APT Vs CAPM
1
Factor Models
The Markowitz mean-variance framework requires having access to many parameters:
If there are n risky assets, with rates of return ri , i = 1, 2, . . . , n, then we must know
2
all the n means (ri ), n variances (σi ) and n(n − 1)/2 covariances (σij ) for a total of
2n + n(n − 1)/2 parameters. If for example n = 100 we would need 4750 parameters, and if n = 1000 we would need 501, 500 parameters! At best we could try to estimate these, but how? In fact, it is easy to see that trying to estimate the means, for example, to a workable level of accuracy is almost impossible using historical (e.g., past) data over time.1 What happens is that the standard deviation of our estimate is too large (for example larger than the estimate itself), thus rendering the estimate worthless. One can bring the standard deviation down only by increasing the data to go back to (say) over a hundred years!
To see this: If we want to estimate expected rate of return over a typical 1-month period we could take n monthly data points r(1), . . . , r(n) denoting the rate of return over individual months in the past, and then average
1 n r(j). n j=1
(1)
This estimate (assuming independent and identically distributed (iid) returns over months) has a mean and standard deviation given by
√
r, σ/ n respectively, where r and σ denote the true mean and standard deviation over
1-month. If, for example, a stock’s yearly expected rate of return is 16%, then the monthly such rate is r = 16/12 = 1.333% = 0.0133. Moreover, suppose that the monthly standard deviation is σ = 0.05 (e.g., a variance of 0.0025). Using (1) for our estimate with n = 12
√
yields a standard deviation for the estimate as 0.05/ 12 = 0.0144, which is larger than the mean! (e.g., confidence intervals would be worthless.) Using √ = 60 (5 years of data), n helps to lower the standard deviation for the estimate to 0.05/ 60 = 0.00645, which is a little below half of what it was, hardly
Factor Models
The Markowitz mean-variance framework requires having access to many parameters:
If there are n risky assets, with rates of return ri , i = 1, 2, . . . , n, then we must know
2
all the n means (ri ), n variances (σi ) and n(n − 1)/2 covariances (σij ) for a total of
2n + n(n − 1)/2 parameters. If for example n = 100 we would need 4750 parameters, and if n = 1000 we would need 501, 500 parameters! At best we could try to estimate these, but how? In fact, it is easy to see that trying to estimate the means, for example, to a workable level of accuracy is almost impossible using historical (e.g., past) data over time.1 What happens is that the standard deviation of our estimate is too large (for example larger than the estimate itself), thus rendering the estimate worthless. One can bring the standard deviation down only by increasing the data to go back to (say) over a hundred years!
To see this: If we want to estimate expected rate of return over a typical 1-month period we could take n monthly data points r(1), . . . , r(n) denoting the rate of return over individual months in the past, and then average
1 n r(j). n j=1
(1)
This estimate (assuming independent and identically distributed (iid) returns over months) has a mean and standard deviation given by
√
r, σ/ n respectively, where r and σ denote the true mean and standard deviation over
1-month. If, for example, a stock’s yearly expected rate of return is 16%, then the monthly such rate is r = 16/12 = 1.333% = 0.0133. Moreover, suppose that the monthly standard deviation is σ = 0.05 (e.g., a variance of 0.0025). Using (1) for our estimate with n = 12
√
yields a standard deviation for the estimate as 0.05/ 12 = 0.0144, which is larger than the mean! (e.g., confidence intervals would be worthless.) Using √ = 60 (5 years of data), n helps to lower the standard deviation for the estimate to 0.05/ 60 = 0.00645, which is a little below half of what it was, hardly