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{\bf Question}

Suppose that a r.v. $X$,  either discrete or continuous, has mean
$\mu$ and variance $\sigma^2$.  Define a function $h(c)$ by

$$h(c)=E[(X-c)^2],\ \ c \in R^1.$$

By noting that $h(c)$ is a quadratic function is $c$, show that
$h(c)$ attains its minimum value,\ $\sigma^2$,\ at $c=\mu$. By
noting $E[(X-\mu)^2] \geq 0$, show that $E(X)^2 \leq E(X^2)$. If
$E(X)^2 = E(X^2)$ what can be said about $X$?


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{\bf Answer}

Note that
\begin{eqnarray*} h(c)& = & E[(X-\mu+\mu-c)^2]\\ & = &
E[(X-\mu)^2+2(X-\mu)(\mu-c)+(\mu-c)^2]\\ & = & \sigma^2 +
(\mu-c)^2 \end{eqnarray*}

from which it is clear that $h(c)$ attains its minimum $\sigma^2$
at $c=\mu$.  $E(X^2) \geq \{E(X)\}^2$ follows obviously from

$var(X)=E(X^2)-\{E(X)\}^2 \geq 0$. If the quantity $var(X)=0$ then
$X$ is constant and so $X$ is non-random.

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