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Let's say you have a variable $X$ with a Poisson distribution ($Po( \lambda )$) and a variable $Y = 0.15 * X$. What is $EY$ and $Y$'s distribution?

I'm almost certain that $EY = EX * 0.15$ but I'm quite unsure about it's distribution. At first I thought it should be $Po(\lambda * 0.15)$ but after a bit of thinking it doesn't seem to be right.

The second approach I tried was to go directly trough distribution. We know that for $Po(\lambda)$ the distribution is: $$P[x=k] = ( {e^{-\lambda} * {\lambda^k \over k!}})$$ so given that $P[Y = k] == P[X = k/0.15]$ (which I believe is true) the distribution of $Y$ should be $Po(\lambda)$ the distribution is: $$P[Y=k] = ( {e^{-\lambda} * {\lambda^{k \over 0.15} \over {k \over 0.15}!}})$$. But then, I'm unable to tell what the $EY$ is (though I still suspect it to be $EX * 0.15$).

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  • $\begingroup$ It no longer has probability on (and only on) each of the non-negative integers, so a scaled Poisson is not Poisson. $\endgroup$ – Glen_b -Reinstate Monica Feb 9 '16 at 0:50
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I think you are right about the mean of the distribution (it will be $0.15 \cdot E(X)$), but the dependency between first moment (mean) and second (variance) ($\lambda = \mu = \sigma^2$ for Poisson) is not invariant for linear transformations so the distribution of $Y$ is not Poisson anymore. You can fit Negative Binomial for $Y$, it gives you a good approximation (in case the task is to model $Y$). If you want to infer parameters (like second and other moments), it should be done directly, using distribution function.

Also you can apply Anscombe transformation to $X$ and work with it like with normal.

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