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Smith is waiting for his two friends Lee and Yang to visit his house. The time until Lee arrives is Exp($\lambda_1$) and the time until Yang arrives is Exp($\lambda_2$). After arrival, Lee stays an amount of time that is Exp($\mu_1$), whereas Yang stays an amount of time that is Exp($\mu_2$). All four random variables are independent. What is the probability that Lee arrives before and departs after Yang?

Attempt:

Denote A=Lee's arrive time; B=Lee's stay time; C=Yang's arrive time; D=Yang's stay time.

Then, this question is basically ask $P(A<C$ & $A+B<C+D).$

I recognize that $A-C<0$, and that the second inequality of the term above can be re-written as $A-C<D-B$.

But then I got stuck. Can I simply calculate $P(0<D-C)?$

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    $\begingroup$ The question is $P(A<C \cap A+B>C+D)$ $\endgroup$
    – kmitov
    Mar 2, 2016 at 6:32

1 Answer 1

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We have

\begin{align} P(A\lt C\cap A+B\gt C+D) &= \int_{a=0}^{\infty} \lambda_1 e^{-\lambda_1 a} \int_{b=0}^{\infty} \mu_1 e^{-\mu_1 b} \int_{c=a}^{a+b} \lambda_2 e^{-\lambda_2 c} P(D\lt a+b-c)\;dc\;db\;da \\ & \\ &= \int_{a=0}^{\infty} \lambda_1 e^{-\lambda_1 a} \int_{b=0}^{\infty} \mu_1 e^{-\mu_1 b} \int_{c=a}^{a+b} \lambda_2 e^{-\lambda_2 c} \left(1-e^{-\mu_2(a+b-c)}\right)\;dc\;db\;da \\ \end{align}

This is a do-able but tedious integral. I don't know of any shortcut method partly because all four parameters, $\lambda_1,\lambda_2,\mu_1,\mu_2$, can be different.

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