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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.

choco_addicted
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BinhDDT
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    Looks okay to me. – drhab Feb 26 '15 at 10:38
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    Me too. If you want to justify some steps then $Pr(X<y,Z>y|Y=y)=Pr(X<y,Z>y)$ by independence and $Pr(X<y,Z>y)=Pr(X<y)P(Z>y)$ again by independence... also note that all this follows by independence and works for any distribution (i.e. the exponential distribution is superfluous). – user103828 Feb 26 '15 at 10:40
  • @drhab: Can we evaluate the above probability as follows

    $\Pr \left[ {X < Y < Z} \right] = \int\limits_0^\infty {\int\limits_0^z {\int\limits_0^y {{f_{X,Y,Z}}\left( {x,y,z} \right)} } } dxdydz\ = \int\limits_0^\infty {\int\limits_0^z {\int\limits_0^y {{f_X}\left( x \right){f_Y}\left( y \right){f_Z}\left( z \right)} } } dxdydz $

    I have tried, but results are different. Do you have any idea?

     – BinhDDT Mar 02 '15 at 02:30
  • That is legal too. If the results don't match then I can find only one explanation for it: in at least one of the evaluations a mistake has been made. – drhab Mar 02 '15 at 09:52

1 Answers1

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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 !

Tuong Nguyen Minh
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