Can you help me understand this derivation? This is finance related so $E[X]$ means the expected outcome (of a gamble) and $u(x)$ is the utility function.
With any well-behaved utility function, when the range of possible outcomes of a gamble is small relative to the risk tolerance of the decision maker, the certainty equivalent of the gamble can be approximated by making use of the local curvature of the utility function.
Consider a gamble $X$ with expected value $E[X]$ and variance $Var(X)$, and a utility curve $u(x)$. From the first two terms of a Taylor series expansion of $u(x)$ around the expected value $E[X]$ we obtain:
$E[u(x)]=u(E[X])+1/2 u^{''}(E[X])Var(X)$ i'm good here
We equate this expected utility to the utility of the certainty equivalent $CE=E[X]+D$, where $D$ is a risk discount.
Again, using a Taylor series expansion, $u(E[X]+D)$ can be approximated by $u(E[X])+u''(E[X])D$. how did we get to this approximation?
Hence the first order approximation for the risk discount is $D = 1/2(u''/u') Var(X)$. how did we get this result?
Therefore:
$CE(X)=E[X]+1/2 (u''/u')Var(X)$ how did we get to this conclusion?
The maximum EU (or maximum CE) principle reduces to a linear function of mean and variance in which the relative weight given the variance is half the local curvature of the utility function.
Thank you.
I found the derivation here: https://quant.stackexchange.com/questions/20628/utility-theory-certainty-equivalent-approximation-formula-derivation
Here is a slightly modified version of that answer:
A second-order expansion near $ \bar{x}=E(x)$ gives:
$u(x)\approx u(\bar{x})+u'(\bar{x})(x-\bar{x})+1/2 u''(\bar{x})(x-\bar{x})^2$
Hence,
$E[u(x)]\approx u(\bar{x}) +\frac{1}{2} u''(\bar{x})Var(x)$
On the other hand, if we let c denote the certainty equivalent and assume it is close to $\bar{x}$, we can use the first-order expansion:
$u(c)\approx u(\bar{x})+u'(\bar{x})(c-\bar{x})$
We transform this equation to get:
$c=\bar{x}+\frac{u(c)-u(\bar{x})}{u'(\bar{x})}$
We can get the value of the numerator in the fraction part of the equation. We replace $E[u(x)]=u(c)$ in the previous formula to get:
$u(c)\approx u(\bar{x})+\frac{1}{2} u''(\bar{x}) Var(x)$
So that
$u(c)-u(\bar{x})\approx\frac{1}{2} u''(\bar{x}) Var(x)$
Therefore:
$c\approx\bar{x}+\frac{1}{2}(\frac{u''(\bar{x})}{u'(\bar{x})}) Var(x)$