If a function of $X$ is uniformly distributed over $(0,1)$, is it the C.D.F. of $X$?

Question

I am self-studying Introduction to Mathematical Statistics by Hogg. On pages 126-127, the textbook reads:

Let the random variable $Y$ by distributed uniformly over the unit interval $0 < y < 1$. Suppose that $F(x)$ is a distribution function of the continuous type which is strictly increasing when $0 < F(x) < 1$.

If we define the random variable $X$ by the relationship $Y = F(X)$, we now show that X has a distribution which corresponds to $F(X)$.

I'm a little confused on what they're trying to prove here. Are they defining $F$ and $X$ such that $F(X)$ is a uniformly distributed random variable in $(0,1)$? If so - why does this imply that $F(X)$ is the C.D.F. of $X$? Can you always say that some function of $X$ is a C.D.F. of $X$ if that function is uniformly distributed from $(0,1)$?

Thank you!

Answer 1

There are two mutually inverse facts.

If we take a r.v. $X$ with C.D.F. $F(x)$, and $F(x)$ is continuous and strictly increasing, then r.v. $Y=F(X)$ is uniformly distributed over $(0,1)$.

In fact, "strictly increasing" for this statement is not required. But let it be.

And the inverse:

If we take a r.v. $Y$ which is uniformly distributed over $(0,1)$ and some C.D.F. $F(x)$ which is continuous and strictly increasing, then r.v. $X=F^{-1}(Y)$ has C.D.F. $F(x)=\mathbb P(X\leq x)$.

We have continuous and strictly increasing function $F$ which satisfies all the requirements to be a C.D.F. of some random variable. Namely, it has zero limit at $-\infty$ and unit limit at $+\infty$.

We want to construct a r.v. $X$ such that $F$ is the C.D.F. of $X$.

To do it, one need to take $Y$ uniformly distributed on $(0,1)$ and $X=F^{-1}(Y)$. The last equality is equivalent to the equality $Y=F(X)$ since $F$ is invertible.

Both facts are checked simply. Take the second one: let $Y$ is uniformly distributed over $(0,1)$ and $F(x)$ is continuous and strictly increasing C.D.F. Define $X=F^{-1}(Y)$. Find C.D.F. of $X$:

$$ F_X(x)=\mathbb P(X\leq x) = \mathbb P(F^{-1}(Y)\leq x) = \mathbb P(Y\leq F(x)) = F(x). $$ since $\mathbb P(Y\leq a)=a$ for $0<a<1$. We conclude that $F(x)$ equals to C.D.F. of r.v. $X=F^{-1}(Y)$.

As an example, if $Y$ is uniformly distributed over $(0,1)$, then

1. $X=-\ln(1-Y)$ is exponentially distributed with $F_X(x)=1-e^{-x}$ ($x>0$). Another words, if $X$ is such that $$Y=1-e^{-X},$$ then $X$ has C.D.F. $F_X(x)=1-e^{-x}$ ($x>0$).
2. If $\Phi(x)$ is C.D.F. of standard normal distribution then $X$ such that $$Y=\Phi(X)$$ has standard normal distribution. One can write $X$ as $X=\Phi^{-1}(Y)$.

And so on.

So, the answer to your question in the header is "Yes". If proper function of $X$ is uniformly distributed, then this function is C.D.F. of $X$.