X and Y are distributed by the same distribution

Question

I know that if $X_1 \tilde~ Poisson(\lambda_1)$ and $X_2 \tilde~ Poisson(\lambda_2)$ then $X+Y \tilde~ Poisson(\lambda_1 + \lambda_2)$

Is this true with any distribution? Or is it merely because here $E(X)=Var(X)=\lambda$

Answer 1

If $X_1$ and $X_2$ are Poisson like you describe, and independent, then the sum $X_1+X_2$ is indeed Poisson.

The analogue fails for almost all other distributions. But it is true in a few important cases. If $X_1$ and $X_2$ are independent and normally distributed, then $X_1+X_2$ has normal distribution.

If $X_1$ has binomial distribution with parameters $n_1$ and $p$, and $X_2$ has binomial distribution with parameters $n_2$ and $p$, and $X_1$ and $X_2$ are independent, then $X_1+X_2$ has binomial distribution with parameters $n_1+n_2$ and $p$.

But for instance if $X_1$ and $X_2$ are independent with exponential distribution, then $X_1+X_2$ does not have exponential distribution.

If $X_1$ and $X_2$ are independent with uniform distribution, then $X_1+X_2$ does not have uniform distribution.

Answer 2

I agree with Andre on the fact that it is not necessary for two independent identical random variable's sum to have the same distribution. Regarding your question if it's because $E(X)= Var (x)=\lambda$, I think, it's not the case. Because in case of normal variables, this relation does not hold, yet, it has property that the sum of two normally distributed random variables add up to normal random variable. Hope this helps. Correct me if I'm wrong.