Find PDF of $Y=-2\ln X$ given that $X$ is uniform on $(0,1)$ Let $X$ have a uniform distribution with p.d.f. $f(x) = 1$, $x$ is in $(0, 1)$, zero elsewhere.
Find the p.d.f. of $Y = -2 \ln X$.
I don't think this is a very difficult question, I just don't really understand what it is asking or where to start. Any help would be very much appreciated. Thank you! 
Update: I did
$$F(y)= P(Y \le y) = P(-2\ln x \le y) = P(\ln x \ge -y/2) = P(x \ge e^{-y/2}).$$
Then $ x=e^{-y/2}$ and $dx/dy =-1/2e^{-y/2}$
Is this all I need to do? Also, I'm not 100% sure why I am using inequalities here- can someone give me a quick explanation? 
 A: you are asked to find the probability distribution of the random variable $Y$ that is related to the random variable $X$ by the relation $Y=-2\ln X$, being $X$ uniform in the interval $[0,1]$. You can solve it considering a change of variable applied to the cumulative function $F(x)$:
$$F(x)=\int_0^x f(x)dx = \int_\infty^y f(y)\Big|\frac{dx}{dy}\Big|^{-1}dy$$
so that
$$f(y) = f(x)\Big|\frac{dx}{dy}\Big|_y$$
and in your situation $x=\exp(-y/2)$, so $\Big|\frac{dx}{dy}\Big|_y = \frac{1}{2}\exp(-y/2)$, so $$f(y)= \frac{1}{2}\exp(-y/2)$$
If you try to integrate $f(y)$ between $\infty$ and $0$ you can verify that it gives you $1$.
A: It was such a long time ago I did this, but this seems to work too. At least in my computer simulations. Maybe it is just a coincidence it works for this particular function, if that is the case, please show me why.
$Y = -2 \ln(X)$ means that if a sample of X is x, then the corresponding sample of Y is $y = -2\ln(x)$
$y = \ln(x^{-2})$ and then $e^y = x^{-2}$ and $e^{-y/2} = x$. We see that since x is between 0 and 1, y will be between 0 and infinity. Now we just need to normalize, i.e. find $k$ so that $\int_0^\infty ke^{-y/2} dy = 1$ and that is an easy exercise in calculus.
A: Let $f(x)=1_{(0,1)}(x)=F_{X}'(x)$. Note that
$$F_{Y}(y)=P(X\geq e^{-y/2})=1-P(X\leq e^{-y/2})=1-F_{X}(e^{-y/2}).$$
Then the density is
\begin{align}
F_{Y}'(y)&=(1-F_{X}(e^{-y/2}))'\\
&=-F_{X}'(e^{-y/2})\frac{\mathrm{d} }{\mathrm{d} y}(e^{-y/2})\\
&=-1_{(0,1)}(e^{-y/2})\cdot \left ( -\frac{1}{2}e^{-y/2} \right )\\
&=1_{(0,\infty)}(y)\frac{1}{2}e^{-y/2}.
\end{align}
A: $ P(-2logX\le x)=P(logX\ge -x/2)=P(X\ge e^{\frac{-x}{2}})=\int_{e^{\frac{-x}{2}}}^1dt=1-e^{\frac{-x}{2}}$
$F_X(x)=1-e^{\frac{-x}{2}}$
hence $f_X(x)=\frac{1}{2}e^{\frac{-x}{2}}$
which is the pdf of exponential with parameter $\frac{1}{2}$
