The continuous type of random variable (rv) X has the following density function
$f(x)=1 $ for $0\lt x\lt 1$ And $f(x)=0 $ for otherwise.
Part-a.
Derive the moment generating function(mgf)of Y, H(b) where $b\lt 1/2$ if $Y=−2ln(X).$
Part-b. Let $Y_1$ and $Y_2$ be the random variables, which have the same density function with Y ( Y given in part a). Suppose that $Y_1$ is independent of Y2. If $Z = Y_1 + Y_2$, compute the density function of Z.
For the part b, as I search, I found the following hint.
However, I am not sure about how to solve these two parts. Please help me to do these. Thanks.
Part a. The MGF of $Y$ is, for any $b<1/2$, \begin{align} M_Y(b) &= E[\exp(-2b \ln(X))]\\ &=E[X^{-2b}]\\ &=\int_0^1 x^{-2b}dx\\ &=\frac{1}{1-2b}. \end{align}
Part b. The MGF of $Z$ satisfies $$M_Z(b)=E[\exp(b(Y_1+Y_2))]=E[\exp(bY_1)]E[\exp(bY_2)]=M_Y(b)^2=\left(\frac{1}{1-2b}\right)^2,$$ where I used that $Y_1$ and $Y_2$ are independent and distributed as $Y$. Thus, $Z$ has the MGF of a chi-square distribution with $r=4$ degrees of freedom. Since the MGF uniquely determines the distribution, $Z$ has density $$f_Z(z)=\frac{1}{4}x\exp(-x/2),\; x>0.$$