# How to arrive at expectation of negative utility function via Taylor series expansion

I'm attempting to follow an author's steps in an argument and having trouble seeing how Taylor series expansion can be applied to give the stated result. The scenario is as follows.

The mid price of a stock evolves according to: $$dS_{u} = \sigma dW_{u}$$ with initial value $S_{t}=s$ and $W_{t}$ being a standard one-dimensional Brownian motion with constant $\sigma$.

Assuming an stock inventory of size $q$ and an initial wealth, in dollars, of $x$, and agent's value function is given as: $$v(x, s, q, t)= \mathbb{E}[-exp(-\gamma(x+qS_{T}))]$$ The author moves to state that this expectation can be written as: $$v(x, s, q, t)= -exp(-\gamma x)exp(-\gamma qs)exp \left(\frac{\gamma^2q^2\sigma^2(T-t)}{2}\right)$$ I believe that a Taylor series expansion is be applied to get this result but I can't follow how this is the case. Can someone help me understand the steps?

I've tried to follow the method outlined in this Wikipedia entry on Taylor and random variables, but the above doesn't quite 'fit'.

To answer your question you need to look at what the distribution is of $S_T$. Since $dS = \sigma dW$ we have that $S_T$ is normal distribution with variance of $\sigma^2 (T)$.

Now you get $E(-exp(-\gamma(x + qS_T))) = -exp(-\gamma x) E (exp(-\gamma q S_T))$. Given that $S_T$ is normal $-\gamma q S_T$ is normal as well. You just scale the variance with $q^2 \gamma^2$. Finally you need to see that an exponent of a normal distribution is log-normal. The mean of a log normal is given by $exp(\mu + 0.5\sigma^2)$ (where $\mu$ is mean of your normal and $\sigma$ is stdev of your normal). You just need to plug in the mean and variance of your scaled normal $\gamma q S_T$.

Hope this helps.

• Thanks. Ok, so no need to use Taylor series expansion at all? So you are suggesting I need to find $E (exp(-\gamma q S_T))$, with $-\gamma q S_T$ being a normal variable. With the mean of a log normal being $exp(\mu + 0.5\sigma^2)$, the result I am given in the text would suggest that the $\mu=-\gamma q s$ and $0.5\sigma^2=\frac{\gamma^2q^2\sigma^2(T-t)}{2}$. Is that a correct interpretation of your answer?
– tfb
Commented Aug 24, 2015 at 21:13
• yes agreed. To summarize: there are 2 steps in this exercise. A) recognize that if dS = sigma dW then that the distribution of S_T is normal B) knowing what the mean is of exp(normal). Commented Aug 24, 2015 at 21:22

First, you need to derive the distribution of $S_T$. You have given $$dS_u=\sigma dW_u$$ Integrate both side:

$$\int_t^T dS_u=\int_t^T dW_u$$ $$S_T-S_t=\sigma\big(W_T-W_t\big)$$ We know that $W_t-W_t \sim N(0, T-t)$, so $$S_T-S_t \sim N(0, \sigma^2 (T-t))$$ $$S_T|S_t\sim N(S_t, \sigma^2(T-t))$$ It is given that initial value of $S_t$ is $s$, so we have $$S_T|S_t\sim N(s, \sigma^2(T-t))$$

Now, comes to main problem, You want $\mathbb{E}[-e^{-\gamma(x+qS_{T})}]$. Everything except $S_T$ is constant and get them out of expectation expression. So, $$\mathbb{E}[-e^{-\gamma(x+qS_{T})}]=-e^{\gamma x}\mathbb{E}[e^{-\gamma qS_{T}}]$$ The expectation of the RHS is just like moment generating function. If random variable $X$ follows normal distribution, with mean $\mu$ and variance $\sigma^2$, its MGF is: $$\mathbb{E}[e^{kX}]=e^{\mu k + \frac12 \sigma^2 k^2 }$$ Assuming $k=-\gamma q$, your last equation will result into \begin{align} \mathbb{E}[-e^{-\gamma(x+qS_{T})}]&=-e^{\gamma x}\mathbb{E}[e^{-\gamma qS_{T}}]\\ &=-e^{\gamma x} e^{-s\gamma q+ \frac12 \sigma^2 (T-t)^2 \gamma^2 q^2} \end{align} This is exactly what you wanted. So, to solve this neither you required Taylor series nor any Log normal distribution.
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