If $X~Poisson(\lambda)$ then does $2X$ also follow Poisson with parameter $2\lambda$

Question

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?

Answer 1

$2X$ 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 $X$.