Stochastic Processes and Financial Mathematics
(part one)

9.4 Long term behaviour: asymmetric case \(\msconly \)

We will now move on to study the simple asymmetric random walk. We’ll recycle some of our previous notation. Let \(S_n=\sum _{i=1}^n\) where the \((X_i)\) are independent and have identical distribution \(\P [X_i=1]=p>\frac 12\) and \(\P [X_1=-1]=1-p<\frac 12\). The case \(p<\frac 12\) can be handled by considering \((-S_n)\) in place of \((S_n)\). Let \(R=\inf \{n\geq 1\-S_n=0\}\) be the first return time of \((S_n)\) to zero. As before, we set \(R=\infty \) if the walk does not return to zero.

Like the symmetric random walk, the asymmetric case sees oscillations, but they are dominated by the drift upwards, so much so that \((S_n)\) might not return to origin even once.

Proof: We will first prove that \(S_n\stackrel {a.s.}{\to }\infty \). In fact, we have already proved this using a martingale convergence argument in Exercise 7.10. We’ll give a different proof here, using the strong law of large numbers. We have \(S_n=\sum _{i=1}^n X_i\), where the \(X_i\) are i.i.d. random variables, so the strong law of large numbers implies that \(\frac {S_n}{n}\stackrel {a.s.}{\to } \E [X_1]\) as \(n\to \infty \). Note that \(\E [X_1]=p(1)+(1-p)(-1)=2p-1\). It follows that there exists (a random variable) \(N\in \N \) such that for all \(n\geq N\) we have \(\frac {S_n}{n}\geq \frac {2p-1}{2}\), which implies \(S_n\geq n\frac {2p-1}{2}\). Since \(2p-1>0\) we therefore have \(S_n\stackrel {a.s.}{\to }\infty \). In particular this means that \(\P [(S_n)\text { visits zero infinitely many times}]=0\).

If \(\P [R<\infty ]\) was equal to \(1\) then, by applying the strong Markov property at each successive return time to zero, we would have that almost surely \((S_n)\) visited the origin infinitely many times. We have shown above that this is not the case. Hence \(\P [R<\infty ]<1\).   ∎

If we take \(T_k\) to be the hitting time of \(k\in \Z \), then it follows immediately from Lemma 9.4.1 that \(\P [T_k<\infty ]=1\) for \(k\geq 0\) and \(\P [T_k<\infty ]<1\) for \(k<0\). In the symmetric case Lemma 9.3.4 showed that \(\E [T_1]=\infty \), but here in the asymmetric case the opposite is true.

Proof: We calculate

\begin{align*} \E [T_1] &= \E [T_1\1_{\{S_1=1\}}]+\E [T_1\1_{\{S_1=-1\}}] \\ &= \E [1\1_{\{S_1=1\}}] + \E [\E [T_1\1_{\{S_1=-1\}}\|\mc {F}_{1}]] \\ &= p + \E [\1_{\{S_1=-1\}}\E [T_1\|\mc {F}_{1}]] \\ &= p + \E [\1_{\{S_1=-1\}}\E [1+T'_1+T''_1\|\mc {F}_{1}]] \\ &= p + \E [\1_{\{S_1=-1\}}\E [1+T'_1+T''_1]] \\ &= p + \E [\1_{\{S_1=-1\}}(1+2\E [T_1])] \\ &= p + (1-p)(1+2\E [T_1]). \end{align*} In the first line of the above we partition on the value of \(S_1\). The first term of the second line follows because if \(S_1=1\) then \(T_1=1\), and the second term uses the relationship between expectation and conditional expectation. The third line follows because \(S_1\in \mc {F}_1\). The fourth line uses that, on the event \(S_1=-1\), \(T_1\) is equal to \(1\) (accounting for the first move \(0\mapsto -1\)) plus two independent copies (\(T'_1\) and \(T''_1\)) of \(T_1\) (accounting for the time to move from \(-1\mapsto 0\) and then \(0\mapsto 1\)). The strong Markov property gives that \(T'_1\) and \(T''_1\) are independent of \(\mc {F}_1\), leading to the fifth line. The remainder of the calculation is straightforward. Rearranging to isolate \(\E [T_1]\), we obtain the required result.   ∎

Note that when \(p\approx 1\) we have \(\E [T_1]\approx 1\), which makes sense since when \(p\) is close to \(1\) it is very likely that the first step of the walk will be upwards. As \(p\searrow \frac 12\) we have \(\E [T_1]\to \infty \), which also makes sense because as \(p\) gets close to \(\frac 12\) we would expect the asymmetric walk to look more and more like the symmetric case, which has \(\E [T_1]=\infty \).