problem 51

(a) We know that, $X^{2} \leq X$ with probability 1. So $E [X^{2}] \leq EX$

\begin{matrix} V (X) & = E [X^{2}] - E {[X]}^{2} & (5.15) \leq μ - μ^{2} & (5.16) \leq \frac{1}{4} taking the minimum of the above quadratic & (5.17) \end{matrix}

(b) I have to show that $V (X) = 1 ∕ 4$ leads to a unique distribution. From (a), $V (X) \leq μ - μ^{2} \leq 1 / 4$ implies that, $μ = 1 ∕ 2$ Now

E [{(X - 1 ∕ 2)}^{2}] = 1 / 4

But

0 \geq {(X - 1 ∕ 2)}^{2} \leq 1 / 4

with probability 1. To get $E [{(X - 1 ∕ 2)}^{2}] = 1 / 4$ , we need ${(X - 1 ∕ 2)}^{2} = 1 / 4$ with probability 1. So,

X = {\begin{matrix} 0 & with prob p 1 & with prob 1 - p \end{matrix}

(5.18)

Using $μ = 1 ∕ 2$ gives us, $p = 1 ∕ 2$ .