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?
Best 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