# How to combine 3 random variables?

I have a homework assignment that is giving me a hard time on the statistics. Lets say you have 3 stocks, all with n expected return (mean) $\mu = 8\%$, a risk (standard deviation) $\sigma = 16\%$ and a correlation coefficient $\rho = 0.3$ between every tow stocks. The homework questions is: If you build a portfolio from three stocks with an equal part form each stock. What is the expected return, and the risk?. The expected return is easy, its the average of 8%,8%,8% which is 8%. But how to I calculate the risk (standard deviation) of all three? There are plenty of examples how to do it with 2 random variables, its: $\sigma_1^2*w_1^2+\sigma_2^2*w2^2-\sigma_1*\sigma_2*w_1*w_2*\rho_{1,2}$ (Where $w_1, w_2$ are the relative weights of the rabdom variable). But what is it for three? My guess is$\sigma_1^2*w_1^2+\sigma_2^2*w_2^2-\sigma_1*\sigma_2*w_1*w_2*\rho_{1,2} + \\ \sigma_2^2*w_2^2+\sigma_3^2*w_3^2-\sigma_2*\sigma_3*w_2*w_3*\rho_{2,3} + \\ \sigma_1^2*w_1^2+\sigma_3^2*w_3^2-\sigma_1*\sigma_3*w_1*w_3*\rho_{1,3} + \\ \sigma_1^2*w_1^2+\sigma_2^2*w_2^2+\sigma_3^2*w_3^2 + \sigma_1*\sigma_2*\sigma_3*w_1*w_2*w_3*\rho_{1,2,3}$

In my question where $\sigma_1=\sigma_2=\sigma_3=16\%$ and $w_1=w_2=w_3=8\%$ and $\rho_{1,2}=\rho_{2,3}=\rho_{1,3}=0.3$ I get a simplified formula: $(\sigma^2*w^2*2-\sigma^2*w^2*\rho_{1,2}) * 3 + \sigma^3*w^3*\rho_{1,2,3}$

I took this idea from the Inclusion–exclusion principle.

Now my questions: Is that even true for random variables? If so, how do I know $\rho_{1,2,3}$? (Please leave an explanation\reference on $\rho_{1,2,3}$ and not just a formula.)

Vocabulary (in case it was not clear how the financial terms fit into statistics):

• random variable = stock
• mean = $\mu$ = expected return of a stock
• standard deviation = $\sigma$ = risk of a stock\portfolio

There is a typo is the formula you used for 2 stocks, the correct formula should contain a "$+$" instead of a "$-$", that is $\sigma_1^2w_1^2+\sigma_2^2w_2^2+\sigma_1\sigma_2w_1w_2\rho_{1,2}$. Now the method is exactly identical for any number of variable. Just write the definition of the variance and expand:
$Var(\sum_{i=0}^n w_i X_i) = Cov(\sum_{i=0}^n w_i X_i, \sum_{j=0}^n w_j X_j)=\sum_{i=0}^n\sum_{j=0}^n Cov(X_i, X_j)$
Bear in mind the expression of the covariance $Cov(X,Y)=E[(X-E[X])(Y-E[Y])]$ which also relates by definition to the correlation coefficient $\rho_{xy}$ as $Cov(X,Y)=\rho_{xy}\sigma_x\sigma_y$.
• OK, so just to be sure: 1) The end forumla would be $\sigma_{final} = \sum_{i=0}^3 \sum_{j=0}^3 \rho_{ij}\sigma_i\sigma_j$ 2) There is no connection to the Inclusion-exclusion principle? – Ramzi Kahil Jan 5 '13 at 12:57
• The correct one is $\sigma_{final}^2 = \sum_{i=0}^3 \sum_{j=0}^3 \rho_{ij}\sigma_i\sigma_j$ indeed. – ThePawn Jan 5 '13 at 13:42