Expectation and variance of the squared distance between $X$ and $Y$

Question

Given that $X$ and $Y$ are two independent univariate random variables sampled uniformly from the unit interval [0,1]. I am trying to find the expected value and the variance of the random variable $Z = \mid X - Y \mid ^2 $ (the squared distance between $X$ and $Y$.

So far, I'm assuming that $X$ and $Y$ are uniformly distributed and this is what I have: \begin{align} Z &= \mid X - Y\mid ^2 \\ Z &= \mid X^2 - 2XY + Y^2 \mid \\ E[Z] &= \mid E[X^2] - 2E[XY] + E[Y^2]\mid \\ &= \mid E[X^2] - 2E[X]E[Y] + E[Y^2]\mid \\ &= \frac{1}{3}\frac{1}{b-a}(b^3-a^3)-2(\frac{a+b}{2})(\frac{a+b}{2})+\frac{1}{3}\frac{1}{b-a}(b^3-a^3) \end{align}

But I am not sure if this is the right approach or the right direction and what is E[Z] = $\mid X - Y\mid ^2$. If so, can I do something similar to the variance?

Answer 1

If I substitute 0 and 1 for a and b respectively to:

\begin{align} &= \frac{1}{3}\frac{1}{b-a}(b^3-a^3)-2(\frac{a+b}{2})(\frac{a+b}{2})+\frac{1}{3}\frac{1}{b-a}(b^3-a^3) \\ &=\frac{1}{6} \end{align}

Would the answer (the expected value of Z) be $\frac{1}{6}$?

Answer 2

Variace is $E[(X-Y)^4] -[E[(X-Y)^2]]^2$

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=$ E(X^4) + E(Y^4) + 6E(X^2)E(Y^2) -4E(X^3)E(Y) - 4E(X)E(Y^3) -\frac{1}{36}$

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= $\frac{2}{5} + \frac{6}{9} - 1 - \frac{1}{36}$

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=$\frac{7}{180}$