Referencing Wei Jiao (2003) p. 8, formula (1.12), for $Ax = b$ set of linear constraints in a portfolio, the solution for the optimum weights to maximize the utility is: $$w^* = \Sigma^{-1}A^T \left( A \Sigma^{-1} A^T \right)^{-1}b + \lambda \Sigma^{-1} \left( \mu-A^T \left( A \Sigma^{-1} A^T \right) A \Sigma^{-1} \mu\right)$$

Based on a similar question, $\lambda$ could be removed by dividing by the sum of the numerator, ie dividing the second component of $w^*$ by the sum of the component. Is this the correct way of eliminating the risk-aversion variable?

  • $\begingroup$ Are you familiar with the method of Lagrange multipliers? $\endgroup$ – Matthew Gunn Nov 1 '18 at 16:46

You cannot eliminate the dependence of a solution on the risk aversion parameter (which this author confusingly calls $\lambda$).

Perhaps a source of confusion?

  • Typically $\lambda$ is used to denote a Lagrange multiplier in Lagrangian optimization, but the author is using $\lambda$ as a risk tolerance parameter. (In your other linked question, $\lambda$ denotes a Lagrange multiplier.)

The author uses a $m \times 1$ vector $\boldsymbol{\gamma}$ as the Lagrange multipliers and the scalar $\lambda$ as the risk tolerance parameter for utility specification $u(\mathbf{w}) = \mu_p - \frac{1}{2 \lambda} \sigma^2_p$ where $\mu_p$ is the expected portfolio return given portfolio weights $\mathbf{w}$ and $\sigma^2_p$ is the variance of the portfolio return.

For a closed forms solution to the optimization problem, the author's goal is to find an expression for the solution that does not use the multipliers. In this case, that means eliminating $\boldsymbol{\gamma}$ (which he does). In your other link, $\lambda$ is used in the typical way as a Lagrange multiplier so a closed form solution eliminates $\lambda$.

The optimization problem by the way is: \begin{equation} \begin{array}{*2{>{\displaystyle}r}} \mbox{maximize (over $\mathbf{w}$)} & \boldsymbol{\mu}'\mathbf{w} - \frac{1}{2 \lambda} \mathbf{w}'\Sigma \mathbf{w} \\ \mbox{subject to} & A \mathbf{w} = \mathbf{b} \end{array} \end{equation}

The solution $\mathbf{w}^*$ will be a function of expected returns $\boldsymbol{\mu}$, covariance matrix $\Sigma$, and risk tolerance $\lambda$.

This is a simple, pretty standard problem, and you can undoubtedly find other people solving it all over the Internet.

Motivation for objective $\boldsymbol{\mu}'\mathbf{w} - \frac{1}{2 \lambda} \mathbf{w}'\Sigma \mathbf{w}$

Let's assume the agents preferences over various lotteries can be represented by:

Let $X$ be some lottery that's normally distributed with mean $\mu$ and variance $\sigma^2$. $$X \sim \mathcal{N}(\mu, \sigma^2)$$

You can show that this lottery $X$ has a certainty equivalent value to our agent given by: $$ c(X) = \mu - \frac{1}{2}a\sigma^2 $$ (Start with $\mathbb{E}[-e^{-aX}] = \frac{1}{\sqrt{2\pi}\sigma}\int_{- \infty}^{\infty}-e^{-ax-\frac{1}{2}\left(\frac{x - \mu}{\sigma}\right)^2}$ and use normal pdf sums to 1 to find certainty equivalent.)

Here, $a$ is Arrow-Pratt coefficient of absolute risk aversion. You can also define risk tolerance $\tau = \frac{1}{a}$. Writing the certainty equivalent with risk tolerance:

$$ c(X) = \mu - \frac{1}{2\tau}\sigma^2 $$

Maximizing expected utility with CARA risk aversion over a normally distributed lottery is equivalent to maximizing the certainty equivalent given above. It's a nice, convenient specification that makes the math easy to work with.

You can of course point out all kinds of deficiencies which would motivate richer specifications:

  • Portfolio returns covary with other variables agents care about.
  • Returns aren't normally distributed
  • CARA has problems: would you insure the risk of a 1,000 loss the same way if your wealth was 2,000 as if your wealth was 2,000,000,000? Probably not.
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  • $\begingroup$ Thank you for your elaboration, it's helped clarify some things. Using a similar framework as described, would it be a similar process to eliminate the risk tolerance variable so that w* is a function of mu, covar_matrix and the Lagrangian multipliers? $\endgroup$ – Cameron Cox Nov 2 '18 at 15:48

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