$X$ and $1 - X$ are identically distributed

Question

I have a two-fold goal for this question. First, I'm trying my hand at making hypotheses and proving them as far as I can. I want to understand the limits of proof, not just the techniques. Second, I'm hoping to find slick ways to prove things.

Suppose that $X$ and $Y = 1 - X$ are identically distributed continuous random variables. It is my hypothesis that both $X$ and $Y$ are uniformly distributed on the unit interval. My attempt to prove it goes something like this:

(On the relation between $F_X(x)$ and $F_X(1 - x)$): Since $X$ and $Y$ are identically distributed, $F_X(x) = P(X<x) = P(Y<x) = P(1 - X < x) = P(1 - x < X) = 1 - P(X < 1 - x) = 1 - F_X(1 - x)$.

(Concerning the density of $X$): Differentiate to show that $f_X(x) = f_X(1 - x)$

(Concerning the range of $X$): Now suppose $x = 1$. Then, clearly, $0 = 1 - x < 1 \leq x = 1$. Since $F_X$ is monotone increasing, we have the relation $1 - F_X(1) = F_X(0) \leq F_X(1) = 1 - F_X(0)$.

This is where I'm getting stuck. I'd like to show that the range is $[0,1]$ (up to measurability). To me, it feels like there aren't enough constraints on $F_X$ to prove that $X$ the range of $X$ is $[0,1]$. Similarly, I'm not sure there are enough constraints to prove that $f_X$ is constant.

What are your thoughts? Is there a slick proof? Even a hint for a property of $F_X$ or $f_X$ I might have missed would be extremely useful.

Answer 1

Your condition says that the density function of $X$ is symmetric about $\frac{1}{2}$, and no more.

If $X=\frac{1}{2}+W$, where $W$ is any random variable which is symmetric about $w=0$, then $X$ and $Y$ will have the same distribution.

Answer 2

Your condition on the density is a sufficient condition for $X$ and $1-X$ to be identically distributed, so if $X$ has a density $ \propto x(1-x), 0 < x < 1 $, then $1-X$ will have the same distribution

Answer 3

Suppose $g$ is any probability density function on the real line.

Let $h(x) = g(1-x)$.

Let $f(x) = \dfrac{h(x)+g(x)}{2}$.

The $f$ is a probability density function. And $$ f(1-x) = \frac{h(1-x)+g(1-x)}{2} = \frac{g(1-(1-x))+h(x)}{2} = \frac{g(x)+h(x)}{2} = f(x). $$

So there are plenty of probability density functions satisfying the equation $f(x)=f(1-x)$.

All of them give probability distributions symmetric about $1/2$.