Sharpe Ratio Aggregation and Annualization

Worked Solution

How to Think About It: This is a practitioner's bread-and-butter question, but the first part has a trap. The instinct is to quote a fixed "range" from $\max_i S_i$ up to the root-sum-of-squares $\sqrt{\sum_i S_i^2}$. That is *not* a universal range: the ten individual Sharpe ratios alone do not pin down the combined Sharpe, because the combination depends on the full correlation structure. With independent alphas you get the familiar quadrature add-up; but with *negatively* correlated alphas, or unconstrained long/short weights, the optimally combined Sharpe can climb *above* $\sqrt{\sum_i S_i^2}$ -- and it grows without bound as the correlation matrix approaches a singular (near-riskless-hedge) configuration. The annualization piece tests whether you understand the i.i.d. assumption hiding behind $\sqrt{252}$.

Quick Estimate: Ten roughly independent models each with $S_i\approx1$ combine to about $\sqrt{10}\approx3.16$ -- the diversification "free lunch." But that $\sqrt{10}$ is the *zero-correlation* value, not a ceiling: introduce a good negative-correlation hedge and the number can be far larger, in principle unbounded.

Approach: Write the optimal combined Sharpe in terms of the Sharpe vector $\mathbf{s}$ and the correlation matrix $R$, identify the special cases the individual SRs do and do not determine, then handle annualization via $\sqrt{T}$ scaling under i.i.d. returns and enumerate why realized performance departs from the naive scaling.

Formal Solution:

(a) "Range" of the combined Sharpe ratio -- the correct statement. For strategies with Sharpe vector $\mathbf{s}=(S_1,\dots,S_n)^{\mathsf T}$ and return *correlation* matrix $R$ (assumed nonsingular), the maximum Sharpe ratio attainable by an optimally weighted (unconstrained) combination is $S_{\text{comb}}^{\star}=\sqrt{\mathbf{s}^{\mathsf T} R^{-1}\mathbf{s}}.$ The individual Sharpe ratios $S_1,\dots,S_n$ alone do not determine a finite universal range -- you also need $R$. The familiar extremes are only *special cases*:

• Uncorrelated alphas ($R=I$): $S_{\text{comb}}^{\star}=\sqrt{\sum_{i=1}^{n}S_i^2}$. If all ten have $S_i=1$, this is $\sqrt{10}\approx3.16$. This is the *zero-correlation* value, not an upper bound.
• Single best, perfectly correlated, long-only: if the alphas are perfectly positively correlated and you are restricted to picking effectively one of them, the best you achieve is $\max_i S_i$ -- no diversification.
• Negative correlation / long-short: off-diagonal negative entries in $R$ make $\mathbf{s}^{\mathsf T}R^{-1}\mathbf{s}$ *exceed* $\sum_i S_i^2$. As $R$ approaches singularity (a near-riskless hedge), $S_{\text{comb}}^{\star}\to\infty$. Concretely, two alphas each with $S=1$ and correlation $\rho=-0.9$ give $S_{\text{comb}}^{\star}=\sqrt{2/(1+\rho)}=\sqrt{20}\approx4.47$, already well above $\sqrt2$.

So the honest answer: the optimum is $\sqrt{\mathbf{s}^{\mathsf T}R^{-1}\mathbf{s}}$; $\sqrt{\sum_i S_i^2}$ is the uncorrelated case and $\max_i S_i$ the perfectly-correlated long-only case, but neither is an unconditional bound -- negative correlation or long/short can exceed $\sqrt{\sum_i S_i^2}$, even unboundedly.

(b) Annualization. $S_{\text{annual}}=S_{\text{daily}}\times\sqrt{252}.$ Over a year the mean return scales linearly with the number of days, $\mu_{\text{annual}}=252\,\mu_{\text{daily}}$, while the standard deviation scales with the square root, $\sigma_{\text{annual}}=\sqrt{252}\,\sigma_{\text{daily}}$ (variances add for independent days). Taking the ratio, $S_{\text{annual}}=\frac{252\,\mu_{\text{daily}}}{\sqrt{252}\,\sigma_{\text{daily}}}=\sqrt{252}\,S_{\text{daily}}.$ The critical assumption is that daily returns are i.i.d. -- independent (no serial correlation) and identically distributed (stable mean and variance). Normality is *not* required.

(c) Why realized performance differs. Any of the following break the i.i.d. assumption or the mean-variance summary the scaling relies on:

1. Serial correlation in returns. Positive autocorrelation (momentum) makes realized annual volatility exceed $\sqrt{252}\,\sigma_{\text{daily}}$, *deflating* the true annualized Sharpe; negative autocorrelation (mean reversion) does the reverse.
2. Time-varying volatility. Vol clustering / GARCH effects mean $\sigma$ is not constant day to day, so a single scaling factor misstates annual risk.
3. Fat tails and skewness. The Sharpe ratio sees only mean and variance; left-skewed, heavy-tailed P&L is understated by the metric.
4. Transaction costs and market impact. Theoretical Sharpe ignores slippage; high-turnover strategies see realized Sharpe eroded.
5. Regime changes. A Sharpe estimated in one regime (e.g. low vol) does not carry into another (e.g. a crisis), so the i.i.d.-based extrapolation fails.

Answer: - (a) The ten individual Sharpe ratios alone do not fix a finite range. The optimally combined Sharpe is $S_{\text{comb}}^{\star}=\sqrt{\mathbf{s}^{\mathsf T}R^{-1}\mathbf{s}}$ (for nonsingular correlation matrix $R$). It equals $\sqrt{\sum_i S_i^2}$ only when the alphas are uncorrelated, and equals $\max_i S_i$ in the perfectly-correlated long-only case; negative correlations or unconstrained long/short can push it *above* $\sqrt{\sum_i S_i^2}$, growing without bound as $R$ nears singularity. - (b) $S_{\text{annual}}=\sqrt{252}\,S_{\text{daily}}$, valid under i.i.d. (no serial correlation, stable distribution). - (c) Realized performance deviates due to serial correlation, time-varying volatility, fat tails/skew, transaction costs/slippage, and regime changes.