Suppose $X$ has density (pdf) $f_X(x) = \frac{2}{3}(x-1)$ for $2 < x < 3$.
a.) Find the pdf of $Y = -\log X$. (Make sure to specify the range and validity of your answer).
b.) Find a montone function $u(x)$ such that the random variable $Z = u(x)$ has a Uniform$(0,1)$ distribution.
Solution a.) We have from Casella and Berger theorem 2.1.8 $$f_Y(y) = \sum_{i=1}^{n}f_X(g^{-1}_i(y))\left|\frac{d}{dy}g_i^{-1}(y)\right|$$ We are given $$Y = -\log X \Rightarrow g^{-1}(y) = e^{-y}$$ Thus \begin{align*} f_Y(y) = f_X(g^{-1}(y))\left|\frac{d}{dy}g^{-1}(y)\right|\\ &= \frac{2}{3}(e^{-y} - 1)e^{-y}\\ &= \frac{2}{3}e^{-y}(e^{-y} - 1) \ \ \ \ \ \ -\log(3) < y < -\log(2) \end{align*} To verify this is correct note that $$\int_{-\log(3)}^{-\log(2)} \frac{2}{3}e^{-y}(e^{-y} - 1) dy = 1$$
I am not sure how to do part b any suggestion is greatly appreciated.
See that $\Pr(Z\leq z)=z$ if $Z$ is uniformly distributed. Now see that: $$ \Pr(Z\leq z)=\Pr(u(X)\leq z)=\Pr(X\leq u^{-1}(z))\\ =\int_2^{ u^{-1}(z)}\frac 23(x-1)dx=\frac{u^{-1}(z)^2}{3}-\frac{2u^{-1}(z)}{3} $$ So we have: $$ \frac{u^{-1}(z)^2}{3}-\frac{2u^{-1}(z)}{3}=z. $$ which means that: $$ u(x)=\frac{x^2}{3}-\frac{2x}{3}. $$