Assume $X_1, X_2, X_3$ are three independent exponential random variables with means $1/A$, $1/B$ and $1/C$ resp. How do we calculate $P(t-X_1<X_2\mid t-X_1<X_3)$?

My try:

\begin{align} &P(X_1>t-X_2\mid X_1>t-X_3)\\ &=\int_{x_3}\int_{x_2}P(X_1>t-x_2\mid X_1>t-x_3)(B e^{-Bx_2})(C e^{-Cx_3})dx_2 dx_3\\ &=\int_{x_3}\int_{x_2}P(X_1>x_3-x_2)(B e^{-Bx_2})(C e^{-Cx_3})dx_2 dx_3\\ \end{align}

I end up at a result which is independent of $t$. What is the right approach?


The belief that, if $(X,U,V)$ is independent, $U$ and $V$ with densities $f_U$ and $f_V$ respectively, then $$P(X\gt U\mid X\gt V)=\iint P(X\gt u\mid X\gt v)\,f_U(u)\,f_V(v)\,\mathrm du\,\mathrm dv,$$ is unfounded. This might be (wrongly) imitating the (true) identity that, if $(X,U)$ is independent, $U$ with density $f_U$, then $$P(X\gt U)=\int P(X\gt u)\,f_U(u)\,\mathrm du.$$


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