From Exponential Distributions to Weibull Distribution (CDF)

Question

Problem: Let $X_1$ and $X_2$ be two independent exponential random variables with the PDFs $f_{X_i}(x_i)={1\over \lambda_i} \exp(-\frac{x_i}{\lambda_i})$ (where $i=1,2$). Also, let $Y=\frac{(X_1)^2 X_2}{a}$.

I want to find $(Y\leq x)$ i.e. $F_Y(x)=\frac{(X_1)^2 X_2}{a} \leq x$.

My attempted sol (1):

$$\eqalign{&=(X_1)^2 \leq \frac{a x} {X_2}\\ &=\int_0^\infty X_1 \leq \sqrt{\frac{a x} {z_2}} \quad f_{X_2}(z_2) dz_2\\ &= {1\over \lambda_2} \int_0^\infty \left(1-\exp\big(-{\sqrt\frac{a x} {z_2 \lambda_1^2}}\big)\right) \exp(-\frac{z_2}{\lambda_2}) \quad dz_2\\ &=1-{1\over \lambda_2} \int_0^\infty \exp\left(-{\sqrt\frac{c} {z_2}}-\frac{z_2}{\lambda_2}\right) dz_2\tag{1}}$$

I know that $\int_0^\infty \exp\left(-{\frac{\beta} {4z_2}}-{z_2 \gamma}\right) dz_2 = \sqrt{β\over\gamma}K_1(\sqrt{\beta\gamma})$ from Table of Integrals, Series and Products, 7th edition - equation §3.324.1]. However, the final form of above equation contains $\sqrt{}$ and therefore cannot be solved by using §3.324.1. So if you guys can comment or provide any kind of help that would be very helpful.

My attempted sol (2):

$$\eqalign{&=(X_1)^2 \leq \frac{a x} {X_2}\\ &=\int_0^\infty X_2 \leq {\frac{a x} {z_1^2}} \quad f_{X_1}(z_1) dz_1\\ &= {1\over \lambda_1} \int_0^\infty \left(1-\exp\big(-{\frac{a x} {z_1^2 \lambda_2}}\big)\right) \exp(-\frac{z_1}{\lambda_1}) \quad dz_1\\ &=1-{1\over \lambda_1} \int_0^\infty \exp\left(-{\frac{c} {z_1^2}}-\frac{z_1}{\lambda_1}\right) dz_1\tag{1}}$$ Once again to the best my knowledge this above equation doesn't submit to any closed form solution. So I am stuck here....

Since, X is exponential r.v with mean $\lambda$, then $X^{1\over\gamma}$ is a Weibull (γ, β) random variable. Can we solve it this way? by using the CDF or pdf of weibull during conditioning?

Any kind of help will be very much appreciated.

Answer 1

We can forget about $\lambda_1$ and $\lambda_2$ because $X_1/\lambda_1$ and $X_2/\lambda_2$ are always exponentially distributed with unit mean and the variable $Y=X_1^2X_2$ is related to $(X_1/\lambda_1)^2(X_2/\lambda_2)$ by a constant factor $\lambda_1^2\lambda_2$. So W. L. O. G, let's consider $\lambda_1=\lambda_2=1$. The probability density of $Y$ is given by

$$p(y)=\int_0^\infty dx_1\int_0^\infty dx_2\,e^{-x_1-x_2}\delta(x_1^2x_2-y) =\int_0^\infty \frac{dx_1}{x_1^2}\,\exp\left(-x_1-\frac{y}{x_1^2}\right),$$

where $y>0$. Therefore, the complementary CDF can be found by

$$P(Y>y)=\int_y^\infty p(y')dy'=\int_0^\infty dx_1\exp\left(-x_1-\frac{y}{x_1^2}\right).$$

Mathematica says the integral gives the Meijer G-function. See http://reference.wolfram.com/language/ref/MeijerG.html. So no further simplification possible, I think. Saddle point approximation exists for large $y$.

Answer 2

As another answer has suggested the probability density function you are working with doesn't seem to have a particularly nice closed form, but in case it is of some use to you we can also calculate the characteristic function of $Y = X_1^2 X_2$, where $\lambda_1 = \lambda_2 = 1$ by \begin{align*} \varphi_Y(s) &= \int_0^{\infty} e^{isy} f(y) dy \\ &= \int_0^{\infty} e^{isy} \int_0^{\infty}\frac{1}{2w^{3/2}}e^{-\sqrt{w}}e^{-y/w}dwdy \\ &=\int_0^{\infty}\frac{1}{2w^{3/2}}e^{-\sqrt{w}}\int_0^{\infty} e^{(is - 1/w)y}dydw \\ &= \int_0^{\infty}\frac{1}{2w^{3/2}}e^{-\sqrt{w}}\frac{w}{1-isw}dw\\ &= \int_0^{\infty}\frac{1}{1 - isu^2}e^{-u}du\\ &=\int_0^{\infty} \frac{e^{-u}}{s^2 u^4 + 1}du +i\int_0^{\infty}\frac{su^2e^{-u}}{s^2u^4 + 1}du. \end{align*} Which summing over the roots of $s^2u^4 + 1 = 0$ is given by

\begin{align*} \varphi_Y(s) &= \frac{1}{4}\left(\sum_{r \,: \,s^2r^4 + 1 = 0} \frac{1}{s^2}\frac{e^{-r} \mbox{Ei}(u)}{r^3}\bigg|_{u=0}^{u=\infty} + \frac{i}{s}\frac{e^{-r}\mbox{Ei}(u)}{r}\bigg|_{u=0}^{u=\infty}\right). \end{align*}

So the general case can then be recovered from $\varphi_Y(\lambda_1^2 \lambda_2/a)$, and you could then approximate the CDF with Edgeworth/Gram-Charlier series or Saddlepoint approximations.