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Given that $X_i\sim Exp(\lambda_i),i\in\mathbb{N}$, find

  1. $P(X_1<X_3)$
  2. $P(X_1+X_2<X_3)$

I know that for 1. $P(X_1<X_3)=P\big(X_1=\min\{X_1,X_3\}\big)=\frac{\lambda_1}{\lambda_1+\lambda_3}$, but I'm stuck for 2. I know that $X_1+X_2$ follows a Gamma distribution, but then how do I compare that with $X_3$?

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The solution by Kavi is very fine. But here is a different look at the problem to use what you have at hand.

We have - by Bayes' - that \begin{align} \Pr\{X_3>X_1+X_2\}&=\Pr\{X_3>X_1+X_2|X_3>X_2\}\Pr\{X_3>X_2\}\\ &+\Pr\{X_3>X_1+X_2|X_3\leq X_2\}\Pr\{X_3\leq X_2\}, \end{align} noting that $\Pr\{X_3>X_1+X_2|X_3\leq X_2\}=0$, the expression simplifies to \begin{align} \Pr\{X_3>X_1+X_2\}&=\Pr\{X_3>X_1+X_2|X_3>X_2\}\Pr\{X_3>X_2\}\\ &=\Pr\{X_3>X_1\}\Pr\{X_3>X_2\}\\ &=\frac{\lambda_1}{\lambda_1+\lambda_3}\frac{\lambda_2}{\lambda_2+\lambda_3}. \end{align} where the second equality follows from the memoryless property of exponential distribution.

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First of all none of these probabilities can be computed without the assumption that $X_i$'s are independent. Under indepedence $P(X_1+X_2<X_3)=\int_0^{\infty} \int_0^{\infty} \int_{x_1+x_2}^{\infty}\lambda_1 \lambda_2 \lambda_3 e^{-\lambda_1 x_1} e^{-\lambda_2 x_2} e^{-\lambda_3 x_3} \, dx_3 \, dx_1 \, dx_2$

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  • $\begingroup$ Can you please explain how you arrived at the above expression of $P(X_1+X_2<X_3)=\int_0^{\infty} \int_0^{\infty} \int_{x_1+x_2}^{\infty}\lambda_1 \lambda_2 \lambda_3 e^{-\lambda_1 x_1} e^{-\lambda_2 x_2} e^{-\lambda_3 x_3} \, dx_3 \, dx_1 \, dx_2$? $\endgroup$ – xynikocy Mar 28 at 7:58
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    $\begingroup$ @xynikocy $\lambda_1 \lambda_2 \lambda_3 e^{-\lambda_1 x_1} e^{-\lambda_2 x_2} e^{-\lambda_3 x_3} $ is the joint density of $(X_1,X_2,X_3)$ so you get the probability by integrating over the region $ \{(x_1,x_2,x_3) \in \mathbb R^{3}:x_1+x_2<x_3\}$. $\endgroup$ – Kavi Rama Murthy Mar 28 at 8:04
  • $\begingroup$ @Graham Kemp Thanks for correcting the silly mistakes. $\endgroup$ – Kavi Rama Murthy Mar 28 at 8:08

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