I am trying to wrap my head around a certain topic in my notes, but it seems very confusing.
Let X a continuous random variable whose distribution function $F_X$ is strictly increasing on the possible values of $X$. Then $F_X$ has an inverse function. [Agreed]
Let $U = F_X(X)$, then for $u \in [0,1]$ We wish to find $F_U(u)$:
Since $F_U(u) = P[U\leq u]:$
$$P[U \leq u] = P[F_X(X)\leq u] \; \; \; \; \ \;\;\;(1)$$
$$P[U \leq F_X^{-1}(u)]\; \; \; \; \ \;\;\;(2)$$
$$F_X(F_X^{-1}(u))=u\; \; \; \; \ \;\;\; (3)$$
I am probably missing something super obvious, but I am confused by the above 3 steps.
From $(1)$ to $(2)$ – I recognise that we defined $U = F_X(X)$, so that's fine.
But the right hand side of the inequality in $(2)$, how exactly does that make sense?
From $(2)$ to $(3)$, I think it is saying this is just the CDF of $U<F_X^{-1}(y)$? IS that correct? And the inverse cancels with the $F_X$ to leave $u$.
I.e., $F_U(u) = u$
Any help in understanding would be great.
Thanks.
Best Answer
It does not. It should be $X$ rather than $U$. $F_X(X)\leq u\iff X\leq {F_X}^{-1}(u)$
$$\begin{align}F_U(u) &= \mathsf P(U\leq u) &&\text{by definition of CDF}\\[1ex] &= \mathsf P(F_X(X)\leq u)&&\text{since }U=F_X(X)\\[1ex]&=\mathsf P(X\leq F_X^{-1}(u)) &\star\\[1ex]&=F_X(F_X^{-1}(u))&&\text{by definition of CDF} \\[1ex] &= u~\mathbf 1_{u\in[0..1]}&&\text{by definition of inversion}\\[3ex]\therefore\qquad U&\sim\mathcal U[0..1] \end{align}$$