Finding Densities for Functions of a Random Variable Let X have pdf $$f(x) = \beta(1-x)^{\beta-1}, 0 < x < 1$$ and $$f(x) = 0, otherwise$$ where $\beta > 0$.
(a) Find the density of $Y = log(1-X)$
(b) Find the density of $Z = X / (1-X)$
(c) Compute the mean and variance of Z when they exist. Determine conditions on $\beta$ such that the mean/variance of Z exists.
A theorem states:
Let X have pdf f(x) and let Y = g(X), where g is a monotone function. Suppose that f(x) is continuous on X and that $g^{-1}(y)$ has a continuous derivative on Y. Then the pdf of Y is given by: $$f(y) = f(g^{-1}(y))|\dfrac{d}{dy}g^{-1}(y)|$$ for all y in Y and 0 otherwise.
Does this apply to this problem? I'm having trouble implementing it.
 A: More correctly, the density functions for the different variables are different, so use subscripts or otherwise to identify them:  $$f_{\small Y}(y) = f_{\small X}(g^{-1}(y))\,\left\lvert\dfrac{\mathrm d g^{-1}(y)}{\mathrm d y}\right\rvert$$
When $g(x)=\log(1-x)$, the inverse is $g^{-1}(y)=1-\mathrm e^{y}$, and hence $$\begin{align}f_{\small Y}(y) &= \lvert-\mathrm e^y\rvert~f_{\small X}(1-\mathrm e^{y})\\&=\end{align}$$
A: One way to think about the density which does nothing but illustrate the comment of Sean Roberson...
Take for example your random variable Y. Remember that the density of Y, denoted $p_{1}$ will satisfy the following equality for a "nice" function h $$\mathbb{E}[h(Y)]=\int_{\mathbb{R}}h(y)p_{1}(y)dy$$.
This means that $$\mathbb{E}[h(Y)]=\mathbb{E}[h(log(1-X))]=\int_{\mathbb{R}} h(log(1-x))f(x)dx$$ can be rewritten as $\int_{\mathbb{R}}h(y)p_{1}(y)dy$. If you want h(y) inside the integrand, you should try a change of variables with log(1-x)=y. What does this give you for $p_{1}(y)$ after calculations?
For b) You will do the same thing as above but with a different change of variables, $$z=\frac{x}{1-x}$$
For c) Use that
Mean of Z=$\mathbb{E}[Z]=\int_{R}zp_{2}(z)dz$ where $p_{2}(z)$ is the density you calculated in b). Then do the same for variance, $Var(Z)=\mathbb{E}[Z^{2}]-(\mathbb{E}[Z])^{2}$
