Show that if $X$ is uniform on $[0, 1)$ then $Y = -k\log X$ is an exponential r.v. such that $⟨y⟩ = k$ Attempt: probability is conserved under change of variables. If $p(x)$ is a probability density then $p(x)dx$ is the probability that there is some $x′$ in $[x, x+dx)$. By change of variables $x$ -> $y$ we have $p(y)dy = p(x)dx$ if the transformation is positive definite.
 A: There are two common methods for finding the density of a change of variable: the Jacobian method and the CDF method. Continuous, smooth density functions uniquely identify random variables, so if the density takes a "known form" you can conclude its form.
A: Here is a derivation for the case $k=2$. 
Let
$y=g(p_i):=-2\ln(p_i)$. Then, $p_i=g^{-1}(y)=e^{-\frac{1}{2}y}$
and the density of $-2\ln(p_i)$ is given by
   $$ f_{-2\ln(p_i)}(y)=f_{p_i}(g^{-1}(y))\left|\frac{\partial}{\partial y}g^{-1}(y)\right|.
    $$
Note
    $$\frac{\partial}{\partial
    y}g^{-1}(y)=-\frac{1}{2}e^{-\frac{1}{2}y}$$ 
and 
$$\left|\frac{\partial}{\partial y}g^{-1}(y)\right|=\frac{1}{2}e^{-\frac{1}{2}y}.$$ 
We have $f_{p_i}(g^{-1}(y))=1\;\forall\;g^{-1}(y)\in[0,1]$ (a standard uniform density). This implies
    $$f_{-2\ln(p_i)}(y)=\frac{1}{2}e^{-\frac{1}{2}y}.$$
 The density of a $\chi^2_R$ random variable is
    $$f_{\chi^2_R}(y)=\frac{1}{2^{R/2}\Gamma(R/2)}y^{\frac{R}{2}-1}e^{-\frac{y}{2}}.$$
    With $R=2$, we get $f_{\chi^2_2}(y)=\frac{1}{2\Gamma(1)}e^{-\frac{y}{2}}.$ Recall that $\Gamma(1)=\int_0^\infty t^{1-1}e^{-t}\;dt=1$. So,
    $$
    f_{\chi^2_2}(y)=\frac{1}{2}e^{-\frac{y}{2}}.
    $$
