1 problem found
The random variable \(X\) has mean \(\mu\) and variance \(\sigma^2\), and the function \({\rm V}\) is defined, for \(-\infty < x < \infty\), by \[ {\rm V}(x) = \E \big( (X-x)^2\big) . \] Express \({\rm V}(x)\) in terms of \(x\), \( \mu\) and \(\sigma\). The random variable \(Y\) is defined by \(Y={\rm V}(X)\). Show that \[ \E(Y) = 2 \sigma^2 %\text{ \ \ and \ \ } %\Var(Y) = \E(X-\mu)^4 -\sigma^4 . \tag{\(*\)} \] Now suppose that \(X\) is uniformly distributed on the interval \(0\le x \le1\,\). Find \({\rm V}(x)\,\). Find also the probability density function of \(Y\!\) and use it to verify that \((*)\) holds in this case.
Solution: \begin{align*} {\rm V}(x) &= \E \big( (X-x)^2\big) \\ &= \E \l X^2 - 2xX + x^2\r \\ &= \E [ X^2 ]- 2x\E[X] + x^2 \\ &= \sigma^2+\mu^2 - 2x\mu + x^2 \\ &= \sigma^2 + (\mu - x)^2 \end{align*} \begin{align*} \E[Y] &= \E[\sigma^2 + (\mu - X)^2] \\ &= \sigma^2 + \E[(\mu - X)^2]\\ &= \sigma^2 + \sigma^2 \\ &= 2\sigma^2 \end{align*} If \(X \sim U(0,1)\) then \(V(x) = \frac{1}{12} + (\frac12 - x)^2\). \begin{align*} \P(Y \leq y) &= \P(\frac1{12} + (\frac12 - X)^2 \leq y) \\ &= \P((\frac12 -X)^2 \leq y - \frac1{12}) \\ &= \P(|\frac12 -X| \leq \sqrt{y - \frac1{12}}) \\ &= \begin{cases} 1 & \text{if } y - \frac1{12} > \frac14 \\ 2 \sqrt{y - \frac1{12}} & \text{if } \frac14 > y - \frac1{12} > 0 \\ \end{cases} \\ &= \begin{cases} 1 & \text{if } y> \frac13 \\ \sqrt{4y - \frac1{3}} & \text{if } \frac13 > y > \frac1{12} \\ \end{cases} \end{align*} Therefore $f_Y(y) = \begin{cases} \frac{2}{\sqrt{4y-\frac{1}{3}}} & \text{if } \frac1{12} < y < \frac13 \\ 0 & \text{otherwise} \end{cases}$ \begin{align*} \E[Y] &= \int_{1/12}^{1/3} \frac{2x}{\sqrt{4x-\frac13}} \, dx \\ &= 2\int_{u = 0}^{u=1} \frac{\frac{1}{4}u +\frac1{12}}{\sqrt{u}} \,\frac{1}{4} du \tag{\(u = 4x - \frac13, \frac{du}{dx} = 4\)}\\ &= \frac{1}{2 \cdot 12}\int_{u = 0}^{u=1} 3\sqrt{u} +\frac{1}{\sqrt{u}} \, du \\ &= \frac{1}{2 \cdot 12} \left [2 u^{3/2} + 2u^{1/2} \right ]_0^1 \\ &= \frac{1}{2 \cdot 12} \cdot 4 \\ &= \frac{2}{12} \end{align*} as required