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http://www.columbia.edu/~gmg2/4706-00/pdf/lect_05nn.pdf

In the above paper there is that given the equation of the single index model,

Let $x_m$ denote the market index portfolio we have $\bar R_m=α_m-β_m \bar R _m$

This equations holds for all values of $\bar R_m$ only if $α_m=0$ and $β_m=1$.

Then $R_m=R_m+∑_{i=1}^{n}x_{mi} e_i $

So $∑_{i=1}^{n}x_{mi} e_i =0$

The paper states that "But this last statement contradicts the assumption that the $e$’s are uncorrelated."

Could anyone explain how this holds?

Anna
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1 Answers1

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The idea is that for a linear combination of random quantities to balance out to zero, positive fluctuations in some of those quantities have to be canceled (on average) by negative fluctuations in some of the others: and that is precisely what it means to be correlated.

The following algebraic demonstration brings rigor to this intuition.

Presumably the $x_{mi}$ are constants (not all of which are zero) and the $e_i$ are non-constant random variables, entailing that they all have positive variance.

Supposing -- in order to test the conclusion about correlation -- that the $e_i$ are uncorrelated, which means nothing more than $\operatorname{Cov}(e_i,e_j)=0$ whenever $i\ne j,$ compute that

$$\begin{aligned}0&=\operatorname{Var}(0)=\operatorname{Var}\left(\sum_{i=1}^n x_{mi}e_i\right) \\ &= \sum_{i=1}^n x_{mi}^2\operatorname{Var}(e_i) + 2\sum_{i\lt j}^n x_{mi}x_{mj}\operatorname{Cov}(e_i,e_j)\\ &= \sum_{i=1}^n x_{mi}^2\operatorname{Var}(e_i). \end{aligned}$$

Because every term in this last expression is non-negative and at least one of them is nonzero, we obtain the contradiction $0 \lt 0.$ This forces us to conclude at least one pair of the $e_i,e_j$ are correlated. Indeed, when the coefficients $x_{mi}$ are also nonnegative (which is usually the case for portfolio compositions), this shows that at least one pair has negative correlation.

whuber
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