Random variables X and Y each have a Poisson ($\lambda$) (i.e: X ~ Poisson ($\lambda$) and Y ~ Poisson($\lambda$).
The first part of the question asks to find the PMF of E(X+Y). I know that (X+Y) ~ Poisson ($\lambda + \lambda$) and my $P_x(k)$ will be:
$P_x(k) = \frac{e^{-\lambda-\lambda} \cdot (\lambda + \lambda)^{k}}{k!} $
The second part of the question asks to now assume that X=Y and to find the PMF of X+Y. Would this then imply that 2X~Poisson$(2\lambda)$? Would my PMF then be:
$P_x(k) = \frac{e^{-2\lambda} \cdot (2\lambda)^{k}}{k!} $
For some reason I feel like I'm doing something wrong. Any help would be much appreciated!
If $X = Y$, then $X + Y = 2X$. What this means is that if you observe, for example, $X = 3$, then $X + Y = 2X = 6$. And what this tells you is that $2X$ cannot be Poisson, because it is impossible for $2X$ to be an odd number.
What you can say, however, is that if $$\Pr[X = x] = e^{-\lambda} \frac{\lambda^x}{x!}, \quad x = 0, 1, 2, \ldots,$$ then $$\Pr[2X = 2x] = \Pr[X = x] = e^{-\lambda} \frac{\lambda^x}{x!}, \quad x = 0, 1, 2, \ldots.$$ So the random variable $2X$ has a "Poisson-like" distribution, but not over the nonnegative integers: instead, its support is over the nonnegative even integers. Another way to write this is $$\Pr[2X = m] = \begin{cases} \displaystyle e^{-\lambda} \frac{\lambda^{m/2}}{(m/2)!}, & m = 0, 2, 4, \ldots, \\ 0, & \text{otherwise.}\end{cases}$$ This makes it explicit that $2X$ is not itself Poisson.