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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!

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  • $\begingroup$ Can you see what implicit assumptions you made? For example you assumed $F^{-1}(a)$ exists. It does, but only for some $a$. Which ones? Is this affected by whether $X$ is a continuous random variable or whether $X$ has infinite support? $\endgroup$
    – Henry
    Commented Jul 29, 2023 at 4:59
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    $\begingroup$ I don't understand the wording of your assertion : a CDF is a function. How a function can "have" a distribution ? $\endgroup$
    – Jean Marie
    Commented Jul 29, 2023 at 5:08
  • $\begingroup$ Every distribution has a CDF, and not every distribution is uniform. $\endgroup$
    – Karl
    Commented Jul 29, 2023 at 5:37
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    $\begingroup$ Did you want to ask the following: "for any r.v. $X$ with CDF $F$, r.v. $F(X)$ has uniform distribution"? Your proof suggests that, although notation used in it is very confusing. This question has positive answer when $F$ is continuous $\endgroup$
    – Esgeriath
    Commented Jul 29, 2023 at 6:34
  • $\begingroup$ You denoted proposition (equality) as $Z$ and after it used $Z$ as a random variable? Can you even understand your proof by yourself? Seems like you tried to use chat bots to help you, while this proposition with proper assumptions can be proved in one line. $\endgroup$
    – perepelart
    Commented Jul 29, 2023 at 6:43

1 Answer 1

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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.
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