# $P(X_1 < X_2 < X_3)$ for Exponential Random Variables

Let $X_1 \sim \exp(\lambda_1)$, $X_2 \sim \exp(\lambda_2)$ and $X_3 \sim \exp(\lambda_3)$ be independent random variables, where the exponential density I'm using is $$f_X(x) = \frac{1}{\lambda}e^{-x/\lambda}$$

I'm trying to calculate $P(X_1 < X_2 < X_3)$ but am stuck. So far, I've rewritten this as $$P(X_1 < X_2 < X_3) = P(X_1 < X_2 \cap X_2 < X_3) = P(Y_1 < 0 \cap Y_2 < 0) \tag{*}$$ where $Y_1 = X_1 - X_2$ and $Y_2 = X_2 - X_3$. I've also worked out the marginal distributions for $Y_1$ and $Y_2$. Here's the pdf for $Y_1$: $$f_{Y_1}(y) = \begin{cases} \dfrac{e^{-y/\lambda_1}}{\lambda_1 + \lambda_2}, \qquad y > 0 \\[6pt] \dfrac{e^{y/\lambda_2}}{\lambda_1 + \lambda_2}, \qquad y \leq 0 \end{cases}$$

But, I need the joint pdf of $(Y_1, Y_2)$ to calculate $(*)$ since the two are not independent. I can write the vector $(Y_1, Y_2)$ as $$\begin{pmatrix} Y_1 \\ Y_2 \end{pmatrix} = \begin{pmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \end{pmatrix} \begin{pmatrix} X_1 \\ X_2 \\ X_3 \end{pmatrix}$$ but this doesn't seem to help me work out the joint distribution as it does for linear combinations of normals.

• Introducing dependence is a bad idea, use instead $$P(X_1<x)=1-e^{-x/\lambda_1}\qquad P(X_3>x)=e^{-x/\lambda_3}$$ hence $$P(X_1<X_2<X_3)=E((1-e^{-X_2/\lambda_1})e^{-X_2/\lambda_3})$$ Can you finish this? – Did Oct 10 '16 at 19:47
• @Did Is the $E$ on purpose? – Cehhiro Oct 10 '16 at 19:49
• @OFRBG Tell me... Of course it is. – Did Oct 10 '16 at 19:51
• @Did Cool, I guess $(1 - e^{-X_2/\lambda_1}) = P(X_1 < X_2 \mid X_2)$? – bcf Oct 10 '16 at 19:51
• Indeed.   – Did Oct 10 '16 at 19:52

$\newcommand{\E}{\operatorname{E}}$Suppose the distribution of $X$ is $e^{-x/\lambda} (dx/\lambda)$ for $x\ge0.$ Then you have these two lemmas:

• If $x\ge0$, then $\Pr(X>x) = e^{-x/\lambda}.$
• If $0\le x_1\le x_2$ then $\Pr(X>x_2\mid X> x_1) = \Pr(X>x_2=x_1) = e^{-(x_2-x_1)/\lambda}.$

Then we have \begin{align} & \Pr(X_1<X_2) = \E(\Pr(X_1<X_2\mid X_1)) = \E(e^{-X_1/\lambda_2}) = \int_0^\infty e^{-x_1/\lambda_2} \left( e^{-x_1/\lambda_1}\ \frac{dx_1}{\lambda_1} \right) \\[6pt] = {} & \int_0^\infty e^{-\left( \frac 1 {\lambda_1} + \frac 1 {\lambda_2} \right) x_1} \left( \frac{dx_1}{\lambda_1} \right) = \frac 1 {\lambda_1}\cdot \frac 1 {\frac 1 {\lambda_1} + \frac 1 {\lambda_2}} = \frac 1 {1 + \frac{\lambda_1}{\lambda_2}} = \frac{\lambda_2}{\lambda_1+\lambda_2}. \end{align} And then \begin{align} \Pr(X_1 < X_2 < X_3) & = \E(\Pr(X_1 < X_2 < X_3 \mid X_1)) \\[6pt] & = \E( \cdots \end{align}

(ok, I'm going to finish this later, but maybe this is enough for the original poster to finish this $\text{off } \ldots)$

• Conditioning on $X_2$, not $X_1$, is simpler, as I explain in my comment. – Did Oct 10 '16 at 21:01
• Perhaps..... ok, I'll try it both ways. You get Beta distributions. $\qquad$ – Michael Hardy Oct 10 '16 at 21:23
• What is Beta and relevant to solve this? – Did Oct 10 '16 at 21:24
• @Did : I'll come back to this later$\ldots\qquad$ – Michael Hardy Oct 10 '16 at 21:26