Transforming random variable $1/\sqrt{U}$

Question

I have this practice question that I can't get my head around. "If U is a standard uniform random variable and $X=1/\sqrt{U}$, derive the distribution function $F_X$ of $X$ and differentiate to find $f_X$." Normally I would follow the procedure that $P(X<x)$ hence $P(1/\sqrt{U}<x)$ and would try to rewrite to express the probability in terms of $U$. However, in this case the only possibility I see to do this is to square both terms which I am not sure if it is allowed in an inequality. I know that E[X] should be 2. So what is the strategy to proceed after $P(1/\sqrt{U}<x)$ to end up with $F_X$?

Answer 1

As you said here, we need to start with computing $\mathbb P(1/\sqrt{U}\le x)$. Notice that given $U\sim U[0,1]$, for all $x< 0$ we have $F_X(x)=\mathbb P(1/\sqrt U\le x)=0$, and for all $x\ge 0$, $$\frac{1}{\sqrt U}\le x\,\,\Leftrightarrow \,\,\frac{1}{U}\le x^2\,\,\Leftrightarrow\,\,U\ge \frac{1}{x^2}.$$ So it follows that (using the fact that $U\sim U[0,1]$) $$F_X(x)=\mathbb P(1/\sqrt U\le x)=\mathbb P(U\ge x^{-2})=\begin{cases}1-x^{-2}, &\text{if } x\ge 1\\0, &\text{if }0\le x<1\end{cases}.$$ Then finding $f_X$ only refers to differentiating $F_x$ .

Answer 2

There is no issue with squaring: if $x$ and $y$ are positive then $x < y$ if and only if $x^2 < y^2$. There might be some issues doing this with negative values. For instance, $-2 < 1$ but $(-2)^2 > 1^2$. But for positive values the square function is increasing (its graph goes up as you move right) so there is no issue.

So you are allowed to say

$$ P(1/\sqrt{U} < x) = P(1/U < x^2) = P(U > 1/x^2). $$