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I have a confusion in the parameter part. I know that the sum of n i.i.d Poisson variates is also a Poisson variate with parameter as the sum of individual parameters. For instance, if

$X_i$~$Poisson(\lambda_i) \Rightarrow \sum_1^nX_i$~$Poisson(\sum_1^n\lambda_i)$. And for a poisson distribution the mean and variance are equal. So for $\sum_1^nX_i$ we have mean = variance = $\sum_1^n\lambda_i$

Now if we take an example $X$~$Poisson(\lambda)$ then what is the distribution of $2X$. So, according to the above theorem we can say $2X$~$Poisson(\lambda+\lambda)$ $\Rightarrow 2X$~$Poisson(2\lambda)$ $\Rightarrow Mean(2X)=Var(2X)=2\lambda$

But if we proceed with the mean and variance formula directly we get

$Mean(2X)=2Mean(X)=2*\lambda =2\lambda$ (this step is fine)

$Var(2X)=2^2Var(X) =4Var(X)=4*\lambda=4\lambda$ (I have confusion in this step) Here the variance is not equal to mean. So how can the distribution be Poisson?

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    $\begingroup$ What is the support of $2X$? $\endgroup$
    – user257566
    Sep 27, 2019 at 5:49
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    $\begingroup$ Support of X is {0,1,2,3,...}. So for 2X it should be {0,2,4,6,8,...} $\endgroup$
    – Azka
    Sep 27, 2019 at 5:50
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    $\begingroup$ That's right! This means that $2X$ certainly cannot be Poisson. The theorem you stated only holds when $X_i$ are independent, which you write in text in your first paragraph, but not in notation in your second paragraph. $\endgroup$
    – user257566
    Sep 27, 2019 at 5:51
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    $\begingroup$ Yes. I got your point. Of course X and X are not independent of each other. Thankyou. $\endgroup$
    – Azka
    Sep 27, 2019 at 5:54
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    $\begingroup$ I am certain this has been asked before but I can't locate a duplicate. $\endgroup$
    – Glen_b
    Sep 27, 2019 at 9:22

1 Answer 1

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$2X$ is not a Poisson RV. First contradiction is its support as @Baer comments, because the support of $2X$ is $\{0,2,4,...\}$ instead of $\{0,1,2,3...\}$, which is the support set for any Poisson RV. Also, the theorem states that the Poisson RVs to be summed need to be independent. In your case, $X$ is not independent of $X$.

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    $\begingroup$ Yes. I understood now. X and X are not independent. $\endgroup$
    – Azka
    Sep 27, 2019 at 5:55

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