Suppose that $X$ and $Y$, two random variables, are both uniformly distributed over $[0,1]$. Let $Z=\frac{1}{2}X+\frac{1}{2}Y$.
I know that in general, $Z$ is not uniform. For instance, $Z$ is not uniform if $X$ and $Y$ are independent.
On the other hand, if $X=Y$, then $Z$ is uniformly distributed over $[0,1]$.
My question: Suppose $Z$ is uniformly distributed over $[0,1]$. Is $X=Y$? In other words, is $X=Y$ the only case where $Z$ is uniform over $[0,1]$?
Best Answer
The answer is YES.
We have $$ \begin{align*} \frac 1 3 &=E[Z^{2}] \\ &=\frac 1 4 E[(X+Y)^{2}]\\ &=\frac 1 4(E[X^{2}]+E[Y^{2}]+2E[XY])\\ &=\frac 1 4(\frac 1 3+\frac 1 3+2E[XY]). \end{align*}$$ This gives $E[XY]=\frac 1 3$. This implies that we have equality in Cauchy-Schwarz inequality: $$E[XY]=\sqrt {E[X^{2}]}\sqrt {E[Y^{2}]}$$ and hence $X$ and $Y$ are constant multiples of each other. But the constant factor has to be $1$ since $X$ and $Y$ have uniform distribution on $[0,1]$. Hence $X=Y$.