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I'm trying to follow a line in a derivation for $P(Z>X+Y)$ where $X,Y,Z$ are independent continuous random variables distributed uniformly on $(0,1)$.

I've already derived the pdf of $X+Y$ using the convolution theorem, but there's a line in the answer that says:

$P(Z>X+Y) = \mathbb{E}[\ P(Z>X+Y\ |\ X+Y )\ ]$ where $\mathbb{E}$ is the expectation.

I'm not familiar with this result. Could anyone give a pointer to a similar result if one exists?

Thanks.

6 Answers 6