Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $X$ and $Y$ are independent identically distributed random variables where

$P(X=k) = P(Y=k) = pq^{k-1}$ where $q = 1-p$.

How do you find $P(X+Y=k)$? Is it acceptable to say that $$P(X+Y=k) = P(\{X=m\} , \{Y=k-m\}) = P(X=m)P(Y=n-m)$$ since$ X, Y$ are independent? I'm not sure if this is legitimate and I'm also not sure whether it's possible to do this without throwing in the extra '$m$' variable.

Any advice would be much appreciated.


The wider question that I'm trying to solve is $P(X=k \mid X+Y=n+1)$ and my method so far requires $P(X+Y = n+1)$ - is there another method?

share|improve this question
You are close. The probability that $X+Y=k$ is the sum $\sum_{m=1}^{k-1}\Pr(X=m)\Pr(Y=k-m)$. –  André Nicolas Dec 18 '13 at 17:44
@AndréNicolas ahhh that makes sense, thank you! –  Taimur Dec 18 '13 at 17:50
You are welcome. Your way of computing the conditional probability should work well. The answer may be a little surprising. –  André Nicolas Dec 18 '13 at 17:54
I just tried the summation you suggested - it looks like the 'm' variable doesn't matter, and I get that P(X+Y=k) = p^2 q^(k-2). However, when I use this to calculate my conditional probability, I just get an answer of 1, which I'm certain is wrong, any idea what I'm doing wrong? Thanks again –  Taimur Dec 18 '13 at 18:17
The sum has $k-1$ terms all of which are equal. So the sum is $(k-1)p^2q^{k-2}$. –  André Nicolas Dec 18 '13 at 19:03

1 Answer 1

By its moment generating function, you can prove that the sum of geometric distribution rv is a negative binomial distribution.

your way is legit. but for a k of large value, you have to calculate a lot.

please refer to this link:Sum of two independent geometric random variables

share|improve this answer
At the moment, not all statements in the question are legit. –  Ragnar Dec 18 '13 at 19:04
where is not legit? –  user116541 Dec 19 '13 at 3:24
The OP states that $P(X+Y=k)=P(X=m)P(y=n-m)$, but that is not true. –  Ragnar Dec 19 '13 at 9:48
why this is not true? I didn't see any logical problem here as long as x and y are iid –  user116541 Dec 20 '13 at 16:58
When $k=2$, you get $P(x+y=2)=P(x=1)P(y=1)$ for $n=2$ and $m=1$, but that's not true, because $P(x+y=2)=P(x=0)P(y=2)+P(x=1)P(y=1)+P(x=2)P(y=0)$. (assuming that $x$ and $y$ are nonnegative integers). –  Ragnar Dec 20 '13 at 17:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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