Question

If X and Y are independent continuous positive random

variables,express the density function of (a) Z=X/Y and (b) Z=XY in
terms ofthe density functions of X and Y. Evaluate these
expressions in thespecial case where X and Y are both exponential
randomvariables.

Answer 1

a)
`Z=\frac XY` has CDF

`F_Z(z)=P(Z\le z)=P(X\le Yz)=\displaystyle\int_{\mathrm{supp}(Y)}P(X\le yz\mid Y=y)P(Y=y)\,\mathrm dy`

`F_Z(z)\displaystyle=\int_{\mathrm{supp}(Y)}P(X\le yz)P(Y=y)\,\mathrm dy`

where the last equality follows from independence of
`X,Y`. In terms of the distribution and density functions of
`X,Y`, this is

`F_Z(z)=\displaystyle\int_{\mathrm{supp}(Y)}F_X(yz)f_Y(y)\,\mathrm dy`

Then the density is obtained by differentiating with respect to
`z`,

`f_Z(z)=\displaystyle(\mathrm d)/(\mathrm dz)\int_{\mathrm{supp}(Y)}F_X(yz)f_Y(y)\,\mathrm dy=\int_{\mathrm{supp}(Y)}yf_X(yz)f_Y(y)\,\mathrm dy`

b)
`Z=XY` can be computed in the same way; it has CDF

`F_Z(z)=P\left(X\le\frac zY\right)=\displaystyle\int_{\mathrm{supp}(Y)}P\left(X\le\frac zy\right)P(Y=y)\,\mathrm dy`

`F_Z(z)\displaystyle=\int_{\mathrm{supp}(Y)}F_X\left(\frac zy\right)f_Y(y)\,\mathrm dy`

Differentiating gives the associated PDF,

`f_Z(z)=\displaystyle\int_{\mathrm{supp}(Y)}\frac1yf_X\left(\frac zy\right)f_Y(y)\,\mathrm dy`

Assuming
`X\sim\mathrm{Exp}(\lambda_x)` and
`Y\sim\mathrm{Exp}(\lambda_y)`, we have

`f_(Z=\frac XY)(z)=\displaystyle\int_0^\infty y(\lambda_xe^(-\lambda_xyz))(\lambda_ye^(\lambda_yz))\,\mathrm dy`

`\implies f_(Z=\frac XY)(z)=\begin{cases}(\lambda_x\lambda_y)/((\lambda_xz+\lambda_y)^2)&\text{for }z\ge0\\0&\text{otherwise}\end{cases}`

and

`f_(Z=XY)(z)=\displaystyle\int_0^\infty\frac1y(\lambda_xe^(-\lambda_xyz))(\lambda_ye^(\lambda_yz))\,\mathrm dy`

`\implies f_(Z=XY)(z)=\lambda_x\lambda_y\displaystyle\int_0^\infty\frac{e^(-\lambda_x\frac zy-\lambda_yy)}y\,\mathrm dy`

I wouldn't worry about evaluating this integral any further unless you know about the Bessel functions.