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1) If $X$, $Y$ are independent Binomial random variables with parameters $(n, p)$ and $(m, p)$, respectively, then $X+Y \sim Binomial(n+m, p)$.

2) If $X$,$Y$ are independent Poisson random variables with parameters $\lambda_1$ and $\lambda_2$,respectively, then $X+Y \sim Poisson (\lambda_1+\lambda_2)$.

How can I prove these with examples ??

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closed as off-topic by heropup, Did, Jimmy R., user91500, user186473 Nov 8 '16 at 10:16

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Some hints:

Note that a $\text{Binomial}(n,p)$ random variable is the sum of $n$ independent $\text{Bernoulli}(p)$ random variables. From this fact, 1) should be easy to prove.

For 2), you have to get your hands dirty and compute the convolution $P(X+Y=k)=\sum_{i=0}^k P(X=i) P(Y=k-i)$. You can find this on this website and elsewhere.

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  • $\begingroup$ I have to show by numbers !! I calculated the probabilities of: X with (n,p), Y with (m,p) and Z (n+m,p), then how can i prove? $\endgroup$ – yarub Nov 8 '16 at 6:37
  • $\begingroup$ @yarub Use the convolution, if you must, $\mathsf P(X+Y=z) = \sum_{x=\max(0, z-n)}^{\min(z, n)}\mathsf P(X=x)\mathsf P(Y=z-x)$ $\endgroup$ – Graham Kemp Nov 8 '16 at 7:06
  • $\begingroup$ OK Did it !! Thanks a lot for your explanation $\endgroup$ – yarub Nov 8 '16 at 8:26

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