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I'm trying to derive the probability $Pr[Y<a+bZ,Y<X,Z<X]$, where $X$, $Y$, and $Z$ are independent exponential distributed random variables with parameters $1/\lambda_x$, $1/\lambda_y$, and $1/\lambda_z$, respectively. What I did is as follows:

$\Pr \left[ {Y < a + bZ,Y < X,Z < X} \right] = \int\limits_0^\infty {\int\limits_0^x {\int\limits_0^{a + bz} {{f_{X,Y,Z}}\left( {x,y,z} \right)dydzdx} } } \\ = \int\limits_0^\infty {\int\limits_0^x {\int\limits_0^{a + bz} {{f_X}\left( x \right){f_Y}\left( y \right){f_Z}\left( z \right)dydzdx} } } \\ = 1 - \frac{{{e^{\frac{{ - a}}{{{\lambda _y}}}}}}}{{{\lambda _z}}}{\left( {\frac{b}{{{\lambda _y}}} + \frac{1}{{{\lambda _z}}}} \right)^{ - 1}} - \frac{1}{{{\lambda _x}}}{\left( {\frac{1}{{{\lambda _x}}} + \frac{1}{{{\lambda _z}}}} \right)^{ - 1}} + \frac{{{e^{\frac{{ - a}}{{{\lambda _y}}}}}}}{{{\lambda _x}{\lambda _z}}}{\left( {\frac{b}{{{\lambda _y}}} + \frac{1}{{{\lambda _z}}}} \right)^{ - 1}}\left( {\frac{1}{{{\lambda _x}}} + \frac{b}{{{\lambda _y}}} + \frac{1}{{{\lambda _z}}}} \right)$

I did simulation to verify the analytic result. What I got is enter image description here.

Do you have any idea where the gap between simulated and analytic curves come from? Thank you very much.

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Your limits of integration don't include the restriction $Y\lt X$. To fix it, the integral needs to be split into three parts and the results added:

$$\int_{x=0}^a \int_{z=0}^{x} \int_{y=0}^{x} {f_X(x) f_Y(y) f_Z(z)\;dy\;dz\;dx} + \int_{x=a}^\infty \int_{z=0}^{\frac{x-a}{b}} \int_{y=0}^{a+bz} {f_X(x) f_Y(y) f_Z(z)\;dy\;dz\;dx} + \int_{x=a}^\infty \int_{z=\frac{x-a}{b}}^{x} \int_{y=0}^{x} {f_X(x) f_Y(y) f_Z(z)\;dy\;dz\;dx}.$$

The upper limit $\frac{x-a}{b}$ comes from finding that value of $z$ such that $y$ can range between $0$ and $x$ without exceeding $a+bz$. It is found by solving $x=a+bz$.

For $x\lt a$ we have $\frac{x-a}{b}\lt 0$ so this range for $x$ needs separating.

Note that these integral limits rely on $a\geq 0$ and $b\geq 1$. If this is not the case, the integrals need reworking.

We now calculate the three definite integrals in turn. For easier notation, we let: $$\alpha=1/\lambda_x \\ \beta=1/\lambda_y \\ \gamma=1/\lambda_z.$$

$$$$

\begin{eqnarray*} I_1 &=& \int_{x=0}^a \int_{z=0}^{x} \int_{y=0}^{x} {\alpha\beta\gamma\; e^{-\alpha x-\beta y-\gamma z}\;dy\;dz\;dx} \\ &=& \int_{x=0}^a \int_{z=0}^{x} \left[ -\alpha\gamma\; e^{-\alpha x-\beta y-\gamma z}\right]_{y=0}^{x} \;dz\;dx \\ &=& \int_{x=0}^a \int_{z=0}^{x} \left( -\alpha\gamma\; e^{-\alpha x-\beta x-\gamma z} + \alpha\gamma\; e^{-\alpha x-\gamma z}\right) \;dz\;dx \\ &=& \int_{x=0}^a \left[ \alpha\; e^{-\alpha x-\beta x-\gamma z} - \alpha\; e^{-\alpha x-\gamma z}\right]_{z=0}^{x} \;dx \\ &=& \int_{x=0}^a \left( \alpha\; e^{-\alpha x-\beta x-\gamma x} - \alpha\; e^{-\alpha x-\gamma x} - \alpha\; e^{-\alpha x-\beta x} + \alpha\; e^{-\alpha x}\right) \;dx \\ &=& \left[ \frac{-\alpha}{\alpha+\beta+\gamma }\; e^{-\alpha x-\beta x-\gamma x} + \frac{\alpha}{\alpha +\gamma}\; e^{-\alpha x-\gamma x} + \frac{\alpha}{\alpha +\beta}\; e^{-\alpha x-\beta x} - e^{-\alpha x}\right]_{x=0}^a \\ && \\ &=& \frac{-\alpha}{\alpha+\beta+\gamma }\; e^{-a(\alpha+\beta+\gamma)} + \frac{\alpha}{\alpha +\gamma}\; e^{-a(\alpha+\gamma)} + \frac{\alpha}{\alpha +\beta}\; e^{-a(\alpha +\beta)} - e^{-a\alpha} \\ &&\qquad + \frac{\alpha}{\alpha+\beta+\gamma } - \frac{\alpha}{\alpha +\gamma} - \frac{\alpha}{\alpha +\beta} + 1. \\ \end{eqnarray*}

$$\\$$

\begin{eqnarray*} I_2 &=& \int_{x=a}^{\infty} \int_{z=0}^{(x-a)/b} \int_{y=0}^{a+bz} {\alpha\beta\gamma\; e^{-\alpha x-\beta y-\gamma z}\;dy\;dz\;dx} \\ &=& \int_{x=a}^{\infty} \int_{z=0}^{(x-a)/b} \left[ -\alpha\gamma\; e^{-\alpha x-\beta y-\gamma z}\right]_{y=0}^{a+bz} \;dz\;dx \\ &=& \int_{x=a}^{\infty} \int_{z=0}^{(x-a)/b} \left( -\alpha\gamma\; e^{-\alpha x-\beta (a+bz)-\gamma z} + \alpha\gamma\; e^{-\alpha x-\gamma z}\right) \;dz\;dx \\ &=& \int_{x=a}^{\infty} \left[ \frac{\alpha\gamma}{b\beta+\gamma}\; e^{-\alpha x-\beta (a+bz)-\gamma z} - \alpha\; e^{-\alpha x-\gamma z}\right]_{z=0}^{(x-a)/b} \;dx \\ &=& \int_{x=a}^{\infty} \left( \frac{\alpha\gamma}{b\beta+\gamma}\; e^{-\alpha x-\beta x-\gamma (x-a)/b} - \alpha\; e^{-\alpha x-\gamma (x-a)/b} - \frac{\alpha\gamma}{b\beta+\gamma}\; e^{-\alpha x-a\beta} + \alpha\; e^{-\alpha x}\right) \;dx \\ &=& \left[ \frac{-\alpha\gamma}{(b\beta+\gamma)(\alpha+\beta+\gamma/b) }\; e^{-\alpha x-\beta x-\gamma (x-a)/b} + \frac{\alpha}{\alpha +\gamma/b}\; e^{-\alpha x-\gamma (x-a)/b} + \frac{\gamma}{b\beta + \gamma}\; e^{-\alpha x-a\beta} - e^{-\alpha x}\right]_{x=a}^\infty \\ && \\ &=& \frac{\alpha\gamma}{(b\beta+\gamma)(\alpha+\beta+\gamma/b) }\; e^{-a(\alpha +\beta)} - \frac{\alpha}{\alpha +\gamma/b}\; e^{-a\alpha} - \frac{\gamma}{b\beta + \gamma}\; e^{-a(\alpha +\beta)} + e^{-a\alpha}. \\ \end{eqnarray*}

$$\\$$

\begin{eqnarray*} I_3 &=& \int_{x=a}^{\infty} \int_{z=(x-a)/b}^{x} \int_{y=0}^{x} {\alpha\beta\gamma\; e^{-\alpha x-\beta y-\gamma z}\;dy\;dz\;dx} \\ &=& \int_{x=a}^{\infty} \int_{z=(x-a)/b}^{x} \left[ -\alpha\gamma\; e^{-\alpha x-\beta y-\gamma z}\right]_{y=0}^{x} \;dz\;dx \\ &=& \int_{x=a}^{\infty} \int_{z=(x-a)/b}^{x} \left( -\alpha\gamma\; e^{-\alpha x-\beta x-\gamma z} + \alpha\gamma\; e^{-\alpha x-\gamma z}\right) \;dz\;dx \\ &=& \int_{x=a}^{\infty} \left[ \alpha\; e^{-\alpha x-\beta x-\gamma z} - \alpha\; e^{-\alpha x-\gamma z}\right]_{z=(x-a)/b}^{x} \;dx \\ &=& \int_{x=a}^{\infty} \left( \alpha\; e^{-\alpha x-\beta x-\gamma x} - \alpha\; e^{-\alpha x-\gamma x} - \alpha\; e^{-\alpha x-\beta x-\gamma(x-a)/b} + \alpha\; e^{-\alpha x-\gamma(x-a)/b}\right) \;dx \\ &=& \left[ \frac{-\alpha}{\alpha+\beta+\gamma}\; e^{-\alpha x-\beta x-\gamma x} + \frac{\alpha}{\alpha +\gamma}\; e^{-\alpha x-\gamma x} + \frac{\alpha}{\alpha+\beta+\gamma/b}\; e^{-\alpha x-\beta x-\gamma(x-a)/b} - \frac{\alpha}{\alpha+\gamma/b}\; e^{-\alpha x - \gamma(x-a)/b}\right]_{x=a}^\infty \\ && \\ &=& \frac{\alpha}{\alpha+\beta+\gamma}\; e^{-a(\alpha +\beta+\gamma)} - \frac{\alpha}{\alpha +\gamma}\; e^{-a(\alpha+\gamma)} - \frac{\alpha}{\alpha+\beta + \gamma/b}\; e^{-a(\alpha +\beta)} + \frac{\alpha}{\alpha+\gamma/b}\; e^{-a\alpha}. \\ \end{eqnarray*}

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  • $\begingroup$ I think it should be: $\int\limits_{x = 0}^\infty {\int\limits_{z = 0}^{\frac{{x - a}}{b}} {\int\limits_{y = 0}^{a + bz} {{f_X}\left( x \right){f_Y}\left( y \right){f_Z}\left( z \right)dydzdx} } } + \int\limits_{x = 0}^\infty {\int\limits_{z = \frac{{x - a}}{b}}^x {\int\limits_{y = 0}^x {{f_X}\left( x \right){f_Y}\left( y \right){f_Z}\left( z \right)dydzdx} } }$ right? $\endgroup$ – BinhDDT Mar 10 '15 at 5:07
  • $\begingroup$ @BìnhVănNguyễn Yes, sorry, you're exactly right. I've made the change. Thanks for the correction. $\endgroup$ – Mick A Mar 10 '15 at 5:18
  • $\begingroup$ Dear, The first probability is given as ${P_1} = 1 - \frac{{{e^{\frac{{ - a}}{{{\lambda _y}}}}}}}{{{\lambda _z}}}{\left( {\frac{b}{{{\lambda _y}}} + \frac{1}{{{\lambda _z}}}} \right)^{ - 1}} - \frac{{{e^{\frac{a}{{b{\lambda _z}}}}}}}{{{\lambda _x}}}{\left( {\frac{1}{{{\lambda _x}}} + \frac{1}{{b{\lambda _z}}}} \right)^{ - 1}} + \frac{{{e^{\frac{a}{{b{\lambda _z}}}}}}}{{{\lambda _x}{\lambda _z}}}{\left( {\frac{1}{{{\lambda _x}}} + \frac{1}{{{\lambda _y}}} + \frac{1}{{b{\lambda _z}}}} \right)^{ - 1}}$ $\endgroup$ – BinhDDT Mar 10 '15 at 8:14
  • $\begingroup$ while the second probability is: ${P_2} = \frac{{{e^{\frac{a}{{b{\lambda _z}}}}}}}{{{\lambda _x}}}{\left( {\frac{1}{{{\lambda _x}}} + \frac{1}{{b{\lambda _z}}}} \right)^{ - 1}} - \frac{1}{{{\lambda _x}}}{\left( {\frac{1}{{{\lambda _x}}} + \frac{1}{{{\lambda _z}}}} \right)^{ - 1}} - \frac{{{e^{\frac{a}{{b{\lambda _z}}}}}}}{{{\lambda _x}}}{\left( {\frac{1}{{{\lambda _x}}} + \frac{1}{{{\lambda _y}}} + \frac{1}{{b{\lambda _z}}}} \right)^{ - 1}} + \frac{1}{{{\lambda _x}}}{\left( {\frac{1}{{{\lambda _x}}} + \frac{1}{{{\lambda _y}}} + \frac{1}{{{\lambda _z}}}} \right)^{ - 1}}$ $\endgroup$ – BinhDDT Mar 10 '15 at 8:17
  • $\begingroup$ However, the sum of them is totally different from the simulated (exact) probability. Can you check it again? Thank you. $\endgroup$ – BinhDDT Mar 10 '15 at 8:18

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