# Calculating mean variance portfolio with risk aversion parameter

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:

$w_t=\frac{x_t}{1_N^Tx_t}$

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:

$w_t=\frac{\Sigma_t^{-1}\mu_t}{1_N\Sigma_t^{-1}\mu_t}$

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

• I don't know why it is not possible to edit the question. Could you please put the sign $in order to convert your equations in LateX form ?Not possible to read them easily for the moment. – optimal control Oct 6 '15 at 19:48 • sorry about that. It is now done! – tyr Oct 6 '15 at 19:53 • I know that there exists a literature about ambiguity etc on these topics but as I am far of this literature, let me know about some stuff about your question. as you are making an optimization, what is your budget constraint ? also where do you use the parameter$w_{t}$? – optimal control Oct 6 '15 at 20:10 • 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. – tyr Oct 6 '15 at 20:29 ## 1 Answer 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:

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