1 problem found
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: \(X \sim U(-1,1)\) \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*}