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I have a following task: Let $X_0=1$ and then in every $t>0$, $X_t$ is $-1$ or $1$. The number of "changes" from $-1$ to $1$ in interval $\Delta$ is described by Poisson distribution with expectation $\lambda\Delta$. What is the expectation of $X_t\cdot X_s$? I was trying this way" $EX_t\cdot X_s=E(X_t|X_s=1)\cdot P(X_s=1)+E(X_t|X_s=-1)\cdot P(X_s=-1)$

mwrooo
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1 Answers1

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Let us assume that $0<s<t$ here.

Observe that $X_{t},X_{s}$ and also $X_{t}X_{s}$ only take values in $\left\{ -1,1\right\} $, leading to:

$$\mathbb{E}X_{t}X_{s}=P\left(X_{t}X_{s}=1\right)-P\left(X_{t}X_{s}=-1\right)=P\left(X_{t}=X_{s}\right)-P\left(X_{t}\neq X_{s}\right)=$$$$P\left(X_{t}=X_{s}\right)-\left[1-P\left(X_{t}=X_{s}\right)\right]=2P\left(X_{t}=X_{s}\right)-1$$

Here $\left\{ X_{t}=X_{s}\right\} $ can be recognized as the event that on the interval $\left(s,t\right)$ the number of changes is even.

Can you take it from here?

drhab
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