I am self-learning undergraduate calculus-based probability. I'd like someone to verify, if my conclusion for the below exercise problem from Intro to Probability, Blitzstein and Hwang, 4.27 is correct.
[BH 4.27] Let $X$ and $Y$ be $Poisson(\lambda)$ random variables and $T=X+Y$. Suppose that $X$ and $Y$ are not independent, and in fact $X = Y$. Prove or disprove the claim that $T \sim Poisson(2\lambda)$ in this scenario.
Solution. (My Attempt)
I cannot come up with a direct counterexample to disprove the above. However, we see that:
\begin{align*} P(T = k) &= P(2X = k)\\ &= P(X = k/2) \\ &= e^{-\lambda}\frac{(\lambda)^{k/2}}{(k/2)!} \quad k\in\{0,2,4,\ldots\} \end{align*}
This density is certainly different from $Poisson(2\lambda)$.