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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?

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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

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

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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

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