Q. 7.4 - A First Course in Probability

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.

Step-by-Step Solution

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 1: Given Information

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

Step 2: 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 $a$.

Step 3: 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,

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

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

Step 4: 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.