Proof that Sharpe ratio of the benchmark is related to the maximal information ratio and Sharpe ratio

Question

I understand the economic logic behind it, that the active portfolio with the highest information ratio will also have the highest Sharpe ratio, but I can't see how $SR_B^2 = SR_P^2 - IR^2 $

Answer 1

This has been shown in Grinold & Kahn (1999), Active Portfolio Management (p. 137ff). First, write $SR_P^2=SR_B^2+IR^2$ as $\left(\frac{f_Q}{\sigma_Q}\right)^2 = \left(\frac{f_B}{\sigma_B}\right)^2 + IR^2$. Also note that the maximum information ratio is related to the portfolio's $Q$ Sharpe as follows: $IR=\frac{\alpha_Q}{\omega_Q}=SR\cdot \frac{\omega_Q}{\sigma_Q}$ where $\omega_Q$ is the residual risk. It is defined as $\omega_Q=\sqrt{\sigma_Q^2-\beta^2_Q \sigma^2_B}$ (see p. 50) where $\beta_Q=\frac{Cov[r_{Q},r_{B}]}{\sigma^2_B}$ is the beta of portfolio $Q$ and benchmark $B$.

Then, it follows that:

$$ \begin{align*} \left(\frac{f_Q}{\sigma_Q}\right)^2 &= \left(\frac{f_B}{\sigma_B}\right)^2 + IR^2 \\ &= \left(\frac{f_B}{\sigma_B}\right)^2 + \left(\frac{f_Q}{\sigma_Q} \right)^2 \left(\frac{\omega_Q}{\sigma_Q} \right)^2 \\ &= \frac{f^2_B}{\sigma^2_B} + \frac{f^2_Q}{\sigma^2_Q} \cdot \frac{\sigma_Q^2-\beta^2_Q \sigma^2_B}{\sigma^2_Q} \\ &= \frac{f^2_B}{\sigma^2_B} + \frac{f^2_Q}{\sigma^2_Q} - \beta^2_B \frac{f^2_Q\sigma^2_B}{\sigma^4_Q} \\ &= \frac{f^2_B}{\sigma^2_B} + \frac{f^2_Q}{\sigma^2_Q} - \left(\frac{f_B\sigma^2_Q}{f_Q\sigma^2_B} \right)^2 \frac{f^2_Q\sigma^2_B}{\sigma^4_Q} \\ &= \frac{f^2_B}{\sigma^2_B} + \frac{f^2_Q}{\sigma^2_Q} - \frac{f^2_B}{\sigma^2_B} \\ &= \frac{f^2_Q}{\sigma^2_Q} \end{align*} $$

Note that the third last step can be solved using the statement about portfolio $Q$'s holdings (which is a mix of benchmark $B$ and managed portfolio $A$) on page 136.