I want to calculate the classic mean variance portfolio (Markowitz) with a risk aversion parameter $\gamma$. I have the following problem where I want to maximize:

$max(x_t) \ \ x_t^T\mu_t - \frac{\gamma}{2}x_t^T\Sigma_tx_t$

Where $\mu$ = mean

$x_t$ = portfolio at time $t$

$\Sigma$ = sample covariance matrix

w(t) is the vector of relative weights in the portfolio, calculated as follows:


1_N is a vector of ones (N elements)

My concern is that in the literature I have seen only solution for the above maximization problem where the risk aversion parameter drops out:


I'm wondering at which point do we account for the risk aversion parameter during the optimization?

  • $\begingroup$ I don't know why it is not possible to edit the question. Could you please put the sign $ in order to convert your equations in LateX form ?Not possible to read them easily for the moment. $\endgroup$ Oct 6 '15 at 19:48
  • $\begingroup$ sorry about that. It is now done! $\endgroup$
    – tyr
    Oct 6 '15 at 19:53
  • $\begingroup$ I know that there exists a literature about ambiguity etc on these topics but as I am far of this literature, let me know about some stuff about your question. as you are making an optimization, what is your budget constraint ? also where do you use the parameter $w_{t}$ ? $\endgroup$ Oct 6 '15 at 20:10
  • $\begingroup$ hi. THanks a lot for your help. I added a couple of more details, I hope it will be easier to understand. (you can refer to Demiguel et al: Optimal Versus Naive Diversification How inefficient is the 1/N portfolio strategy) Regarding the budget constraint, I am not so sure, I think there is the usual constraint applies here too that the sum of weights should add up to 1. $\endgroup$
    – tyr
    Oct 6 '15 at 20:29

Here's how you would account for risk aversion.

First of all, I think this problem is usually set up using weights as the variables (see here http://www.math.ku.dk/~rolf/CT_FinOpt.pdf, page 141). If you want to use dollar weights, you will have to transform the result.

The optimization problem is:

$$ \max_w \quad w'\mu −\frac{\gamma}{2} w'\Sigma w$$

Weights add up to 1:

$$ w'e = 1 $$

Where you have $N$ assets, $w$ and $\mu$ are $N\times1$ vectors, and $\Sigma$ is an $N\times N$ matrix. $e$ is an $N\times1$ vector of ones.

The Lagrangian for the problem is:

$$ \max_w \quad w'\mu −\frac{\gamma}{2} w'\Sigma w + \lambda (w'e - 1)$$

To optimize, we take the gradient of the objective function, and set each element equal to zero. So we will have an $N\times 1$ vector equation:

$$0 = \mu - \gamma \Sigma w + \lambda e$$

This can be solved for $w$:

$$w = \left(\gamma \Sigma\right)^{-1} \left(\mu + \lambda e\right)$$

Plugging this back into the constraint, we get:

$$ \left(\left(\gamma \Sigma\right)^{-1} \left(\mu + \lambda e\right)\right)'e = 1$$

The transpose reverses the order of the terms, and has no effect on the symmetric $\Sigma^{-1}$ matrix:

$$ \left(\left(\mu' + \lambda e'\right)\frac{\Sigma^{-1}}{\gamma} \right)e = 1$$

This can be simplified to:

$$ \mu' \Sigma^{-1} e + \lambda e' \Sigma^{-1} e = \gamma$$

Solving for $\lambda$, we get

$$ \lambda = \frac{\gamma - \mu' \Sigma^{-1} e}{e'\Sigma^{-1} e} $$

So the overall formula for the weights is:

$$w = \left(\gamma\Sigma\right)^{-1} \left(\mu + \frac{\gamma - \mu' \Sigma^{-1} e}{e'\Sigma^{-1} e} e\right)$$


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