What is the probability of $X I have been trying to derive the probability $Pr[X<Y<Z]$, where $X$, $Y$, and $Z$ are independent and follow exponential distribution with parameters $\lambda_x$, $\lambda_y$, and $\lambda_z$, respectively.
What I did is as follows:
$\Pr \left[ {X < Y < Z} \right] = \Pr \left[ {X < Y,Y < Z} \right]\\
 = \int\limits_0^\infty  {\Pr \left[ {X < y,Z > y} \right]{f_Y}(y)dy} \\
 = \int\limits_0^\infty  {\Pr \left[ {X < y} \right]\Pr \left[ {Z > y} \right]{f_Y}\left( y \right)dy}$
Is it right or wrong?
If it is wrong, how can we solve this problem?
Thank you very much.
 A: You can integrate the joint pdf over the region of integration in 3!=6 different ways:
$\begin{gathered}
  \int\limits_{z = 0}^{z = \infty } {\int\limits_{y = 0}^{y = z} {\int\limits_{x = 0}^{x = y} {{f_{X,Y,Z}}\left( {x,y,z} \right)dxdydz} } }  \hfill \\
  \int\limits_{z = 0}^{z = \infty } {\int\limits_{x = 0}^{x = z} {\int\limits_{y = x}^{y = z} {{f_{X,Y,Z}}\left( {x,y,z} \right)dydxdz} } }  \hfill \\
  \int\limits_{y = 0}^{y = \infty } {\int\limits_{x = 0}^{x = y} {\int\limits_{z = y}^{z = \infty } {{f_{X,Y,Z}}\left( {x,y,z} \right)dzdxdy} } }  \hfill \\
  \int\limits_{y = 0}^{y = \infty } {\int\limits_{z = y}^{z = \infty } {\int\limits_{x = 0}^{x = y} {{f_{X,Y,Z}}\left( {x,y,z} \right)dxdzdy} } }  \hfill \\
  \int\limits_{x = 0}^{x = \infty } {\int\limits_{z = x}^{z = \infty } {\int\limits_{y = x}^{y = z} {{f_{X,Y,Z}}\left( {x,y,z} \right)dydzdx} } }  \hfill \\
  \int\limits_{x = 0}^{x = \infty } {\int\limits_{y = x}^{y = \infty } {\int\limits_{z = y}^{z = \infty } {{f_{X,Y,Z}}\left( {x,y,z} \right)dzdydx} } }  \hfill \\ 
\end{gathered} $
You may be able to find a close form expression among these. I haven't try it myself yet !
