If I assume that $X(t)$ is a compound poisson process, how can it be found what $Cov(X(s), X(t))$ is? I have seen this over and over in books, but they only state it as fact. It is stated to be $\lambda E[Y]min(s,t)$, but I cannot quite seem to get the computation down. Many thanks if someone could guide me in the right direction.

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    $\begingroup$ What is $Y$? If it refers to the distribution of the size of the jumps, then your formula should involve the expectation of $Y^2$. After all, for $s=t$ it must reduce to the variance of $X(t)$, which is $\lambda t E[Y^2]$. $\endgroup$
    – whuber
    Commented Apr 16, 2015 at 15:14

1 Answer 1


$Cov(X(s),X(t))$ are overlapping in timeline...

So now, we need to get it non-overlapping in timeline.

For $s<t$, $Cov(X(s),X(t))=Cov(X(s),X(t)-X(s)+X(s))=Cov(X(s),X(t)-X(s))+Cov(X(s),X(s))=\lambda{s}E(Y^2)$

Now here, "$Cov(X(s),X(t)-X(s))+Cov(X(s),X(s))$" see first term is independent so equals zero, second term is variance at timeline 's'.

For $t<s$.... It will be $Cov(X(t),X(s))=Cov(X(t),X(s)-X(t)+X(t))=Cov(X(t),X(s)-X(t))+Cov(X(t),X(t))=\lambda{t}E(Y^2)$

Hence, consider $min(s,t)$

Note:-as @whuber's comment says y is not defined here, Y considered as size of individual output.


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