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Let $X,Y$ be discrete independent random variables with $X,Y \geq 0$. Then $P(X+Y=n) = \underset{k=0}{\overset{n}{\sum}} P(X=n-k)P(Y=k).$

(I am sure this is very very easy and I am just overlooking something very simple)

$\textbf{Proof:}$ Trying to show right hand side is the same as left hand side.

$\sum_{k=0}^n P(X=n-k)P(Y=k) = \sum_{k=0}^n P(X=n-k,Y=k)$ since $X,Y$ independent, (At this point I am just trying things to see if I can get the conclusion) $ = \sum_{k=0}^n P(X+Y=n,Y=k)$ $ = \sum_{k=0}^n P(X+k=n)$ $ = P(X=n) + P(X=n-1)+ \cdots +P(X=0)$ $ = F_X(n)$

Could anyone help with this?

*Thank you for your responding André and Dilip.

It is very clear now.

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    Ok thank you, I will remember this for the future ;)2012-11-30

1 Answers 1

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We need more, for the result as it is stated. We need to assume that $X$ and $Y$ can only take on non-negative integer values.

If this is the case, how could $X+Y=n$?

We could have $X=n$ and $Y=0$. By independence, the probability that $X=n$ and $Y=0$ is $\Pr(X=n)\Pr(Y=0)$.

We could have $X=n-1$ and $Y=1$. By independence, the probability that $X=n-1$ and $Y=1$ is $\Pr(X=n-1)\Pr(Y=1)$.

We could have $X=n-2$ and $Y=2$. By independence, the probability that $X=n-2$ and $Y=2$ is $\Pr(X=n-2)\Pr(Y=2)$.

And so on!

Finally, we could have $X=0$ and $Y=n$. By independence, the probability that $X=0$ and $Y=n$ is $\Pr(X=0)\Pr(Y=n)$.

The above events are (pairwise) disjoint, and they are the only way we can have $X+Y=n$. So to find $\Pr(X+Y=n)$, we need to add up all the above probabilities. That's exactly what the formula $\Pr(X+Y=n)=\sum_{k=0}^n \Pr(X=n-k)\Pr(X=k)$ says.

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    Ok very c$l$ear thank $y$ou.2012-11-30