P.D.F of a mapped function of a uniformly distributed random variable I have a random variable U which is uniformly distributed over [0,1]; Now $$X=-2logU$$
Then what would be the P.D.F. of X?
I know that P.D.F of U is 1 for [0,1] and 0 otherwise so the limits of X would be [0, $\infty]$ but how to determine its PDF(probability density function).
 A: *

*I would recommend using a change of variable on
$$Y = -\log U.$$
People also like to use the cdf method $P(Y<y) = P(-\log U < y)$ to find the distribution. 
The distribution of this is

 $\text{Exp}(1)$


*Notice that now your problem has become
$$X = 2 Y.$$
Using properties possibly shown in class or in the textbook, recognize that the distribution of $X$ is

 $\text{Exp}(1/2)$

Yes, you could have done a change of variable from the beginning or the cdf method on $P(Y< y) = P(-2\log U <y)$ from the beginning, but it's good to keep in mind everything you have learned up to now.
A: Assuming that is the natural logarithm; $\;X=-2\log_\mathsf e(U)\;$ iif $\;U=\mathsf e^{-X/2}\;$.
Then by change of variables:
$$f_X(x) = f_U(\mathsf e^{-x/2})\left\lvert\dfrac{\operatorname{d} \mathsf e^{-x/2}}{\operatorname{d} x}\right\rvert$$

If you have not yet encountered this application of the chain rule we can show it the long way.
$$\begin{align} f_X(x) & =\frac{\operatorname d}{\operatorname dx} \mathsf P(X\leq x)
\\[1ex] & = \frac{\operatorname d}{\operatorname dx} \mathsf P(-2\log_\mathsf e U\leq x)
\\[1ex] & = \frac{\operatorname d}{\operatorname dx} \mathsf P(U\geq \mathsf e^{-x/2}) 
\end{align}$$

 $$\begin{align} & = \frac{\operatorname d}{\operatorname dx} (1-F_U(\mathsf e^{-x/2})) \\[1ex] & = \frac{\operatorname d}{\operatorname dx} (1-\mathsf e^{-x/2})\;\mathbf 1_{\mathsf {exp}(-x/2)\in(0;1)} \\[1ex] & =  -\frac{\operatorname d(\mathsf e^{-x/2})}{\operatorname dx}\;\mathbf 1_{x\in(0;\infty)} \\[1ex] & =  \tfrac12\mathsf e^{-x/2}\;\mathbf 1_{x\in(0;\infty)} \end{align}$$



 Thus $X\sim\mathcal{Exp}(1/2)$

A: All you need is that $\log$ is a strictly increasing transformation. Using that fact, we have: for $x\geq 0$,
$$
\Pr(X\leq x)=\Pr(-2\log U\leq x)=\Pr(\log U\geq -x/2)=\Pr(U\geq e^{-x/2})=1-e^{-x/2}.
$$
Thus, for $x\geq 0$, the pdf of $X$ at $x$ is
$$
\frac{d}{dx}(1-e^{-x/2})=\frac{1}{2}e^{-x/2}.
$$
