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Let $X$ be a exponentially distributed random variable(time interval) with mean of $u$

And $Y$ be a exponentially distributed random variable (time interval)with mean of $\lambda$

And $s$ be a constant value

Suppose $X$ is split by a uniformly distributed time epoch ($\Omega$). to find distribution and expected value of $\Omega$ which is between $0$ and $X$ I already used: \begin{align} \nonumber f_{\Omega}(\omega)&=\int_{t=0}^{\infty}f_{\Omega_i|T}(\omega|t) f_{T}(t)dt \\ &=\int_{t=0}^{\infty}\frac{1}{t} \frac{1}{u} e^{-t/u}dt \end{align} \begin{align} \label{eq:Es-w2} \nonumber \mathbb{E}[\Omega] &= \int_{\omega=0}^{\omega=\infty} \omega f_{\Omega}(\omega)d \omega\\ \nonumber &=\int_{\omega=0}^{\omega=\infty} \omega \int_{t=0}^{\infty}\frac{1}{t} \frac{1}{u} e^{-t/u}dt d\omega \\ \nonumber &= \int_{t=0}^{\infty} \int_{\omega=0}^{\omega=t} \frac{\omega}{t} \frac{1}{u} e^{-t/u}d\omega dt = \int_{t=0}^{\infty} \frac{t}{2u} e^{-t/u} dt =u/2,\\ \end{align} Now I want to know what is probability of $p(\Omega>Y+s)$?

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Assuming $X, Y$ are independent and you mean that $\Omega|X = x \sim \text{Uniform}(0, x)$. Then for $s > 0$,

$$ \begin{align} \Pr\{\Omega > Y + s\} & = \int_0^{\infty}\int_0^{\infty} \Pr\{\Omega > y + s|X = x\} \frac {1} {\mu} e^{-\frac {x} {\mu}} \frac {1} {\lambda} e^{-\frac {y} {\lambda}} dydx \\ & = \frac {1} {\mu\lambda}\int_s^{\infty}e^{-\frac {x} {\mu}} \int_0^{x-s} \left(1 - \frac {y + s} {x}\right) e^{-\frac {y} {\lambda}}dydx \\ & = \frac {1} {\mu}\int_s^{\infty}e^{-\frac {x} {\mu}} \left[ \left(1 - \frac {s} {x}\right) \left(1 - e^{-\frac {x - s} {\lambda}}\right) + \frac {1} {x} e^{-\frac {x - s} {\lambda}}(\lambda + x - s) - \frac {\lambda} {x} \right]dx \\ \end{align}$$

This will end up with exponential integral so I do not proceed further from here.

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