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I am given $X_t^2$ to be checked, where $X_t$ is Poisson process. How do I check? What are the properties that need to be checked?

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up vote 3 down vote accepted

Let $Y_t$ be a Poisson process. For what values of $k$ is $P(Y_t=k)\ne 0$?

For what values of $k$ is $P(X_t^2=k)\ne 0$?

Added: The above hint leads to probably the simplest way to see that $X_t^2$ is not a Poisson process. However, almost all of the characteristic features of the Poisson process fail for $X_t^2$.

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Are you asking me for a clearer task or are these the answers I should ask myself to help me solve the problem? – Andrius Naruševičius May 22 '12 at 10:36
@AndriusNaruševičius: It should make you able to solve the problem. Recall that if $(X_t^2)$ was indeed a Poisson proces, then $X_t^2\sim \text{po}(\lambda t)$, for some $\lambda>0$. Can this be true if $X_t$ is Poisson distributed? – Stefan Hansen May 22 '12 at 11:59
I get $E[X^2_t]=\lambda^2+\lambda$ Does that mean that $X^2_t\approx po((\lambda^2+\lambda)t)$ and thus not equal to $po(\lambda t)$ and thus not a Poisson process? (unless $\lambda=0$) – Andrius Naruševičius May 22 '12 at 12:03
There are many ways to see it. Your expectation calculation, by itself, is not one of them, the parameter could be, as you point out, $\lambda^2+\lambda$. Poisson does not necessarily mean Poisson parameter $\lambda$. My hint was for the simplest way to an answer. A Poisson process takes on values $0$, $1$, $2$, $3$, $4$, and so on all with non-zero probability. But if $X$ is Poisson, $X^2$ cannot take on value $2$ (or $3$, or $5$). There are many other ways we could handle showing that $X_t^2$ is not Poisson. – André Nicolas May 22 '12 at 13:17
Your mean calculation is one of them. If it is Poissom, with parameter $\lambda^2+\lambda$, then the probability that $X^2=0$ should be what? But what is it? Or prove non-additivity. Almost every special property of Poisson will fail for $X_t^2$. However, the one-line proof is the one that uses the hint of my answer. Most simply, if $Y$ is Poisson, $P(Y=2)>0$. But $P(X^2=2)=0$, since $X^2$ only takes on perfect square values. – André Nicolas May 22 '12 at 13:21

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