Question

Let $X$be a random variable having finite expectation $\mu$and variance ${\sigma }^2$, and let $g(*)$be a twice differentiable function. Show that

$E\left[g\right(X\left)\right]\approx g\left(\mu \right)+\frac{g^{\text{'}\text{'}}\left(\mu \right)}{2}{\sigma }^2$

Hint: Expand $g(·)$in a Taylor series about $\mu$. Use the first

three terms and ignore the remainder.

Short answer

The random variable is showed as $E\left[g\right(X\left)\right]\approx g\left(\mu \right)+\frac{g^{\text{'}\text{'}}\left(\mu \right)}{2}{\sigma }^2$having finite expectation.

Step-by-step solution

Given Information

The finite expectation of variance and twice differentiable function $g(·)$.

Explanation

If $n⩾0$is an integer and $g$is a function which is $n$times continuously differentiable on the closed interval $[a,x]$and $n+1$times differentiable on the open interval $(a,x)$, then we have:

$g\left(x\right)=g\left(a\right)+\frac{g^{\text{'}}\left(a\right)}{1!}(x-a)+\frac{g^{\text{'}\text{'}}\left(a\right)}{2!}(x-a{)}^2+\dots +\frac{g^{\left(n\right)}\left(a\right)}{n!}(x-a{)}^n+R_n$

The remainder term $Rn$depends on $x$and is small if $x$is close enough to$a$.

Explanation

Expand $g(·)$in Taylor polynomial:

$g\left(x\right)=g\left(\mu \right)+\frac{g^{\text{'}}\left(\mu \right)}{1!}(x-\mu )+\frac{g^{\text{'}\text{'}}\left(\mu \right)}{2!}(x-\mu {)}^2+R_n$

Expected value of both side is,

$\left[Eg\right(x\left)\right]=g\left(\mu \right)+0+\frac{g^{\text{'}\text{'}}\left(\mu \right)}{2!}{\sigma }^2+R_n\left({\sigma }^2\right),E(x-\mu )=0$

$\left[Eg\right(x\left)\right]\approx g\left(\mu \right)+\frac{g^{\text{'}\text{'}}\left(\mu \right)}{2!}{\sigma }^2$.

Final answer

The random variable is showed as $\left[Eg\right(x\left)\right]\approx g\left(\mu \right)+\frac{g^{\text{'}\text{'}}\left(\mu \right)}{2!}{\sigma }^2$having finite expectation.