# Show that this random variable is uniformly distributed in $\left(0,1\right)$

Let $X$ be a continuous random variable with distribution function $F_{X}\left(x\right)$ and let $Y=F_{X}\left(x\right)$, show that $Y\sim U\left(0,1\right)$, where $U$ is the uniform distribution.

Since the density function $f_{X}\left(x\right)$ can be seen as the derivative of $F_{X}\left(x\right)$, this problem looks like an obvious application of the Change of Variable Theorem. Nevertheless, the statement of the theorem requires that the function in which the random variable is evaluated (composed) to be strictly monotonic (strictly increasing or strictly decreasing) and, as it is well known, distribution functions are non-decreasing only.

So my question is: can the Change of Variable Theorem somehow be applied in this case?

Thanks in advanced for any help provided.

Note: Sorry if there are any gramatical errors in the redaction, English is not my first language.

• This looks similar to this question. Commented Sep 12, 2016 at 6:38
• Thank you Jimmy, I will ask my teacher if the Distribution function is strictly monotonous.
– eNR
Commented Sep 12, 2016 at 6:44
• msm. $F_{X}\left(x\right)$. is the c.d.f. of $X$.
– eNR
Commented Sep 12, 2016 at 6:45
• That $F_X$ is strictly increasing is not necessary. That $f_X$ exists is not guaranteed. But $Y=F_X(X)$, not $Y=F_X(x)$ (which is absurd).
– Did
Commented Sep 12, 2016 at 7:05

As Did wrote, let $Y=F_X(X)$ and note that $F_X$ need not be strictly increasing for the following arguments to holds. If we define the generalized inverse of $F_X$ by $F_X^{\leftarrow}(y) =\inf \{ x \in \mathbb{R}: F_X(x)\geq y \}$, then try to prove the following steps
1. If $F_X$ is continuous then (iff actually holds) $F_X^{\leftarrow}$ is strictly increasing on $[\inf\{\text{Range}(F_X)\},\sup\{\text{Range}(F_X)\}]=[0,1]$.
2. Use 1. to argue that $P(F_X(X)\leq x)=P(F^{\leftarrow}_X(F_X(X))\leq F^{\leftarrow}_X(x))$, for any $x\in[0,1]$.
3. Show that $F^{\leftarrow}_X(F_X(y))\leq y$ and determine for what values of $y$ one has $F^{\leftarrow}_X(F_X(y))<y$. Use this to show that $P(F_X^{\leftarrow}\circ F_X(X)=X)=1$ (or look at this).
4. Show that if $F_X$ is continuous, then $F_X(F^{\leftarrow}_X(x))=x$ for any $x\in[0,1]$.
5. Use 2.,3. and 4. together to prove that $F_X(X)\sim \text{Unif}(0,1)\iff P(F_X(X)\leq x)=x, \, \, \forall x\in[0,1]$.