Topics: Capital asset pricing model, Arithmetic mean, Rate of return Pages: 6 (1317 words) Published: October 31, 2013

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
n j=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 a great improvement. To get√ standard the
deviation down to about one-tenth of the mean would require that 0.05/ n = 0.00133 or about n = 1413 corresponding to data for 117 years!
As such, it would seem imperative to derive simpler models that are not so data intensive, and which capture enough of reality to make them useful.


Factor models

A (linear) factor model assumes that the rate of return of an asset is given by r = a + b1 f1 + · · · + bk fk + e,


where the fj , j = 1, . . . , k, are k ≥ 1 random variables (rvs) called factors, a and the bj are constants and e is a mean zero “error” term rv assumed uncorrelated with the factors, E(e) = 0 and E(efj ) = E(e)E(fj ) = 0, j = 1, . . . k. The factors themselves are allowed to be correlated and are meant to simplify and reduce the amount of randomness required in an analysis of our assets. When k = 1 we call the model a single-factor model 1

This is called the historical blur problem.


and when k ≥ 2 we call it a multi-factor model. We use the notation f j = E(fj ), σfj = 2
var(fj ), σej = var(ej ) throughout our discussion.
The factors are chosen by the modeler and depend upon the type of assets being considered. For example, for stocks, factors might be selected from among the stock market average, dividend yield of the S&P 500’s Composite common stock, a measure of the risk of corporate bonds, interest rate variables, and various macroeconomic factors that capture the state of the economy such as employment rate, monthly growth rate in industrial production, monthly change in inflation rate, monthly growth rate in consumption and in disposable income. (See for example Ericsson and Karlsson, (2003) “Choosing factors in a multi-factor pricing model”, Stockholm School of Economics, http://econpapers.hhs.se/paper/hhshastef/0524.html.)

When there are n risky assets indexed by 1 = 1, 2 . . . , n (stocks say), then there are n equations for the model, one for each asset;

ri = ai + b1,i f1 + · ·...
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