# Homework Help: Computing an optimum portfolio for a CARA utility function

1. Oct 29, 2013

### TaPaKaH

1. The problem statement, all variables and given/known data
$u(x)=1-e^{-ax}$.
Random variable $Y\in\mathbb{R}^d$ is a multivariate normal distribution with mean vector $m$ and invertible covariance matrix $\Sigma$.
Task: Find $\xi^*\in\mathbb{R}^d$ that maximises $\mathbb{E}u(\xi\cdot Y)$ over $\xi$.

2. Relevant equations.
For $Y$ above, its density function is $$f_Y(x)=\frac{1}{(2\pi)^{d/2}|\Sigma|^{1/2}}e^{-\frac{(x-m)^T\Sigma^{-1}(x-m)}{2}}$$

3. The attempt at a solution
Do I get it right that in order to find $\xi^*$ I need to solve $\frac{d}{d\xi_i}\mathbb{E}u(\xi\cdot Y)=0$ for $i=1..d$ and then check whether all $\frac{d}{d\xi_i^2}\mathbb{E}u(\xi\cdot Y)$ are negative?

2. Oct 29, 2013

### Ray Vickson

Yes, for $F(\vec{\xi}) = E u (\vec{\xi} \cdot \vec{Y})$ you need to set
$$\frac{\partial F}{\partial \xi_i} = 0, \: i = 1, \ldots, d,$$
but the second-order test is more complicated that what you indicated: you need to check that the $d \times d$ Hessian matrix
$$H(\vec{\xi}) = \left[ \begin{array}{cccc} \partial^2 F/ \partial \xi_1 ^2 & \partial^2 F / \partial \xi_1 \partial \xi_2 & \cdots & \partial^2 F/ \partial \xi_1 \partial \xi_d\\ \partial^2 F /\partial \xi_2 \partial \xi_1 & \partial^2 F/ \partial \xi_2 ^2 & \cdots & \partial^2 F/ \partial \xi_2 \partial \xi_d\\ \vdots & \vdots & \ddots & \vdots \\ \partial^2 F /\partial \xi_d \partial \xi_1 &\partial^2 F /\partial \xi_d \partial \xi_2 & \cdots & \partial^2 F/ \partial \xi_d ^d \end{array}\right]$$
is negative-definite at the optimal value $\vec{\xi} =\vec{\xi^*}$.

However, before embarking on any of that I urge you to first evaluate $F(\vec{\xi})$ as an explicit formula; believe it or not it is not too bad!

Last edited: Oct 29, 2013
3. Oct 29, 2013

### TaPaKaH

This is what I got so far:
for $g(\xi)=\mathbb{E}e^{-\xi\cdot Y}$ which we are now looking to minimise,
$$g(\xi)=\int_{\mathbb{R}^d}e^{-\xi\cdot x}f_Y(x)dx=\frac{1}{(2\pi)^{d/2}|\Sigma|^{1/2}}\int_{\mathbb{R}^d}e^{-\xi\cdot x}e^{-(x-m)^T\Sigma^{-1}(x-m)/2}dx$$
$$\frac{\partial g}{\partial\xi_i}(\xi)=\frac{-1}{(2\pi)^{d/2}|\Sigma|^{1/2}}\int_{\mathbb{R}^d}x_ie^{-\xi\cdot x}e^{-(x-m)^T\Sigma^{-1}(x-m)/2}dx=-\mathbb{E}(Y_ie^{-\xi\cdot Y})$$
$$\frac{\partial^2g}{\partial\xi_i\xi_j}(\xi)=\mathbb{E}(Y_iY_je^{-\xi\cdot Y})$$
Indeed, it doesn't seem to look too bad, but now I am at slight loss at how to compute the derivatives.

4. Oct 29, 2013

### Ray Vickson

This is not what I meant: I suggested you derive an explicit formula for $F(\xi)$, and that means calculating the expectation first, by actually doing the d-dimensional integrations! YES: believe it or not, it is quite easy. You just need to use a number of fundamental facts about multivariate normal and univariate normal random variables, and these are available via a Google search on the keywords 'multivariate normal distribution'.

5. Oct 29, 2013

### Office_Shredder

Staff Emeritus
It might be instructive to first solve this in the case where $\Sigma$ is a diagonal matrix.