# Transformation of a probability distribution - Random variable function of a CDF?

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 $$F_U(u) = P[U\leq u]:$$ $$P[U \leq u] = P[F_X(X)\leq u] \; \; \; \; \ \;\;\;(1)$$ $$= P[X \leq F_X^{-1}(u)]\; \; \; \; \ \;\;\;(2)$$ $$= F_X(F_X^{-1}(u))=u\; \; \; \; \ \;\;\; (3)$$

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 the LHS is 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 $$U? IS that correct? And the inverse cancels with $$F_X$$ to leave $$u$$.

I.e., $$F_U(u) = u$$

• Eq 2 should be P(X<= ...) Oct 23 '19 at 4:32
• I'm quite certain this is covered by answers already on site. Oct 23 '19 at 5:06
• Search for 'inverse CDF method'. Many hits. Oct 23 '19 at 8:18
• Perhaps a more effective search is Probability Integral Transform.
– whuber
Oct 23 '19 at 15:42

First, for $$F_X^{-1}$$ to exist, $$F_X$$ must be strictly increasing and continuous.
Second, if $$F_X(X)\le u$$, then, $$F_X$$ being increasing, $$F_X^{-1}$$ is also increasing, hence $$F_X^{-1}(F_X(X))\le F_X^{-1}(u)$$ by applying $$F_X^{-1}$$ to both sides of the inequality.
Third, since $$F_X^{-1}$$ is the inverse function, $$F_X^{-1}(F_X(X))=X$$ hence the event $$U\le u$$ is the same as the event $$X\le F_X^{-1}(u)$$ which has probability$$F_X(F_X^{-1}(u))=u$$
In conclusion, $$U=F_X(X)$$ is thus distributed as $$\mathcal U(0,1)$$ when $$X\sim F_X$$.