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Let's consider the very basic of a Mean-Variance Portfolio:

$$ \text{max}_{x} (1-\lambda)\sum_i^n\mu_ix_i-\lambda\sum_i^n\sum_j^n x_i Q_{ij}x_j $$ $$\text{ s.t. }\sum_i^nx_i=1 \text{ , } x_i \geq 0 \text{ (No shorting) }$$ where $\lambda, \mu, Q$ is the risk averse parameter, expected return vector and variance-covariance matrix for all the assets.

I Interpret $\sum_i^n\sum_j^n x_i Q_{ij}x_j$ as the variance of my protfolio, am I correct?

Let's assume I invest 10\$ and my first constraint is normalizing the 10\$ to 1 so I have to multiply my results by 10. Let's assume I use software to find the optimal portfolio with the weights $x_1...x_n$ so the value of asset $i$ is $x_i*100\%$ of my total investment.

My expected return is then: $10\$\sum_i^n\mu_ix_i$

QUESTION: What is the variance (risk) of the portfolio then? Becuase of the non-linearity I cant simply multiply by 10.

My first thought is of course to multiply my weights $x_i$ with 10 and simply compute the variance by $\sum_i^n\sum_j^n x_i Q_{ij}x_j$:

EDIT

But consider this case: I have a low $\lambda$ (I am "risk loving") so my solution to the problem is to invest all my money in asset $c$ where the expected return is $\mu_c=23\%=0.23$ and variance of the return is $Q_{c,c}=0.9$. This suggests that my variance is $10^2*0.9=90$ while expected to have a profit of 2.3\$. Am I correct? My concern is then that my variance or risk is so much higher initial investment so it makes me question if my math is correct.

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  • $\begingroup$ $\sum_i^n\sum_j^n x_i Q_{ij}x_j$ is the variance of portfolio return. So it makes sense to simply multiply by $10 (portfolio value) to get the actual variance. $\endgroup$ – raymkchow Oct 3 '18 at 2:52
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    $\begingroup$ Hi: $var(ax) = a^2 \times var(x)$ so you need to square the dollar value. but the variance is then the variance of the profit and loss. not the variance of the return. $\endgroup$ – mark leeds Oct 3 '18 at 6:25
  • $\begingroup$ Dear @markleeds I have now edited my question. Will you please have a look $\endgroup$ – Kim Oct 3 '18 at 10:39
  • $\begingroup$ Hi: your variance is approximately four times your mean so I guess it makes sense note that it's variance on the dollars returned. $\endgroup$ – mark leeds Oct 3 '18 at 13:31

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