I've been given the following problem:
If a random variable $X$ has probability density function $f_X (x)$ and cumulative distribution function $F_X (x)$, prove that $U = F_X (x) \sim$ uniform$(0,1)$
Ordinarily to solve a problem like this, I'd use the following method:
$F_U (u) = Pr(U\leq u) = Pr(g(X) \leq u) = Pr(x \leq g^{-1} (u)) = F_X (g^{-1} (u))$
And then differentiate w.r.t $u$ to get $f_U (u)$. However in this case, I am not sure whether I can find $g^{-1}$ because $F_X (x)$ doesn't seem to necessarily be 1-to-1 even over the range of values $X$ can take. (I can see that $F_X (x)$ is increasing but not necessarily strictly increasing)
How should I choose $g^{-1}$ in this case? Or is there a better method to solve this specific problem?
You don't need $\ F_X\ $ to be strictly increasing for the result to hold, but you do need it to be continuous, which it will be if it has a density function. If $\ F_X\ $is continuous, then $\ P\big(F_X(X)\le y\big)=P\big(F_X(X)<y\big)\ $, because $\ \big\{x\ \big|\ F_X(x)=y\big\}=[a, b]\ $ for some $\ a,b\ $ with $\ b\ge a\ $, and $\ P\big(F_X(X)=y\big)=$$P(a\le X\le b)=$$F_X(b)-F_X(a)=$$0\ $. But $\ F_X(X)<y\Leftrightarrow X\in F_X^{-1}\big([0, y)\big)\ $, and $\ F_X^{-1}\big([0, y)\big)\ $ is an interval of the form $\ (-\infty,a)\ $ where $\ F_X(a)=y\ $. Putting all this together, we have \begin{align} P(U\le y)&=P\big(F_X(X)\le y\big)\\ &=P\big(X\in (-\infty, a)\big)\\ &=F_X(a)\\ &=y\ . \end{align}