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Let $X$ and $Y$ be exponentially distributed with parameters $\lambda$ and $\mu$, respectively. Find $P(X > Y)$ by conditioning.

I have read this answer: Comparing two exponential random variables

But my exercise says that I can do it with conditioning expected value (not a double integral). I I was wondering if anyone knows how to do it this way. I can't figure it out. I tried conditioning on the lower bound of $Y$. The answer should be $\mu/(\lambda + \mu)$.

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$P ( Y \lt X) = \mathbb{E} I ( Y \lt X) = \mathbb{E} \mathbb{E} \left[ I ( Y \lt X) \mid X \right] = \mathbb{E} \text{cdf}_{Y} (X) = \mathbb{E} I (X > 0) (1 - e^{- \mu X}) = \mathbb{E}(1 - e^{- \mu X}) = \\ 1 -\mathbb{E} e^{- \mu X} = 1 -\int_0^\infty e^{- \mu x} \lambda e^{- \lambda x} dx = 1 - \frac{\lambda}{\lambda + \mu} = \frac{\mu}{\lambda + \mu} $

Denis Korzhenkov
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