$X_1$, $X_2$, $X_3$ are random variables distributed following non-identically independent exponential distribution. The PDF $X_i$, $f_{X_i}(x)$=$\frac{1}{\Omega_i}\exp(\frac{x}{\Omega_i}), i=1,...,3$. I want to calculate the CDF

$Y=\frac{aX_1}{X_2(1+b X_3)}$. I was wondering that if it possible to do calculation as follows: $F_Y(y)= \int \limits_{0}^{\infty}\{ \int\limits_{0}^{\infty} \Pr\{aX_1<x_2(1+bx_3)\}f_{X_2}(x_2)\}f_{X_3}(x_3)dx_2 dx_3$.

If anyone knows the method to calculate, Please give me a hint! Many thanks for your help


Let $CDF_{X_{1}}(z)$ denote the cumulative distribution function for $X_{1}$ evaluated at the point $z$, so that $CDF_{X_{1}}(z) = Pr(X_{1}\leq z)$.

Your current setup for $F_{Y}(y)$ is off slightly. You should get: $$ F_{Y}(y) = Pr(Y\leq y) = Pr(\frac{aX_{1}}{X_{2}(1+bX_{3})} \leq y) $$ $$ = Pr(X_{1} \leq \frac{y}{a}X_{2}(1+bX_{3})) = CDF_{X_{1}}(\frac{y}{a}X_{2}(1+bX_{3})). $$

It should be straightforward to write down the CDF for this exponential random variable. So then you need to compute: $$ F_{Y}(y) = \int_{0}^{\infty}\int_{0}^{\infty} CDF_{X_{1}}(\frac{y}{a}x_{2}(1+bx_{3}))[f_{X_{2}}(x_{2})dx_{2}][f_{X_{3}}(x_{3})dx_{3}].$$

  • $\begingroup$ Thanks indeed for your help $\endgroup$ – Tran Tam Mar 20 '12 at 8:21
  • $\begingroup$ I have tried to find the closed-form expression. Unlucky, it does not converge. Do you know any condition that make above integral converge ? $\endgroup$ – Tran Tam Mar 29 '12 at 21:37
  • $\begingroup$ When I do the computation by hand, I am able to easily perform the integral over the variable $x_{2}$, and then the resulting integral over $x_{3}$ is of the form $\int_{0}^{\infty}\frac{\alpha}{\epsilon + \beta x_{3}}\cdot{}\exp{(\frac{-x_{3}}{\Omega_{3}})}$, for constants $\alpha$, $\epsilon$, and $\beta$. According to Wolfram integrator, an integral of this type requires the Exponential Integral function Ei() to express the solution. $\endgroup$ – ely Mar 30 '12 at 0:24
  • $\begingroup$ Thanks for your hints. I will try it again. $\endgroup$ – Tran Tam Apr 1 '12 at 19:00

In general, you have the right approach, but there is a little mistake in your final expression. In fact, it is more complicated than what you have.

$$F_Y(y)=Pr\left\{Y<y\right\}=\int_0^\infty \int_0^\infty Pr \left\{ \frac{aX_1}{x_2(1+bx_3)}<y \right\} f_{X_2}(x_2)f_{X_3}(x_3)dx_2dx_3$$

Now, if $a$ and $b$ are positive, then you can rewrite it as you did (you just forgot to include $y$ in your expression):

$$=\int_0^\infty \int_0^\infty Pr \left\{ aX_1<yx_2(1+bx_3) \right\} f_{X_2}(x_2)f_{X_3}(x_3)dx_2dx_3$$

$$=\frac{1}{\Omega_2}\frac{1}{\Omega_3}\int_0^\infty \int_0^\infty (1-e^{-\frac{yx_2}{a\Omega_1}(1+bx_3)})e^{-\frac{x_2}{\Omega_2}}e^{-\frac{x_3}{\Omega_3}}dx_2dx_3$$

$$=\int_0^\infty \int_0^\infty (1-e^{-\frac{y\Omega_2z_2}{a\Omega_1}(1+b\Omega_3z_3)})e^{-z_2-z_3}dz_2dz_3$$

and you can easily calculate it till the final closed form expression.

However, if $a$ is negative, then you must have switched the inequality:

$$Pr \left\{ aX_1<yx_2(1+bx_3) \right\}=Pr \left\{ X_1>\frac{y}{a}x_2(1+bx_3) \right\}=e^{-\frac{yx_2}{a\Omega_1}(1+bx_3)}$$ for negative $y$.

If $b$ is negative, then you must split the initial integral into parts when $(1+bx_3)$ is positive ($x_3<-1/b$), and when it is negative ($x_3>-1/b$).

  • $\begingroup$ Many thanks for your dedication. $\endgroup$ – Tran Tam Mar 20 '12 at 8:21
  • $\begingroup$ I have tried to find the closed-form expression. Unlucky, it is not converged. $\endgroup$ – Tran Tam Mar 29 '12 at 21:36

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