When a New Asset Improves a Portfolio’s Sharpe Ratio
Summary
The question asks why an asset should be added when its Sharpe ratio exceeds the existing portfolio’s Sharpe ratio multiplied by the asset’s correlation with that portfolio. One answer expresses portfolio Sharpe as a weighted combination of asset Sharpe ratios, with weights reflecting each asset’s contribution to portfolio risk. It then considers a Sharpe-optimal portfolio and the effect of increasing the allocation to a candidate asset. If the marginal change is positive, the candidate improves the portfolio’s Sharpe ratio, yielding the stated inclusion condition.
A second answer reaches the condition through a small-allocation approximation: compare the incremental change in expected return with the corresponding change in volatility, using a first-order expansion. The treatment is a local criterion for adding an asset, rather than a full portfolio construction procedure. The differential argument relies on small changes, while the first derivation assumes an already Sharpe-optimal portfolio and uses a particular weight representation. The cited discussion points to a more detailed derivation elsewhere.
Key ideas
- An asset’s Sharpe ratio is compared with the portfolio Sharpe ratio scaled by its correlation with the portfolio.
- The inclusion condition follows from whether a small allocation raises the portfolio’s Sharpe ratio.
- A marginal-return and marginal-volatility comparison provides an approximate derivation.
- The rule is a local test and does not by itself determine the full optimal portfolio weights.
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Full text
# Question about adding new investment A to portfolio B
# Question about adding new investment A to portfolio B
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
## Answer by oronimbus (score 6)
https://quant.stackexchange.com/a/76391
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:
$$-S_{P^*}+\frac{1}{\rho_{n, P}}S_n \geq 0 \longleftrightarrow \boxed{S_n \geq \rho_{n, P} S_{P^*}}$$
## Answer by Arshdeep (score 4)
https://quant.stackexchange.com/a/76390
$SR=r/vol$ 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 $e^2=0$)Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.