STEP Project — 2017.3 Question 13

We have

\begin{array}{l} V (x) & = E [{(X - x)}^{2}] \\ = E (X^{2} - 2 xX + x^{2}) \\ = E (X^{2}) - 2 x E (X) + x^{2} \\ = σ^{2} + μ^{2} - 2 xμ + x^{2} . \end{array}

Therefore, if $Y = V (X)$ , then

\begin{array}{l} E (Y) & = E (V (X)) \\ = E (σ^{2} + μ^{2} - 2 Xμ + X^{2}) \\ = σ^{2} + μ^{2} - 2 μ E (X) + E (X^{2}) \\ = σ^{2} + μ^{2} - 2 μ^{2} + μ^{2} + σ^{2} \\ = 2 σ^{2} . \end{array}

Let $X \sim U [0, 1]$ , we have $μ = E (X) = \frac{1}{2}$ , and $σ^{2} = Var (X) = \frac{1}{12}$ . Therefore,

\begin{array}{l} V (x) & = \frac{1}{12} + \frac{1}{4} - x + x^{2} \\ = x^{2} - x + \frac{1}{3} . \end{array}

The c.d.f. of $X$ is $F$ , defined as

P (X \leq x) = F (x) = {\begin{matrix} 0, & x \leq 0, \\ x, & 0 < x \leq 1, \end{matrix}

Let the c.d.f. of $Y$ be $G$ , we have $G (y) = P (Y \leq y)$ .

Since $V ([0, 1]) = [\frac{1}{12}, \frac{1}{3}]$ , we must have $G (y) = 0$ for $y \leq \frac{1}{12}$ and $G (y) = 1$ for $y > \frac{1}{3}$ .

For $y \in (\frac{1}{12}, \frac{1}{3}]$ , we have

\begin{array}{l} G (y) = P (Y \leq y) & = P (V (X) \leq y) \\ = P ({(x - \frac{1}{2})}^{2} + \frac{1}{12} \leq y) \\ = P (| x - \frac{1}{2} | \leq \sqrt{y - \frac{1}{12}}) \\ = P (\frac{1}{2} - \sqrt{y - \frac{1}{12}} \leq x \leq \frac{1}{2} + \sqrt{y - \frac{1}{12}}) \\ = F (\frac{1}{2} + \sqrt{y - \frac{1}{12}}) - F (\frac{1}{2} - \sqrt{y - \frac{1}{12}}) \\ = (\frac{1}{2} + \sqrt{y - \frac{1}{12}}) - (\frac{1}{2} - \sqrt{y - \frac{1}{12}}) \\ = 2 \sqrt{y - \frac{1}{12}} . \end{array}

Therefore, the p.d.f. of $y$ , $g$ satisfies that for $y \in (\frac{1}{12}, \frac{1}{3}]$ ,

g (y) = G^{'} (y) = \frac{1}{\sqrt{y - \frac{1}{12}}}

and $0$ everywhere else.

Hence, we have

\begin{array}{l} E (Y) & = \int_{ℝ} yf (y) d y \\ = \int_{\frac{1}{12}}^{\frac{1}{3}} \frac{y}{\sqrt{y - \frac{1}{12}}} d y \\ = \int_{y = \frac{1}{12}}^{y = \frac{1}{3}} 2 y d \sqrt{y - \frac{1}{12}} \\ = {[2 y \sqrt{y - \frac{1}{12}}]}_{\frac{1}{12}}^{\frac{1}{3}} - 2 \int_{\frac{1}{12}}^{\frac{1}{3}} \sqrt{y - \frac{1}{12}} d y \\ = {[2 y \sqrt{y - \frac{1}{12}} - \frac{4}{3} {(y - \frac{1}{12})}^{\frac{3}{2}}]}_{\frac{1}{12}}^{\frac{1}{3}} \\ = 2 \cdot \frac{1}{3} \cdot \frac{1}{2} - \frac{4}{3} \cdot \frac{1}{8} \\ = \frac{1}{6} . \end{array}

Also, $2 σ^{2} = 2 \cdot \frac{1}{12} = \frac{1}{6} = E (Y)$ , so the formula we derived holds in this case.