which of the following measurements can be found for a portfolio by simply taking the weighted average of the individual components.? 1beta II return III.standard deviation A. 1 and II B. II and III c. I and III D. I,II,and III

A

i know but why?

Because you need to know the covariances as well and even with that, the average is not a simple weighted average.

beta-port = w1*beta1 + w2 * beta2 sqrt(w1^2*sd1^2 + w2^2 * sd2^2 + 2*w1*w2*rho12*sd1*sd2) rho12*sd1*sd2 = cov12 only time it is a weighted average of the std devs of the individual std devs is when rho12 = 1

well, the simplest way to do this is remember that return is always a weighted avg, and unless correlation is +1 (perfectly positively correlated), std dev includes cov, so its not a weight avg if you remember beta, its cov(p, m)/var(m), p - portfolio, m - market lets do an easy case, say p=ax+by so beta = cov(p,m)/var(m), now cov will be as follows, cov(p,m) = E[(p-E§)(m-E(m))] = E[(ax+by-E(ax+by))(m-E(m))] = E[(a(x-E(x))+b(y-E(y)))(m-E(m))] = aE[(x-E(x))(m-E(m))] + bE[(y-E(y))(m-E(m))] take this and divide by var(m), then you have abetaX + bbetaY, so weighted avg… hope I didn’t miss anything, I haven’t touched this stuff in 2 months…

apparently, cpk beat me to it

gotta be fast out here.

cov(p,m) = E[(p-E§)(m-E(m))] = E[(ax+by-E(ax+by))(m-E(m))] = E[(a(x-E(x))+b(y-E(y)))(m-E(m))] = aE[(x-E(x))(m-E(m))] + bE[(y-E(y))(m-E(m))] I must confess I’m not very good at typing this…

FYI: the reason portfolio beta and return are both weighted averages is that return is a “linear” function of beta. So, weighting beta is equivalent to weighting returns. But that darn covariance screws things up for std. deviation.