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.