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Computing an optimum portfolio for a CARA utility function

  1. Oct 29, 2013 #1
    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?
     
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  3. Oct 29, 2013 #2

    Ray Vickson

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    Yes, for ##F(\vec{\xi}) = E u (\vec{\xi} \cdot \vec{Y})## you need to set
    [tex] \frac{\partial F}{\partial \xi_i} = 0, \: i = 1, \ldots, d,[/tex]
    but the second-order test is more complicated that what you indicated: you need to check that the ##d \times d## Hessian matrix
    [tex] 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] [/tex]
    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
  4. Oct 29, 2013 #3
    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.
     
  5. Oct 29, 2013 #4

    Ray Vickson

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    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'.
     
  6. Oct 29, 2013 #5

    Office_Shredder

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    It might be instructive to first solve this in the case where [itex] \Sigma[/itex] is a diagonal matrix.
     
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