |M|ϵX|M|. Since we have this inequality for any ϵ>0, in fact X=|M| almost surely.

    • (d) We have shown that (Xn) is an increasing sequence and Xna.s.X, and since each |Mn|0, we have also Xn0. Hence, the monotone convergence theorem applies to Xn and X, and therefore E[Xn]E[X].

    • (e) From the monotone convergence theorem, writing in terms of Mn rather than Xn, we obtain

      (A.11)E[infkn|Mn|]E[|M|].

      However, since infkn|Mn||Mn|C=supjNE[|Mj|], the monotonicity property of expectation gives us that, for all n,

      (A.12)E[infkn|Mn|]C.

      Combining (A.11) and (A.12), with the fact that limits preserve weak inequalities, we obtain that E[|M|]C, as required.

  • 7.15

    • (a) All the X~in+1 have the same distribution, and are independent by independence of the Xin+1 and Cin+1. Thus, Z~n is a Galton-Watson process with off-spring distribution G~ given by the common distribution of the Xin+1.

    • (b) By definition of X~in+1 we have Xin+1X~in+1. Hence, from (7.9), if ZnZ~n then also Zn+1Z~n+1. Since Z0=Z~0=1, an easy induction shows that ZnZ~n for all nN.

    • (c) We have

      f(α)=E[X~in+1]=P[G=0]P[C=1]+i=1iP[G=i]=αP[G=0]+i=1iP[G=i]. Recall that each Xin+1 has the same distribution as G. Hence,

      μ=E[Xin+1]=E[G]=i=1iP[G=i]=f(0).

      Since μ<1 we therefore have f(0)<1. Moreover,

      f(1)=P[G=0]+i=1iP[G=i]i=0P[G=i]=1.

      It is clear that f is continuous (in fact, f is linear). Hence, by the intermediate value theorem, there is some α[0,1] such that f(α)=1.

    • (d) By (c), we can choose α such that f(α)=1. Hence, E[X~in+1]=1, so by (a) Z~n is a Galton-Watson process with an offspring distribution that has mean μ^=1. By Lemma 7.4.6, Z~ dies out, almost surely. Since, from (b), 0ZnZ~n, this means Zn must also die out, almost surely.

Chapter 8
  • 8.1 We have U1<1 so, for all ωΩ, Xn(ω)=Un(ω)0 as n. Hence Xn(ω)0 almost surely as n. Further, |Xn|1 for all nN and E[1]=1<, so the dominated convergence theorem applies and shows that E[Xn]E[0]=0 as n.

  • 8.2 We check the two conditions of the dominated convergence theorem. To check the first condition, note that if n|X| then X𝟙{|X|n}=X. Hence, since |X|<, as n we have Xn=X𝟙{|X|n}X almost surely.

    To check the second condition, set Y=|X| and then E[|Y|]=E[|X|]< so YL1. Also, |Xn|=|X𝟙{|X|n}||X|=Y, so Y is a dominated random variable for (Xn). Hence, the dominated convergence theorem applies and E[Xn]E[X].

  • 8.3

    • (a) We look to use the dominated convergence theorem. For any ωΩ we have Z(ω)<, hence for all nN such that n>Z(ω) we have Xn(ω)=0. Therefore, as n, Xn(ω)0, which means that Xn0 almost surely.

      We have |Xn|Z and ZL1, so we can use Z are the dominating random variable. Hence, by the dominated convergence theorem, E[Xn]E[0]=0.

    • (b) We have

      E[Z]=1xf(x)dx=1x1dx=[logx]1=

      which means ZL1 and also that

      E[Xn]=E[𝟙{Z[n,n+1)}Z]=1𝟙{x[n,n+1)}xf(x)dx=nn+1x1dx=[logx]nn+1=log(n+1)logn=log(n+1n). As n, we have n+1n=1+1n1, hence (using that log is a continuous function) we have log(n+1n)log1=0. Hence, E[Xn]0.

    • (c) Suppose that we wanted to use the DCT in (b). We still have Xn0 almost surely, but any dominating random variable Y would have to satisfy Y|Xn| for all n, meaning that also YZ, which means that E[Y]E[Z]=; thus there is no dominating random variable YL1. Therefore, we can’t use the DCT here, but we have shown in (b) that the conclusion of the DCT does hold: we have that E[Xn] does tend to zero.

      We obtain that the conditions of the DCT are sufficient but not necessary for its conclusion to hold.

  • 8.4 Note that {T=n}={Tn}{Tn1}. If T is a stopping time then both parts of the right hand side are elements of Fn, hence so is {T=n}.

    Conversely, if {T=n}Fn for all n, then for in we have {T=i}FiFn. Hence the identity {Tn}=i=1n{T=i} shows that {Tn}Fn, and therefore T is a stopping time.

  • 8.5 We can write {R=n}={Si0 for i=1,,n1}{Sn=0}={Sn=0}(i=1nΩ{Si=0}). By exercise 8.4 we have {Si=0}Fn for all in, hence {R=n}Fn for all n.

  • 8.6 We can write

    {S+Tn}=k=0nj=0nk{Sj}{Tk}

    Since S and T are both stopping times we have {Tn},{Sn}Fn for all nN, which means that {Tj},{Sk}Fn for all j,kn. Hence {S+Tn}Fn and therefore S+T is also a stopping time.

  • 8.7 If AFS then A{Sn}Fn. Since T is a stopping time we have {Tn}Fn, hence

    A{Sn}{Tn}Fn.

    As ST we have {Tn}{Sn}, hence A{Tn}Fn.

  • 8.8

    • (a) We have

      T=inf{n>0;Bn=2}.

      Note that Tn if and only if Bi=1 for all i=1,2,,n1. That is, if and only if we pick a red ball out of the urn at times i1,2,,n1. Hence,

      P[Tn]=1223n1n=1n. Therefore, since P[T=]P[Tn] for all n, we have P[T=]=0 and P[T<]=1.

    • (b) We note that T is a stopping time, with respect to Ft=σ(Bi;in), since

      {Tn}={a red ball was drawn at some time in}=i=1n{Bi=2}.

      We showed in Section 4.2 that Mn was a martingale. Since Mn[0,1] the process (Mn) is bounded. By part (a) for any nN we have P[T=]P[Tn]=1n, hence P[T=]=0 and P[T<]=1. Hence, we have condition (b) of the optional stopping theorem, so

      E[MT]=E[M0]=12.

      By definition of T we have BT=2. Hence MT=2T+2, so we obtain E[1T+2]=14.

  • 8.9

    • (a) We showed in exercise 7.8 that, almost surely, there was a point in time at which the urn contained either all red balls or all black balls. Hence, P[T<]=1.

      Moreover, almost surely, since the urn eventually contains either all red balls or all black balls, we have either MT=1 (all red balls) or MT=0 (all black balls).

    • (b) We note that T is a stopping time, since

      {Tn}={for some in we have Mi{0,1}}=i=0nMi1({0,1}).

      In exercise 7.8 we showed that Mn was a martingale. Since Mn[0,1] in fact Mn is a bounded martingale, and we have P[T<] from above, so conditions (b) of the optional stopping theorem apply. Hence,

      E[MT]=E[M0]=rr+b.

      Lastly, since MT is either 0 or 1 (almost surely) we note that

      E[MT]=E[MT𝟙{MT=1}+MT𝟙{MT]=0}]=E[𝟙{MT=1}]+0=P[MT=1]. Hence, P[MT=1]=rr+b.

  • 8.10

    • (a) We use the filtration Fn=σ(Ni;in). We have PnmFn and since 0Pn1 we have also that PnL1. Also,

      E[Pn+1|Fn]=E[Pn+1𝟙{the nth ball was red}|Fn]+E[Pn+1𝟙{the nth ball was blue}|Fn]=E[Nn12mn1𝟙{the nth ball was red}|Fn]+E[Nn2mn1𝟙{the nth ball was blue}|Fn]=Nn12mn1E[𝟙{the nth ball was red}|Fn]+Nn2mn1E[𝟙{the nth ball was blue}|Fn]=Nn12mn1Nn2mn+Nn2mn12mnNn2mn=Nn2mn=Pn. Here we use taking out what is known (since NnmFn), along with the definition of our urn process to calculate e.g. E[𝟙{the nth ball was red}|Fn] as a function of Nn. Hence (Pn) is a martingale.

    • (b) Since 0Pn1, the process (Pn) is a bounded martingale. The time T is bounded above by 2m, hence condition (b) for the optional stopping theorem holds and E[PT]=E[P0]. Since P0=12 this gives us E[PT]=12. Hence,

    P[(T+1)st ball is red]=P[NT+1=NT+1]=i=12m1P[NT+1=NT+1|T=i]P[T=i]=i=12m1m12miP[T=i]=E[PT]=12. as required.

Chapter 9
  • 9.1

    • (a) We have that T is a stopping time and (from Section 4.1) Mn=Sn(pq)n is a martingale. We want to apply the optional stopping theorem to (Mn).

      We have E[T]< from Lemma 9.1.1 and for all n we have

      |Mn+1Mn||Sn+1Sn|+|pq|=1+|pq|.

      Thus, condition (c) for the optional stopping theorem applies and E[MT]=E[M0]. That is,

      E[ST(T)(pq)]=E[S0(0)(pq)]=E[S00(pq)]=0.

      We thus have E[ST]=(pq)E[T] as required.

    • (b) We have

      E[ST]=E[ST𝟙{T=Ta}+ST𝟙{T=Tb}]=E[a𝟙{T=Ta}+b𝟙{T=Tb}]=aP[T=Ta]+bP[T=Tb] Putting equations (9.5) and (9.6) into the above, and then using part (a) gives us

      E[T]=1pq(a(q/p)b1(q/p)b(q/p)a+b1(q/p)a(q/p)b(q/p)a).

  • 9.2

    • (a) We have P[Ta<]=P[Tb>]=1 from Lemma 9.3.2. Hence also P[T=Ta< or T=Tb<]=1. Since these two possibilities are disjoint, in fact P[T=Ta]+P[T=Tb]=1.

      For the second equation, we will use that (Sn) is a martingale. We want to apply the optional stopping theorem, but none of the conditions apply directly to (Sn). Instead, set Mn=SnT and note that Lemma 8.2.4 implies that (Mn) is a martingale. By definition of T we have |Mn|max{|a|,|b|}, hence (Mn) is bounded. We have P[T<] from above, hence conditions (b) of the optional stopping theorem apply to (Mn). We thus deduce that

      E[MT]=E[M0]=E[S0]=0.

      Hence

      E[MT]=E[MT𝟙{T=Ta}+MT𝟙{T=Tb}]=E[MTa𝟙{T=Ta}]+E[MTb𝟙{T=Tb}]=E[STa𝟙{T=Ta}]+E[STa𝟙{T=Ta}]=E[a𝟙{T=Ta}]+E[b𝟙{T=Ta}]=aP[T=Ta]+bP[T=Ta] as required. In the above, the first line partitions on the value of T (using what we already deduced). The third line follows from the second because TaT, hence MTa=STaT=STa, and similarly for Tb.

    • (b) Solving the two equations (this is left for you) gives that P[T=Ta]=bba and P[T=Tb]=aba.

    • (c) From Exercise 4.4 we have that Sn2n is a martingale. Again, none of the conditions of optional stopping directly apply, but Xn=SnT2(nT) is a martingale by Theorem 8.2.4. Similarly to above, we have |Mn|max{a2+|a|,b2+|b|}, so (Xn) is bounded. Hence conditions (b) of the optional stopping theorem apply and

      E[XT]=E[X0]=E[S020=]=0.

      Thus E[ST2]=E[T], which gives

      E[T]=E[ST2𝟙{T=Ta}+ST2𝟙{T=Ta}]=E[a2𝟙{T=Ta}+b2𝟙{T=Ta}]=a2P[T=Ta]+b2P[T=Tb]=a2bb1+b2(a)ba=ab. Here, the second line follows because when T=Ta we have ST=STa=a, hence also STa2=a2, and similarly for STb2. The fourth line uses part (b).

  • 9.3

    • (a) Note that movements up at by +2, whilst movements down are by 1. In order for Sn to return to zero at time n, it must have made twice as many movements down as up. In particular n must therefore be a multiple of 3, so P[S3n+1=0]=P[S3n+2=0]=0.

      For S3n, we can count the possible ways to have n movements up and 2n movements down by choosing n upwards movements out of 3n movements, that is (3nn). Taking the probabilities into account and then using (9.7) we obtain

      P[S3n=0]=(3nn)(13)n(23)2n=(3n)!n!(2n)!22n33n6πn(3n/e)3n8πn(n/e)n(2n/e)2n22n33n=32πn.

    • (b) The argument that E[G]<, under the assumption P[R<]=1, is exactly the same as before.

      We need to modify the next part, beginning with (9.9) which is replaced by the result in part (a). Note that that S3n now plays the role of what was previously S2n, being the possible returns to zero. Having (a) introduces an extra factor of 32, but using the same method as before we can still deduce that E[G]=. We thus arrive at a contradiction in the same way.

  • 9.4

    • (a) We have E[Xi]=(p)(1)+(1p2)(12)=5p23. The condition p(56,1] implies that E[Xi]>0. Writing μ=E[Xi], it follows from the strong law of large numbers that (almost surely) there exists NN such that for all nN we have Snnμ2. Hence Sna.s..

    • (b) We calculate

      E[T1]=E[T1𝟙{S1=1}]+E[T1𝟙{S1=1}]+E[T1𝟙{S1=2}]=E[1𝟙{S1=1}]+E[E[T1𝟙{S1=1}|F1]]+E[E[T1𝟙{S1=2}|F1]]=p+E[𝟙{S1=1}E[T1|F1]]+E[𝟙{S1=2}E[T1|F1]]=p+E[𝟙{S1=1}E[1+T1+T1|F1]]+E[𝟙{S1=2}E[1+T1+T1+T1|F1]]=p+E[𝟙{S1=1}E[1+T1+T1]]+E[𝟙{S1=1}E[1+T1+T1+T1]]=p+E[𝟙{S1=1}(1+2E[T1])]+E[𝟙{S1=1}(1+3E[T1])]=p+1p2(1+2E[T1])+1p2(1+3E[T1]). In the first line of the above we partition on the value of S1. The first term of the second line follows because if S1=1 then T1=1, and the other terms use the relationship between expectation and conditional expectation. The third line follows because S1F1.

      The fourth line uses that, on the event S1=1, T1 is equal to 1 (accounting for the first move 01) plus two independent copies (T1 and T1) of T1 (accounting for the time to move from 10 and then 01). Similarly, on the event S1=1, T1 is equal to 1 plus three independent copies of T1. Note that it is crucial here that (Sn) can only jump upwards by 1 in each step of time, meaning that it cannot go above 1 without first being equal to 1.

      The strong Markov property gives that T1,T1 and T1 are independent of F1, leading to the fifth line. The remainder of the calculation is straightforward. We thus obtain

      E[T1]=p+1p2(2+5E[Tt]).

      Solving this equation gives E[T1]=25p3.

  • 9.5 Recall that Sn even when n even, and odd when n is odd. Hence T1 is odd. From the fact given in the question, there exists NN such that for all nN we have P[T1=2n1]14πn3/2. Hence

    E[T1]n=N(2n1)P[T1=2n1]14πn=N2n1n3/212πn=N1n1/2=

    as required.

  • 9.6

    • (a) Note that Mn(θ)=i=1n(eθXi/coshθ). Since XimFn for all in, SnmFn for all n by Proposition 2.2.6. Since |Sn|n we have |Mn|eθn(coshθ)n<, hence MnL1. We have also that

      E[Mn+1|Fn]=(i=1neθXicoshθ)E[eθXn+1coshθ|Fn]=MnE[eθXn+1coshθ]=Mn. Here we use the taking out what is known rule, the fact that Xn+1 is independent of Fn and the relationship between conditional expectation and independence. To deduce the final line we note that E[eθXi/coshθ]=12(eθ+eθ)/coshθ=1.

      Note: This is a more general case of the martingale from Exercise 4.1.

    • (b) Condition (1) fails because Tm is not bounded – to see this, note that for any NN there is a positive probability that (Sn)n=0N remains below zero, in which case TmN. Condition (b) fails because (Mn) is unbounded. Condition (c) fails because we have E[T1]= by Lemma 9.3.4, and TmT1 so also E[Tm]=.

    • (c) By Lemma 8.2.4 the process Ln(θ)=MnT(θ) is a martingale, and by definition of T we have that emθ(coshθ)nLn(θ)emθ(coshθ)n. Since coshθ1, we thus have |Ln(θ)|emθ+emθ, so the martingale (Lnθ) is bounded. By Lemma 9.3.2 P[T<]=1, so conditions (b) of the optional stopping theorem apply to (Ln(θ)) and T, hence

      E[eθST(coshθ)T]=E[eθS0(coshθ)0]=1.

      By Lemma 9.3.2 we have P[T<]=1, hence almost surely either T=Tm or T=Tm. Hence,

      1=E[eθSτ(coshθ)T𝟙{T=Tm}+eθSτ(coshθ)T𝟙{T=Tm}]=E[emθ(coshθ)Tm𝟙{T=Tm}]+E[emθ(coshθ)Tm𝟙{T=Tm}]=emθE[1(coshθ)Tm𝟙{T=Tm}]+emθE[1(coshθ)Tm𝟙{T=Tm}]=(emθ+emθ)E[1(coshθ)Tm𝟙{T=Tm}] In the above, the final line follows by symmetry about zero, which implies that the two expectations are in fact equal. Thus E[𝟙{T=Tm}(coshθ)Tm]=12cosh(mθ), for all mN. With this in hand we can calculate

      E[1(coshθ)T]=E[𝟙{T=Tm}(coshθ)Tm]+E[𝟙{T=Tm}(coshθ)Tm]=12cosh(mθ)+12cosh(mθ)=1cosh(mθ). In the first line we partition on whether T is equal to Tm or Tm. In the second line we apply the formulae deduced above at both m and m, and to deduce the final line we recall that cosh(x)=cosh(x).

      Note: Tm and 𝟙{T=Tm} are not independent, but if they were then it would be possible to write a simpler solution. Check that you didn’t fall into this trap.

  • 9.7 We have show in Lemma 9.5.1 that (Sn) will almost surely return to the origin in finite time. By applying the strong Markov property at each return to the origin, we obtain that (Sn) will return to the origin infinitely often.

    Fix some zZ2. There is a path from the origin to z, and with positive probability the walk will take that path, hence P[Sn=z for some nN] is positive. By the strong Markov property, this means the probability that (Sn) visits z in between any two successive visits to the origin, is also positive. By the strong Markov property, whether this occur is independent, for each successive pair of returns to the origin. Hence we have infinitely many trials to try and visit z, each with the same positive success probability. This will result in infinitely many visits to z, almost surely.

    This holds for any zZ2, which completes the argument.

  • 9.8

    • (a) If L= then the set {n;Sn=0} much be infinite, meaning that (Sn) has returned infinitely many times to the origin. However, this has probability zero; by the strong Markov property, after each visit there is a probability P[R=]>0 of not returning again, independently of the past; eventually that will happen.

    • (b) L is not a stopping time. One way to see this is to note that {L=0}={R=}, which depends on the whole path taken by the walk, so is not an element of Fn for any NN.

    • (c) The event {L=2n} is the event that S2n=0, but the walk is never zero after that time. Hence

      P[L=2n]=E[𝟙{L=2n}]=E[𝟙{S2n=0}𝟙{Sk0 for all k>2n}]=E[E[𝟙{S2n=0}𝟙{Sk0 for all k>2n}|F2n]]=E[𝟙{S2n=0}E[𝟙{Sk0 for all k>2n}|F2n]]=E[𝟙{S2n=0}P[R=]]=P[R=]P[S2n=0]. Note that we have used the Markov property to deduce the fifth line.

      Recall that Sn is even if and only if n is even. Taking the above equation and summing over n, noting that n=0P[L=2n]=1 we obtain

      1=P[R=]n=0P[S2n=0]=P[R=]E[G]. The required result follows by noting that P[R<]=1P[R=].

    • (d) During the proof of Lemma 9.3.4, whilst we were arguing by contradiction, we noted that if p=P[R<]<1 then G would have a Geometric(p) distribution. In part (c) we’ve discovered the mean of this geometric distribution. We also came close, for the same reason, during the proofs of Lemmas 9.5.1 and 9.5.2, and in Remark 9.4.2. In all of those cases, all we needed was to note that E[G] would be finite – we didn’t need to know its value.

  • 9.9 Let Br={zZ3;|z|<n}. Note that Br is a finite set.

    Consider some zZ3. If (Sn) visits z then, the first time that it does, we may apply the strong Markov property. From then on the walk will behave independently of its past, again moving one unit of space in a random direction on each step of time. Apply part (c) of Exercise 9.9, the expected number of returns to z is finite, hence there will almost surely be a last return time to z.

    We can do the above for any zBN. Since Br is finite, this means that almost surely there is a last return time to the set Br. That is, there exists some random variable NN such that SnBr for all nN. Hence |Sn|r for all nN. Since this holds for any rN, we have |Sn|a.s..