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

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?

• What is the support of $2X$? Sep 27, 2019 at 5:49
• Support of X is {0,1,2,3,...}. So for 2X it should be {0,2,4,6,8,...}
– Azka
Sep 27, 2019 at 5:50
• 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. Sep 27, 2019 at 5:51
• Yes. I got your point. Of course X and X are not independent of each other. Thankyou.
– Azka
Sep 27, 2019 at 5:54
• I am certain this has been asked before but I can't locate a duplicate. Sep 27, 2019 at 9:22

$$2X$$ 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 $$X$$.