(a) Let $X$ be uniformly distributed on $(0, 1)$. Find the density of $Y = -2 \log X$.

Solution:

1. Transformation Relationship: Given $X \sim \text{Uniform}(0, 1)$, the PDF of $X$ is:

   \begin{cases} 1, & 0 < x < 1, \\ 0, & \text{otherwise}. \end{cases}$$ The transformation is $$Y = -2 \log X$$. Solving for $$X$$ in terms of $$Y$$: $$X = e^{-Y/2}.$$

2. Finding the Jacobian: The derivative of $X = e^{-Y/2}$ with respect to $Y$ is: $\frac{dX}{dY} = -\frac{1}{2} e^{-Y/2}.$ The absolute value of the Jacobian is: $\left|\frac{dX}{dY}\right| = \frac{1}{2} e^{-Y/2}.$

3. Changing the PDF: For transformations, the PDF of $Y$ is: $f_Y(y) = f_X(x) \cdot \left|\frac{dX}{dY}\right|, \quad \text{where } x = e^{-y/2}.$

   Substituting $f_X(x) = 1$ (since $0 < x < 1$) and $\left|\frac{dX}{dY}\right| = \frac{1}{2} e^{-y/2}$:

   \begin{cases} \frac{1}{2} e^{-y/2}, & y > 0, \\ 0, & \text{otherwise}. \end{cases}$$

Thus, $Y \sim \text{Exponential}(\lambda = 1/2)$ with PDF: [ f_Y(y) = \frac{1}{2} e^{-y/