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I've found a ton of sources that mention the classic rule of

"If the Sharpe ratio of the new asset is greater than the Sharpe ratio of the existing portfolio times the correlation of the existing portfolio with the new asset, then you should include it."

My question is WHY?... I understand the logic behind it but I haven't been able to find any mathematical proof of this.

I tried doing it with fundamental formulas... SR(Before) < SR(After) but I couldn't do anything relevant

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The proposition is intuitive but the proof of this is not so straight forward in my opinion. The paper Benhamou & Guez (2021), Computation of the marginal contribution of Sharpe ratio and other performance ratios gives a more detailed derivation of the problem (see Appendix 0.3). I believe the inequality is part of the CFA course material so I doubt that this is the first ever reference (but simply the one I know of).

The Portfolio Sharpe $S_P$ ratio is a combination of asset Sharpe Ratio's $S_i$ weighted by the inverse of asset correlation to the portfolio $P$. Recall that the numerator of the Sharpe ratio is just the sum of asset weights times returns.

$$S_P=\sum_{i=1}^nw_i\frac{r_i}{\sigma_P}=\sum^n_{i=1}\frac{w_i\rho_{i, P}\sigma_i}{\sigma_P}\frac{1}{\rho_{i,P}}\frac{r_i}{\sigma_i}=\sum_{i=1}^n\theta_i\frac{1}{\rho_{i, P}} S_i$$

Here $\theta_i$ acts as a weight factor. Assuming your goal is to maximize the portfolio Sharpe Ratio, then we can write the following optimization routine for the SR optimal portfolio $P^*$ plus new asset $n$:

$$\text{maximize} \ \ (1-\theta_n)S_{P^*} + \theta_n \frac{1}{\rho_{n, P}} S_n \ \ \ s.t. 0\leq\theta_n \leq 1$$

The optimal solution for $\theta_n$ is not equal to zero if the slope is positive, i.e. we take the first order derivative and we recover:

$$-S_{P^*}+\frac{1}{\rho_{n, P}}S_n \geq 0 \longleftrightarrow \boxed{S_n \geq \rho_{n, P} S_{P^*}}$$

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$SR=r/vol$ so $d(SR)=dr/vol - r/vol^2 * dvol = 0$ so that $dr/r=dvol/vol$. This is the 'indifference condition'.

Let's say you add 'e~0' amount of the asset resulting in

$dr=e*(r_{new}-r_p)$ and

$d(var)=-2*e*Var(r_p)+2*e*rho*Vol(r_{new})*Vol(r_p)$

Also you have $d(vol)/vol=1/2*d(Var)/Var(r_p) $

So the indifference condition is: $(r_{new}-r_p)/r_p=0.5*[-2+2*rho*Vol(r_{new})/Vol(r_p)]$

which on simplification is your desired condition.

I've only used differential approximations/ Taylor expansions everywhere along with definitions of variance and volatility. (like $e^2=0$)

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    $\begingroup$ this appears to be a popular CFA / Chegg question: chegg.com/homework-help/questions-and-answers/… , chegg.com/homework-help/questions-and-answers/… ... :) $\endgroup$ Commented Aug 15, 2023 at 10:44
  • $\begingroup$ Okay, I don't understand what to make of that. I'm not involved in either of those circles. $\endgroup$
    – Arshdeep
    Commented Aug 15, 2023 at 11:35
  • $\begingroup$ In the last equation, where does the $\frac{1}{Var(r_p)}$ in the second summand go? $\endgroup$ Commented Aug 15, 2023 at 12:44
  • $\begingroup$ @Hans-PeterSchrei You divide the $d(var)$ equation with it, and see that var is square of vol. $\endgroup$
    – Arshdeep
    Commented Aug 15, 2023 at 12:45
  • $\begingroup$ Ah, missed the division, thanks for responding so quickly. $\endgroup$ Commented Aug 15, 2023 at 12:47

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