# Prove or Disprove : Scalar Mutliple of Poisson Random Variable is again Poisson

Given 2 random independent variables $$X_1\sim Poisson(\lambda_1), X_2\sim Poisson (\lambda_2)$$. I want to find out if $$V = 2X_1 + X_2 \sim Poisson(2\lambda_1 + \lambda_2)$$ or to show that it's false.

This is true iff $$2X_1\sim Poisson(2\lambda_1)$$

So by definition: $$P(2X_1=y) = \sum_{x\in R_{X_1}:2x=y}P(X_1 = x)$$

This gives a sum over the even numbers, where do i go from here?

## 1 Answer

if $$X \sim poisson (\lambda)$$ then $$kX \not \sim poisson(k.\lambda)$$ .Because note that $$Var[kX] = k^2 \lambda$$ and $$Var[poisson(k.\lambda)] = k.\lambda$$ so $$k^2 \lambda = k. \lambda$$ and since for valid poisson distribution $$\lambda>0$$. Then constant should satisfy $$k(k-1) = 0$$ from which only possible only value is $$k=1$$

We proved that $$kX \not \sim poisson (k.\lambda)$$ unless $$k = 1$$ that doesn't imply that it is not poisson distribution of some other parameter.but it cannot be even that because following .let $$2X = Y$$ and if $$Y \sim poisson(\lambda^{*})$$

$$P[Y = y] = P[2X = y] = P[X = y/2] = \frac{e^{-\lambda_1}.\sqrt{\lambda_1}^y}{(y/2)!} = \frac{e^{-\lambda_{*}} \lambda_{*}^{y} }{y!} \hspace{1cm} forall \, y \geq 0$$

put $$y=0$$ from which we have that $$\lambda_{*} = \lambda_1$$ so $$2X$$ is not poisson r.v