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I have the standard $3$-period Grossman and Miller model with $2$ outside traders and $M$ market makers.

I'm told:

  • $W_t^{(1)}, W_t^{(2)}, W_t^{(m)}$ is the wealth of the first outside trader, second outside trader and market maker at time $t$.
  • $B_t^{(1)}, B_t^{(2)}, B_t^{(m)}$ is the cash of the first outside trader, second outside trader and market maker at time $t$.
  • $x_t^{(1)},x_t^{(2)},x_t^{(m)}$ is the shares of the security of the first outside trader, second outside trade and market maker at time $t$.
  • $\tilde{p}_t$ is the unknown security price at time $t$ that follows a normal distribution.

I'm then asked:

Let $x$ be the number of share holdings, $\tilde{p} \sim N(\mu, \sigma^2)$ be the security price and $W=x \cdot \tilde{p}$ be the wealth. Show that $E(U(W)) = -e^{(((\lambda^2\sigma^2)/2)x^2)-\lambda\mu x}$

So I know that the negative exponential utility function is: $U(W)=-e^{-\lambda W}$, in a previous part of this question I formulated the utility optimization problem for the first outside trader, second outside trader and market maker, but I'm not sure if these help (if they will be of use I can upload them).

This is where I'm stuck, I'm not sure how I can find $E(U(W))$ given what I know and what is provided in the question.

Any help would be greatly appreciated, even if it's only a hint to get me started.

Thanks in advance

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Deriving the expected utility comes from an application of the moment generating function of a Normal random variable:

Let $U(W)=-e^{-\lambda W}$ and $W=x \cdot \bar{p}$ where $\bar{p} \sim N(\mu, \sigma^2)$ be given as above and then further note that $W$ is Normally distributed with $W \sim N(\mu \cdot x, x^2 \sigma^2)$.

See that:

\begin{align} \mathbb{E}\left[U(W)\right] &= \mathbb{E}\left[-e^{-\lambda W}\right]\\ &=-\mathbb{E}\left[e^{-\lambda W}\right], \end{align}

is very similar to the m.g.f. of a Normal random variable.


Remember that the moment generating function of a Normal random variable with distribution $x \sim N(\mu, \sigma^2)$ is given by (see here or here for formula):

$$\mathbb{E}\left[e^{tx}\right] = e^{t\mu+\frac{\sigma^2t^2}{2}}$$


Continuing from above, you can apply the moment generating function to the expected utility, which yields the desired solution:

\begin{align} \mathbb{E}\left[U(W)\right] &= \mathbb{E}\left[-e^{-\lambda W}\right]\\ &=-\mathbb{E}\left[e^{-\lambda W}\right]\\ &=-e^{-\mu x\lambda + \frac{x^2\sigma^2\lambda^2}{2}}\\ &=-e^{\frac{\lambda^2\sigma^2x^2}{2} - \lambda\mu x}, \end{align} where I have realigned everything in the last equation, in order to fit the solution in your question.

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  • $\begingroup$ @CharlieP Hi. Feel free to comment on my solution. Did my answer help or is it not what you're looking for? I am always willing to explain more in detail, if that is what you need. $\endgroup$
    – Pleb
    Jan 13, 2022 at 7:18
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    $\begingroup$ This is perfect, thanks a lot $\endgroup$
    – Charlie P
    Jan 13, 2022 at 18:41
  • $\begingroup$ Glad I could help :-) $\endgroup$
    – Pleb
    Jan 13, 2022 at 18:51

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