# Efficient Frontier #66, P 385, Fig 4

If a portfolio has two holdings that have zero correlation with one another, why isn’t the EF line perfectly straight (page 385)? I ask because if you look at page 399, the capital asset line, which a risky asset is mixed with a Rf asset, the line is perfectly straight and the correction is also zero. Maybe this will sink in if I read it again. (btw, after derivatives, I thought I was done with all the heavy lifting!)

Wait till you get to Treynor Black!

two asset case, A - first asset (mu1,sigma1), B - second asset (mu2, sigma2), correlation 0. Now let’s look at p(w)=w*A+(1-w)*B, has expected return of w*mu1+(1-w)*mu2 and, since correlation is zero, expected volatility of sqrt(sigma1^2*w^2+(sigma2)^2*(1-w)^2). Clearly, efficient frontier is not a line. Now let’s assume that B is the risk free asset (mu2 = RFR, sigma2 = 0). We get p(w)=w*A+(1-w)*B, has expected return of w*mu1+(1-w)*mu2 and expected volatility of sqrt(sigma1^2*w^2+(sigma2)^2*(1-w)^2)=w*sigma1. In this case efficient frontier is a line: Ret = (Sigma/simga1)*mu1+(1-Sigma/sigma1)*mu2=Sigma[(mu1-mu2)/sigma2]+mu2.

The EF line with two risky assets is calculated from two assets with positive Std. Dev’s The CAL is calculated from one asset with a positive std Dev and one asset with a std dev of 0. Std Dev of port w/ 2 risky assets = (w1^2*Std Dev1^2 + w2^2*Std Dev2^2)^1/2 Std Dev of port w/ 1 risky asset = (w1^2*std Dev1^2)^1/2 = w1*Std Dev When you reduce the equation for only 1 risky asset the relationship becomes linear. There’s only a curve when both assets are risky assets (and the correlation is >< 1).

Can you please explain conceptually, without using formulas? maratikus Wrote: ------------------------------------------------------- > two asset case, A - first asset (mu1,sigma1), B - > second asset (mu2, sigma2), correlation 0. > > Now let’s look at p(w)=w*A+(1-w)*B, has expected > return of w*mu1+(1-w)*mu2 and, since correlation > is zero, expected volatility of > sqrt(sigma1^2*w^2+(sigma2)^2*(1-w)^2). Clearly, > efficient frontier is not a line. > > Now let’s assume that B is the risk free asset > (mu2 = RFR, sigma2 = 0). We get > p(w)=w*A+(1-w)*B, has expected return of > w*mu1+(1-w)*mu2 and expected volatility of > sqrt(sigma1^2*w^2+(sigma2)^2*(1-w)^2)=w*sigma1. > In this case efficient frontier is a line: Ret = > (Sigma/simga1)*mu1+(1-Sigma/sigma1)*mu2=Sigma[(mu1 > -mu2)/sigma2]+mu2.

(delete)

Hank Moody Wrote: ------------------------------------------------------- > Can you please explain conceptually, without using > formulas? > Due to diversification, volatility of a portfolio is less than weighted average of volatilities of assets. That leads to non-linear behavior. For example, consider two assets A with mean 5%, sigma 5% and B with mean 10%, sigma 10%. If weight is 50%, mean of the portfolio would be 0.5*5%+0.5*10% = 7.5%, volatility would be sqrt(0.5^2*5%^2+0.5^2*10%^2)=0.5*5%*sqrt(1+4)=5.59% < 7.5% which is the average of volatilities. Efficient Frontier is concave (because vol(w*A+(1-w)*B)

diversification makes the relationship non-linear, that helps for me maratikus.

actually, with the help of Excel I found the answer. In a zero correlation setting, if the standard deviation for one of the assets is zero, the line is straight. If SD is other than zero, the line has a curve. The shape (curved or straight) has nothing to do with diversification because if you diversify with a risk free asset there is no minimum variance portfolio – variance is purely linear. In order to gain a “sweet spot,” both assets have to have at least some volatility. Thank you, FinNinja. It hadn’t quite sunk in when I first read your post.