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} \leq μ - μ^{2} \leq \frac{1}{4} (taking the maximum of the above quadratic) \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 \leq {(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 \end{matrix}

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

The distribution of $X$ is Bernoulli. More specifically, $X \sim Bern (1 ∕ 2)$ .