# Expected Utility and $\log$

I've just started reading about expected utility and utility functions and have the following question.

$\textbf{Question:}$ An investor has an initial wealth of 100 and a utility function of the form: \begin{align} U(w) = \log(w) \end{align} What is their expected utility?

\begin{align} \end{align}

Upon one of the slides I found on the web it states the following:

$\textbf{Calculating Expected Utility}$
1. When the choice variable $x$ is constant, then $E(U(x)) = U(x)$.
2. When the choice variable $x$ is a random variable, then $E(U(x))$ is driven by the PDF of $x$.
3. If $x$ has $k$ outcomes, each with probability $p_k$, then \begin{align} E(U(x)) = \sum_{1}^{k} p_i U(x_i) \end{align}

Since I'm told the initial wealth is 100 does this simply mean the expected utility is $E(U(100)) = \log(100)$?

Apologies if this is trivial - I'm just starting out.

All help is appreciated.
John

• it seems a very strange question. Usually you are given the data as above plus a distribution of returns. In the absence of such a distribution, we only have $log(100).$ – Mark Joshi May 23 '15 at 0:14

This is a strange question. Usually questions about expected utility involve some uncertainty about the future wealth of the investor. If there is no uncertainty in the outcome and the investor is not doing anything which might change his or her future wealth then the expectation of utility is a constant, that is $E[U(w)] = U(w)$.

I am assuming the probability of gaining an amount $x$ is $p$. Then the probability of losing an amount $y$ is $1-p$.

$E[U(W)]=pU(100+x)+(1-p)U(100-y)$

$=p\log{(100+x)}+(1-p)\log{(100-y)}$

$=p\log{(\frac{100+x}{100-y})}-\log{(100-y)}$

It should resemble something like that. I just gave you an overview, because I think a lot of details are missing in your question, never mind because you are new to this as you said.

In order to start a problem like this, you need a table

$$\begin{array}{c|lcr} probabilty & \text{Gain} & \text{W_0+x} & \text{U(W_0+x)} \\ \hline p & x & 100+x & \log({100+x}) \\ 1-p & -y & 100-y & \log({100-y}) \\ \end{array}$$

Your table could contain more probabilities depending on the question, but the sum of the probabilities must be equal to $1$.