Use MGFs to determine whether $X + 2Y$ is Poisson, for $X, Y$ i.i.d. $\sim\mathcal P(\lambda)$.
So, I know that the MGF of $X+Y$ is the product of the two respective MGFs. Thus, we get: $ X+2y = X + Y + Y$, which we can then multiply the MGF of $X * Y * Y$, which = $e^{3λ(e^t -1)}$. However, I don't think I did this calculation right. Anyone help me?
Since $X$ and $Y$ are independent, so are $X$ and $2Y$. The MGF of $X+2Y$ evaluated at $t$ is $$\exp(\lambda(e^{t}-1))\exp(\lambda(e^{2t}-1)) = \exp(\lambda(e^{2t}+e^{t}-2))$$
This is not the MGF of any Poisson distributed random variable, so $X+2Y$ is not Poisson.
To prove that $\exp(\lambda(e^{2t}+e^{t}-2))$ is not the MGF of a Poisson variable, not that it's asymptotically $\displaystyle \sim e^{\lambda e^{2t}}$ while the MGF of a Poisson variable would be $\displaystyle\sim e^{\beta e^{t}}$ (which is a contradiction, except when $\lambda=0$, but this is irrelevant).