Problem source

The random variable $X$ is uniformly distributed on the interval $[-1,1]$. Find $\E(X^2)$ and $\var (X^2)$.
A second random variable $Y$, independent of $X$, is also uniformly distributed on $[-1,1]$, and $Z=Y-X$. Find $\E(Z^2)$ and show that  $\var (Z^2) = 7 \var (X^2)$.

Solution source

$X \sim U(-1,1)$

\begin{align*}
\E[X^2] &= \int_{-1}^1 \frac12 x^2 \, dx \\
&= \frac{1}{6} \left [ x^3 \right]_{-1}^1 \\
&= \frac{1}{3}
\end{align*} 

\begin{align*}
\E[X^4] &= \int_{-1}^1 \frac12 x^4 \, dx \\
&= \frac{1}{10} \left [ x^5 \right]_{-1}^1 \\
&= \frac{1}{5}
\end{align*} 

\begin{align*}
\var[X^2] &=\E[X^4] - \E[X^2]^2 \\
&= \frac{1}{5} - \frac{1}{9} \\
&= \frac{4}{45}
\end{align*} 

\begin{align*}
\E(Z^2) &= \E(Y^2 - 2XY+Z^2) \\
&= \E(Y^2) - 2\E(X)\E(Y)+\E(Z^2) \\
&= \frac{1}{3} - 0 + \frac{1}{3} \\
&= \frac{2}{3}
\end{align*}

\begin{align*}
\E[Z^4] &= \E[Y^4 -4Y^3X+6Y^2X^2-4YX^3+X^4] \\
&= \E[Y^4]-4\E[Y^3]\E[X]+6\E[Y^2]\E[X^2]-4\E[Y]\E[X^3]+\E[X^4] \\
&= \frac{1}{5}+6 \frac{1}{3} \frac13 + \frac{1}{5} \\
&= \frac{2}{5} + \frac{2}{3} \\
&= \frac{16}{15}
\end{align*}

\begin{align*}
\var(Z^2) &= \E(Z^4) - \E(Z^2) \\
&= \frac{16}{15} - \frac{4}{9} \\
&= \frac{28}{45} \\
&= 7 \var(X^2)
\end{align*}