# Find $P(X_1+X_2<X_3)$

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

1. $$P(X_1
2. $$P(X_1+X_2

I know that for 1. $$P(X_1, but I'm stuck for 2. I know that $$X_1+X_2$$ follows a Gamma distribution, but then how do I compare that with $$X_3$$?

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.
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
• Can you please explain how you arrived at the above expression of $P(X_1+X_2<X_3)=\int_0^{\infty} \int_0^{\infty} \int_{x_1+x_2}^{\infty}\lambda_1 \lambda_2 \lambda_3 e^{-\lambda_1 x_1} e^{-\lambda_2 x_2} e^{-\lambda_3 x_3} \, dx_3 \, dx_1 \, dx_2$? – xynikocy Mar 28 at 7:58
• @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 $\{(x_1,x_2,x_3) \in \mathbb R^{3}:x_1+x_2<x_3\}$. – Kavi Rama Murthy Mar 28 at 8:04