Prove that $Z = \frac{X_1}{X_2}$, has an F-distribution Let $X_1, X_2$ be independent random variables following density law $f(x) = e^{-x} , 0 < x < \infty$, Show that
$Z = \frac{X_1}{X_2}$, has an F-distribution.
I thought of solving this by solving for the mgf of $Z$ but then I remembered that mgf of F-distribution doesn't exist, so I couldn't do it that way. How to approach questions like these in general?
 A: Let's not try to generalize about "questions like these,"
but to answer the specific question at hand, using the definition of an F-distribution in terms of chi-squared distributions, as mentioned in a Comment by @whuber.
[While moment generating functions and bivariate
transformations with Jacobians are often useful, they
are not needed in this particular problem.]
$\mathsf{Chisq}(\nu=2)\equiv\mathsf{Exp}(\mathrm{rate}=1/2).$
So multiply the exponential random variable $X_1$ in the numerator by a constant $c$ so that $cX_1 \sim 
\mathsf{Chisq}(2).$ Then $\frac{X_1}{X_2}=\frac{cX_1/2}{cX_2/2}\sim \mathsf{F}(2,2).$
This can be illustrated by a simulation in R.
set.seed(116)
x1 = rexp(10^6);  x2 = rexp(10^6)
f = x1/x2
summary(f)
     Min.   1st Qu.    Median      Mean   3rd Qu.      Max. 
  0.00083   0.31794   1.11853   6.20911   3.21720 201.36394 

$\mathsf{F}(2,2)$ is a highly skewed distribution. For clarity, the histogram of simulated values below omits a (relatively) few observations beyond 100. The red curve is the density function of $\mathsf{F}(2,2).$

hist(f[f<100], prob=2, br=100, ylim=c(0,.5),
     col="skyblue2", main="F(2,2)")
 curve(df(x,2,2), add=T, lwd=2, col="red")

