CAPM - Beta of zero and its implications on diversification

Question

I don't know if this is the right forum in which to ask this question, but here goes. I'm working through Luenberger's Investment Science. The form of CAPM model given in the book is

$$\bar{r}_i - r_f = \beta_i (\bar{r}_M - r_f)$$

It says that if your asset has a $\beta$ of zero, then $\bar{r}_i = r_f$. The book says that "The reason for this [$\bar{r}_i = r_f$] is that the risk associated with an asset that is uncorrelated with the market can be diversified away. If we had many such assets, each uncorrelated with the others and with the market, we could purchase small amounts of them, and the resulting total variance would be small."

I agree with Luenberger's statement, but what bothers me is the part "if we had many such assets". If we don't, $\bar{r}_i = r_f$ still holds! What would be the explanation then? Is there some theory behind CAPM that would say a statement like the following: "if there exists one asset $i$ with $\beta_i = 0$, then there exist infinitely many assets with $\beta = 0$, that are uncorrelated with each other". This is a strong and sort of ridiculous statement, but it would imply the description quoted from Luenberger - I could make a portfolio with such assets whose variance approaches zero.

Any help wrapping my head around this concept would be greatly appreciated.

Answer 1

There is a rationale here and it has to do with the connection between diversification and the number of assets in a portfolio.

Suppose we purchase an equal-weighted portfolio of n stocks. The variance of the return is then:

$\sigma_{p}^2$ = $\sum$ $\sum$ $w_i$$w_j$Cov($R_i$,$R_j$)

In the above notation, the sigmas are summing over $i$ and $j$ respectively each element of the variance-covariance matrix of asset returns.

Since each weight is 1/n, then this equation is equal to:

1/$n^2$ * $\sum$ $\sum$ Cov($R_i$,$R_j$).

Let's say the average variance across all stocks is $\sigma_{avg}^2$ and the average covariance between among all pairs of stocks is $Cov_{avg}$.

Then this equation simplifies to the sum of two products (see Investments, Bodie, Kane, and Marcus for the derivation):

$\sigma_{p}^2$ = $(1/n)$ * $\sigma_{avg}^2$ + $[(n-1)/n]$ * $Cov_{avg}$.

Let's analyze the first product. As the number of stocks increases the contribution of the variance of the individual stocks becomes very small because the term $(1/n)$ * $\sigma_{avg}^2$ approaches zero as $n$ becomes large.

Now let's analyze the second product. As the contribution of the average covariance across strocks to the portfolio variance stays non-zero because the term $[(n-1)/n] $ * $Cov_{avg}$ has a limit of $Cov_{avg}$.

So as the number of assets increases the portfolio variance is approximately equal to the average covariance.

In your above example, if there are many such assets then the contribution to variance is vary small. Also, if the assets are "uncorrelated with others and the market" then the second term is also small. Therefore the variance is nearly zero when both the conditions identified by Luenberger are met.

Here is a more intuitive explanation. A Beta of 0 does not imply zero variance, securities still have idiosyncratic risk (i.e. a random component of return not explained by systematic exposure). A risk-free investment is still less risky than a security with a beta exposure of zero although they both have the same expected return. However, investing in a large portfolio with a beta exposure of zero is far less riskier than investing in a single security with a beta exposure of zero.

Adding the idiosyncratic error term 'e' to the formula you identify captures this idea of diversifying away residual risk. 'e' is a random variable whose expectation is 0 but variance is greater than zero. Only as one builds larger portfolios is this 'e' term increasingly diversified away (i.e. the variance of the error term approaches zero):

$$\bar{r}_i - r_f = \beta_i (\bar{r}_M - r_f) + e $$

Put another way, if the number of securities is large (and weights are small), the covariances become the most important driver of portfolio variance. Consider the variance-covariance matrix of the asset returns. For an 'n' asset portfolio there are 'n' variances (the diagonal), and n*(n-1)/2 covariances (the off-diagonal triangle). A portfolio of two stocks has 2 variances and 1 covariance (i.e. variances dominate). A large portfolio of 1,000 stocks has 1,000 variance terms but 499,950 co-variance terms (i.e. covariance dominates).

Answer 2

In the CAPM, $E(r_i) = r_f$, if $\beta_i=0$ holds ALWAYS.

BUT: recall the calculation of $\beta$:

$\beta_i=\cfrac{\sigma_{iM}}{\sigma_M^2}$

and

${\sigma_{iM}}=\Sigma_jw_j\sigma_{ij}$

if your asset is uncorrelated with all other assets in the market, the last expression simplifies to:

${\sigma_{iM}}=\Sigma_jw_j\sigma_{ij}=w_i\sigma_i^2$

so for $\beta_i$ to be 0, you need the weight $w_i$ to be 0, which implies the asset is not in the market portfolio, i.e. it does not exist...

But obviously, the smaller the weight of the asset, the closer $\beta_i$ comes to zero.

So in the case mentioned in the book, it's like you divide a million $ into an ever increasing number of positions, yes, you get close to that. There would be other cases, e.g. 2 uncorrelated stocks in the market portfolio, one with 99.9% weight, and the other will have a beta close to zero... actually, the rest of the market portfolio does not matter at all for an asset that is not correlated with anything - it's only its weight in the market portfolio and its variance that determines its beta.

P.S.: Of course with positive correlation with other assets, beta would be larger. With negative correlation with other assets, beta might approach zero faster when reducing the weight, but as long as there is no asset or subportfolio perfectly negatively correlated with $i$, the variance will not disappear completely for a non-zero weight. And finally, if there is a perfect hedge for the asset, then we have a case where it makes sense to remind oneself that any point on the efficient frontier could potentially be representing more than one portfolio...(recall short sales allowed)

Answer 3

A possible way is to think at the Single Index Model (SIM). In this case if an asset $i$ is uncorrelated with the market, then it is uncorrelated with each other. Thus the condition ".. each uncorrelated with the others and with the market ..." boil down in the second request. Remember that in the SIM $σ_i,j=β_iβ_jσ^2_m$. In this framework, if the SIM hold, we have $r_it - r_f = a_i + β_i(r_mt- r_f) + e_it$ with $a_i = 0$ for each $i$ and $β_i>0$ for most of $i$. Exactly for the last condition the market is (the only) common risk factor. Thus we can have a few stock with $β_i=0$ while it is impossible to have "... many such assets". In effect if we have many asset with $β_i=0$ simply the SIM tend to vanish and in end the SIM not hold. Or $r_it - r_f = a_i + e_it$ thus $E[r_it] = a_i + r_f$ but in equilibrium $a_i =0$ and effectively if we take in ptf so many asset we tend to achieve, almost surely, the $rf$ rate. However in the real word the $σ_(avg)$, in term of suggested by Ram Ahluwalia, never tend to zero but, in well diversified ptf, tend to a positive, and not too small, value. For this reason common risk factors exist ... probably in more complicated form than SIM. Or, in other word, the diversification cannot completely eliminate the risk. Or, again, many such [ $β_i=0$]assets can not exist. The case in which $β_i<0$ is, a bit more tricky, but similar.