3
$\begingroup$

In Finding Alphas, Chapter 1, Introduction to Alpha Design, the authors state:

An alpha can be represented as a matrix of securities and positions indexed by time. The value of the matrix corresponds to positions in that particular stock on that particular day. Positions in stock change daily; the daily changes are traded in the securities market. The alpha produces returns, and returns have variability. The ratio of return to standard deviation (variability) of the returns is the information ratio of the alpha. It so happens that the information ratio of the alpha is maximized when alpha stock positions are proportional to the forecasted return of that stock.

My emphasis added. This statement is provided without proof. I feel like there is a Grinold-Kahn style proof of this, but I am unable to find it. How does one prove this claim?

$\endgroup$
1
  • $\begingroup$ The book uses uncommon notions where an alpha is a strategy and the information ratio is the return/volatility (=SR without rf). So the statement is just that when you perfectly predict the movement, the metric goes to infinity. $\endgroup$
    – Martin
    Commented Feb 26, 2023 at 13:39

2 Answers 2

0
$\begingroup$

Disclaimer: I don't have any of these books, and I don't know for sure what the author is trying to say.

It sounds vaguely like Markowitz portfolio theory applied to returns relative to some benchmark instead of relative to the risk free rate?

Refresher on classic Markowitz portfolio theory.

I will use bold letters to denote vectors. $\boldsymbol{R}$ is a random vector denoting returns of risky assets and $r_f$ is the risk free rate.

  • Define $\boldsymbol{\mu}_f = \operatorname{E}[\boldsymbol{R} - r_f]$
  • Define covariance matrix $\boldsymbol{\Sigma}_f = \operatorname{Var}(\boldsymbol{R} - r_f)$

Portfolio weights for the tangency portfolio, the portfolio with the highest Sharpe ratio, are given by:

$$\mathbf{w} = \left( \frac{1}{\boldsymbol{1}' \boldsymbol{\Sigma}_f^{-1} \boldsymbol{\mu}_f}\right)\boldsymbol{\Sigma}_f^{-1} \boldsymbol{\mu}_f$$

For example, see derivation here.

A guess of what that the book is trying to talk about?

I don't have the book, and this is heavily extrapolation based upon that short passage.

  • Let $R_b$ be a random variable denoting the return of some benchmark $b$.
  • The author may be using alpha not in the Jensen's alpha sense (or stochastic discount factor alpha sense) but is calling returns above some benchmark $\boldsymbol{R} - R_b$ alpha?
  • An information ratio $\frac{\operatorname{E}[R_a - R_b]}{\operatorname{Var}(R_a - R_b)}$ is just a Sharpe ratio relative to some benchmark $R_b$ instead of the risk free rate $r_f$.
  • Define $\boldsymbol{\mu}_b = \operatorname{E}[\boldsymbol{R} - R_b]$ and $\boldsymbol{\Sigma}_b = \operatorname{Var}(\boldsymbol{R} - R_b)$. Then portfolio weights for the maximum information ratio portfolio would be the same $\mathbf{w} = \left( \frac{1}{\boldsymbol{1}' \boldsymbol{\Sigma}_b^{-1} \boldsymbol{\mu}_b}\right)\boldsymbol{\Sigma}_b^{-1} \boldsymbol{\mu}_b$.

Portfolio weights aren't proportional to $\boldsymbol{\mu}_b$ though. They aren't a scalar $\lambda$ times $\boldsymbol{\mu}_b$. You apply the linear transformation $\left( \frac{1}{\boldsymbol{1}' \boldsymbol{\Sigma}_b^{-1} \boldsymbol{\mu}_b}\right)\boldsymbol{\Sigma}_b^{-1}$ to $\boldsymbol{\mu}_b$ .

$\endgroup$
0
$\begingroup$

In the Treynor Black Model the weights $w_i$ are proportional to the Alpha (the above market expected performance) divided by the idiosyncratic (or residual) variance of the stock: $w_i=\frac{\alpha_i}{\sigma_i^2}$.

So the claimed result is not true unless: the $\sigma_i^2$ are all the same and the holding time is so short that the "forecasted return" is the same as the alpha (e.g. for 1 day the "expected return on the market" is so small that $\alpha_i=R_i-R_{CAPM}$ may be very close to $R_i$, the expected return on the particular stock i from our model). (Or perhaps by "forecasted return" they meant above normal forecasted return).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.