Question

Let $X\sim \text{Unif}(0,4)$ and, given $X=x$, $Y\mid X=x\sim \text{Unif}\!\left(0,\dfrac{x^{2}}{4}\right)$. Compute $E(Y^{2})$ (rounded off to three decimal places).

Hint:

For nested uniforms, compute inner moments using $E(U^2)=a^2/3$ and then average over the outer variable.

Answer 1

Correct Answer: 1.065

By the law of total expectation, \[ E(Y^{2})=E\!\left[E\!\left(Y^{2}\mid X\right)\right]. \] If $U\sim\text{Unif}(0,a)$, then $E(U^{2})=a^{2}/3$. Hence \[ E\!\left(Y^{2}\mid X=x\right)=\frac{1}{3}\left(\frac{x^{2}}{4}\right)^{2}=\frac{x^{4}}{48}. \] Therefore, \[ E(Y^{2})=\frac{1}{48}E(X^{4}),\qquad X\sim\text{Unif}(0,4)\Rightarrow E(X^{4})=\frac{1}{4}\int_{0}^{4}x^{4}\,dx =\frac{1}{4}\cdot\frac{4^{5}}{5}=51.2. \] Thus \[ E(Y^{2})=\frac{51.2}{48}=1.\overline{06}7 \approx \boxed{1.067}. \]