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 ??


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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, Did, Jimmy R., user91500
If this question can be reworded to fit the rules in the help center, please edit the question.


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.

  • $\begingroup$ 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

Not the answer you're looking for? Browse other questions tagged or ask your own question.