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I am investigating a simple model related to the random walk (Just that you work with the product here).

What you do is setting $S_0=1$ and $S_{k}=u\cdot S_{k-1}$ with some probability $p$ and $S_{k}=d\cdot S_{k-1}$ with some probability $q=1-p$. Those choices are of course independent. Therefore $S_n=\prod _{k=1}^n X_k$ with $X_k \in \{u,d\}$

Now the question is $\operatorname{Cov}(S_n,S_l)=?$


Also I would like to know a way how to test some formula I found for this with mathematica if that exists. For example I generated two tables $A,B$ that simulate $S_n$ for $n=10$ (with $100.000$ values each). Now $\text{Variance}(A)$ gives me (almost) the same result in mathematica as $\text{Variance}(B)$ or $\text{Covariance}(A,A)$. But even plugging in $\text{Covariance}(A,B)$ I start getting really tiny values, is that correct?

Now I will provide some own thoughts I did about this problem:

Certainly you have $P(S_n=u^k d^{n-k})=\binom{n}{k} p^k (1-p)^{n-k}$.

With this you can easily see that

$E(S_n)=\sum _{k=0}^n u^k d^{n-k} P(S_n=u^k d^{n-k}) = (d q+p u)^n$

and simillary

$E(S_n^2)=(d^2 q+p u^2)^n$

This also provides the variance, but the covariance seems very tough (at least for me).

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Your computations are correct. The computation of the covariance is best approached by writing $S_{n+k}$ as $S_{n+k}=S_nT_k$ where $T_k$ is independent of $S_n$ and distributed like $S_k$. Hence, using the fact that $E(S_n)=m^n$, $E(S_k)=m^k$ and $E(S_n^2)=v^n$ with $m=pu+(1-p)d$ and $v=pu^2+(1-p)d^2$, one gets $ \mathrm{Cov}(S_n,S_{n+k})=E(S_n^2)E(S_k)-E(S_n)^2E(S_k)=(v^n-m^{2n})m^k. $

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    You are right I fixed it. The problem is that $S_k$ was not defined how I thought it was.2011-05-20