# Mean-Variance portfolio: How do I compute the variance when the portfolio is normalized

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. •$\sum_i^n\sum_j^n x_i Q_{ij}x_j$is the variance of portfolio return. So it makes sense to simply multiply by$10 (portfolio value) to get the actual variance. – raymkchow Oct 3 '18 at 2:52
• 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. – mark leeds Oct 3 '18 at 6:25
• Dear @markleeds I have now edited my question. Will you please have a look – Kim Oct 3 '18 at 10:39
• Hi: your variance is approximately four times your mean so I guess it makes sense note that it's variance on the dollars returned. – mark leeds Oct 3 '18 at 13:31