Find the probability of $P(t<X_1<X_2)$ where $X_i\sim \exp(\rho_i)$ and independent from each other.

Question

Let $X_i\sim \exp(\rho_i)$ independent from each other.

How can I show that the following hold,

$P(t<X_1<X_2)=\int_t^\infty P(x<X_2)f_{X1}(x)dx=\frac{\rho_1}{\rho_1+\rho_2}e^{-(\rho_1+\rho_2)t} $

Answer 1

$X_i\sim exp(\rho_i)$, $\rho_i>0$, so $X_i$ has density $f_{X_i}(x)= \rho_i e^{-\rho_i x}$ for $x>0$ and zero otherwise.

Hence, the cdf is ${\bf P}(X_i\leq x)=F_{X_i}(x)=\int_0^x f_{X_i}(t)dt = \int_0^x \rho_i e^{-\rho_i t}dt = 1- e^{-\rho_i x}$ for $x>0$ and zero otherwise.

Now, \begin{eqnarray*} {\bf P}(t<X_1<X_2) &=& \int_0^{\infty}{\bf P}(t<X_1<X_2|X_1=x)f_{X_1}(x)dx\\ &=& \int_0^{\infty}{\bf P}(t<x<X_2|X_1=x)f_{X_1}(x)dx\\ &=& \int_0^{\infty}{\bf P}(t<x<X_2)f_{X_1}(x)dx\\ &=& \int_t^{\infty}{\bf P}(t<x<X_2)f_{X_1}(x)dx\\ &=& \int_t^{\infty}{\bf P}(x<X_2)f_{X_1}(x)dx\\ &=& \int_t^{\infty}e^{-\rho_2 x}\rho_1e^{-\rho_1 x}dx\\ &=& \rho_1\int_t^{\infty}e^{-(\rho_1+\rho_2) x}dx\\ &=& \rho_1 \frac{e^{-(\rho_1+\rho_2) x}}{-(\rho_1+\rho_2)}\Big|_t^{\infty}\\ &=& \frac{\rho_1}{\rho_1+\rho_2} e^{-(\rho_1+\rho_2) t}\\ \end{eqnarray*}