Transormation of two exponential RV

Question

$X_{1}$ and $X_{2}$ are IID exponentially distributed with parameter $\lambda=3$. Calculate $X_{1}/X_{2}$.

If $Y_{1}=X_{1}/X_{2}$ and $Y_{2}=X_{2}$ then distribution: $$g(y_{1},y_{2})=9\ y_{2}\ exp(-3(y_{1}y_{2}+y_{2}))=9\ y_{2}\ exp(-3y_{1}y_{2})exp(-3y_{2})$$

Then I try to calculate marginal density $g(y_{1})$ but it is impossible (or at least it can take few hours,and many unconventional subsitutions).

Is it even possible to calculate such marginal density or we have to use some "trick" or property of exponential distribution?

Answer 1

Why not just compute the cdf and take a derivative:

$P(\frac{X_1}{X_2} < a) = P(X_2 > \frac{1}{a}X_1) = \int_0^\infty P(X_2 > \frac{1}{a}x)f_{X_1}(x)dx = \int_0^\infty e^{-3x/a} 3e^{-3x} dx = \frac{a}{a+1}$

Take a derivative with respect to a and you have the pdf of $f_{\frac{X_1}{X_2}}(a)$

Answer 2

Heuristically, if $X_1,X_2$ are independent and have one sign (as in this case) you should expect $f_{X_1/X_2}(r)=\int_{-\infty}^\infty f_{X_1}(rx) f_{X_2}(x) dx$, because if $X_1/X_2=r$ and $X_2=x$ then $X_1=rx$.

Plugging in the relevant densities in this case gives $\int_0^\infty 9 e^{-3(rx)} e^{-3x} dx$. This integral is of course quite easy to calculate, but can you prove that this formula is correct? (Probably the best way to carry out this proof is to pass through the CDF $F_{X_1/X_2}$.)