Proof: Any CDF (Cumulative Distribution Function) has a uniform distribution

Question

I am trying to prove the following mathematical statement: Any Cumulative Distribution Function (CDF) Has A Uniform Probability Distribution.

Here is my attempt to prove this:

In general, we can define a CDF as the probability of some Random Variable $X$ being less than or equal to some amount:

$$F(X = a) = Pr(X \leq a )$$

Suppose we now denote this above relationship by $Z$. This means that we can write:

$$Z: F(X = a) = Pr(X \leq a)$$

This means that we can replace $X$ by $Z$ and write:

$$F(Z = a) = Pr(Z \leq a)$$

Since $Z = F(X = a )$, we can replace $Z$ with $F(X)$ to write:

$$F(Z = a) = Pr[F(X) \leq a]$$

Now, by using the inverse function relationship, we can write the above as:

$$F(Z = a) = Pr[ X \leq F^{-1}(a)]$$

Since $F^{-1}(a)$ is a constant number, this is like saying - what is the CDF evaluated at the point $ F^{-1}(a)$?

Thus, we can write:

$$F(Z = a) = F[F^{-1}(a)] = a$$

As we have it , $F(Z = a) = a$ . This means that the Cumulative Probability Function of $Z$ itakes on a constant value of $a$ when evaluated at the point $a$. And this is only possible in the case of a Uniform Distribution. Thus, we have shown that any Cumulative Probability Function must have a Uniform Distribution.

Is my proof correct?

Thanks!

Answer 1

I will emphasize some problematic places. I assume that you wanted to prove that

For any random variable $X$ with monotonic and continuous CDF $F_X(\cdot)$, random variable $F_X(X)$ has uniform distribution.

At first, you write

$F(X = a) = Pr(X \leq a )$

But $F_X(\cdot)$ is a function from $\mathbb{R}$ to $[0,1]$, so it cannot take event $\{X=a\} = \{w \in \Omega : X(w)=a\}$ as its argument. I guess you just wanted to write $F_X(a)\triangleq\mathbb{P}(X \leq a)$. Note that $a \in \mathbb{R}$ is an argument of the function $F_X$ and not event $X=a$.

Suppose we now denote this above relationship by $Z$. This means that we can write: $Z:F(X=a)=Pr(X≤a)$

It looks like you denoted this proposition as $Z$ (it's sometimes useful in the mathematical logic), while maybe you wanted to denote as $Z := F_X(a)$. But $Z := F_X(a)$ is a number, it has very primitive CDF and after it you repeat the error with the argument of the $F_X$.

If you want to prove it from zero by yourself, maybe these hints will be useful :

1. CDF of the random variable $X$ is a function $F_X : \mathbb{R} \to [0,1]$. For every $x \in \mathbb{R}~~~ F_X(x) \triangleq \mathbb{P}(X \leq x).$ So it "takes as input" any real number and "returns as output" probability that $X \in (-\infty, x]$.
2. To find the CDF of the random variable $F_X(X)$ at first you can think, why is it a random variable? Maybe you need to recall that is a function composition.
3. After it you can recall that is an invertible function and which conditions $F_X$ must satisfy to be an invertible function to have the inverse function $F^{-1}_X(\cdot)$. Also if you want to prove this proposition in the more general form, it can be useful to read about generalized inverse distribution function.