Unit 13, lesson 103

Signal evaluation and performance attribution

How the information coefficient, a strategy return and the Sharpe ratio differ, how serial dependence enters Sharpe ratio aggregation over several periods, and how factor attribution splits a mean return.

  • Video 103
  • Signals, costs and capacity

A backtest reports a monthly Sharpe ratio of 0.10.1, and its summary multiplies by 12\sqrt{12} to quote an annual figure of about 0.350.35. Whether that scaling is right depends on how the monthly returns covary, and the strategy’s mean return still has to be split into the part its factor exposures explain and the part they leave unexplained. This lesson separates three objects that are easily blurred, the correlation between a signal and its later target, the return of a fully specified trading rule, and the attribution of that return to a declared factor model. It derives the exact variance of a sum of covariance-stationary returns, uses it to aggregate the Sharpe ratio, and states what the sampling uncertainty of an estimated Sharpe ratio depends on.

Key terms

Information coefficient

Definition Fix a horizon hh and a declared collection of signal and target pairs, such as all assets of a universe at one date or the successive dates of one asset. Each signal ss is issued at a date tt using only the information Ft\mathcal{F}_t available then, and its target YY, typically the excess return over (t,t+h](t, t+h], is realized afterwards. For random variables ss and YY with finite, strictly positive variances, the information coefficient is the Pearson correlation

IC⁡=Corr⁡(s,Y)=Cov⁡(s,Y)Var⁡(s)Var⁡(Y).\begin{aligned} \operatorname{IC} &= \operatorname{Corr}(s, Y) \\[0.4em] &= \frac{\operatorname{Cov}(s, Y)}{\sqrt{\operatorname{Var}(s)\operatorname{Var}(Y)}} . \end{aligned}

From nn observed pairs (si,yi)(s_i, y_i) the sample information coefficient is

IC⁡^=SsySss Syy,Ssy=∑i=1n(si−sˉ)(yi−yˉ),\begin{aligned} \widehat{\operatorname{IC}} &= \frac{S_{sy}}{\sqrt{S_{ss}\, S_{yy}}}, \\[0.4em] S_{sy} &= \sum_{i=1}^{n} (s_i - \bar s)(y_i - \bar y), \end{aligned}

where SssS_{ss} and SyyS_{yy} are the corresponding sums of squared deviations. It is defined when neither the signals nor the targets are all equal, and the divisor of the sample covariance cancels. By the Cauchy-Schwarz inequality ∣IC⁡∣≤1\lvert \operatorname{IC} \rvert \le 1, with IC⁡=1\operatorname{IC} = 1 exactly when Y=a+bsY = a + bs for constants aa and b>0b > 0, almost surely. For constants a>0a > 0 and c>0c > 0 and any bb and dd, the correlation of as+bas + b and cY+dcY + d equals Corr⁡(s,Y)\operatorname{Corr}(s, Y), so the coefficient has no units and ignores the scale of both variables. The rank information coefficient is the Pearson correlation of the ranks, Spearman’s correlation, which measures monotone rather than linear association and is a different statistic. A signal built from data published after its issue date measures look-ahead, not prediction.

In plain words The information coefficient asks whether higher signals tend to be followed by higher targets, on a scale from −1-1 to 11 that ignores units. It belongs to a declared set of pairs and a horizon, and it says nothing yet about positions, sizes or costs.

Example In a constructed sample, the signals −1-1, 00 and 11 are paired with the targets −2-2, 00 and 22. Both means are 00, the sum of products is 44, and the sums of squares are 22 and 88, so the sample coefficient is 4/16=14/\sqrt{16} = 1. A second constructed sample shows how the two coefficients can disagree. With the signals 11, 22, 33, 44 and the next-period returns 0.010.01, 00, −0.01-0.01, 0.050.05, the Pearson coefficient is

0.0555⋅0.002075≈0.540,\frac{0.055}{\sqrt{5 \cdot 0.002075}} \approx 0.540,

while the returns have the ranks 33, 22, 11, 44 and the rank coefficient is 1/5=0.21/5 = 0.2. The Pearson value rests mostly on the one large return of the asset with the highest signal.

Strategy return

Definition A strategy return needs a trading rule that turns the information Ft\mathcal{F}_t into holdings. Over the period (t,t+1](t, t+1], let the rule hold the weights wi,tw_{i,t} in the assets i=1,…,ni = 1, \dots, n, chosen with the information Ft\mathcal{F}_t and measured as fractions of a declared capital At>0A_t > 0, and keep the rest of the capital in the money-market account, which earns a simple rate rtfr^{f}_{t} over the period that is known at tt. Assume the positions are held unchanged over the period, short sale proceeds earn the same rate, and there are no trading costs or financing spreads. If asset ii has the simple return ri,t+1r_{i,t+1}, the end-of-period wealth is At(1+Rt+1p)A_t(1 + R^{p}_{t+1}) with

Rt+1p=rtf+∑i=1nwi,t (ri,t+1−rtf).\begin{aligned} &R^{p}_{t+1} = r^{f}_{t} \\[0.4em] &\quad + \sum_{i=1}^{n} w_{i,t}\,\bigl(r_{i,t+1} - r^{f}_{t}\bigr). \end{aligned}

The strategy’s excess return is therefore Rt+1=∑iwi,t ri,t+1eR_{t+1} = \sum_i w_{i,t}\, r^{e}_{i,t+1}, the weighted sum of the asset excess returns ri,t+1e=ri,t+1−rtfr^{e}_{i,t+1} = r_{i,t+1} - r^{f}_{t}. This is a gross return. The net return also subtracts the cost of moving from one set of weights to the next. Over several periods the additive excess return ∑tRt\sum_t R_t and the compounded wealth return ∏t(1+Rtp)−1\prod_t (1 + R^{p}_t) - 1 are different quantities.

In plain words A signal earns money only through a rule that says what to hold and how much. The strategy return is the return of those holdings on a stated capital, and two rules built on the same signal can have very different returns.

Example Turn the first sample of the previous card into the rule with weights w=(−0.5,0,0.5)w = (-0.5, 0, 0.5), short the asset with the lowest signal and long the asset with the highest. The weights add to 00, so the whole capital stays in the money-market account and the excess return is ∑iwirie\sum_i w_i r^{e}_i. If the targets −2-2, 00 and 22 are excess returns in percent, the strategy’s excess return is (−0.5)(−0.02)+0.5⋅0.02=0.02(-0.5)(-0.02) + 0.5 \cdot 0.02 = 0.02. If they are −0.2-0.2, 00 and 0.20.2 percent, the information coefficient is still 11 but the excess return is 0.0020.002, ten times smaller, and no cost has been charged yet. Additive and compounded returns also differ. Two periods of 1010 percent and −10-10 percent add to 00, while wealth changes by the factor 1.1⋅0.9=0.991.1 \cdot 0.9 = 0.99, a compounded return of −1-1 percent.

Sharpe ratio

Definition Let RR be the excess return of a strategy over a stated period, such as a month, with E[R2]<∞\mathbb{E}[R^2] < \infty and Var⁡(R)>0\operatorname{Var}(R) > 0, where expectations are taken under the physical measure P\mathbb{P} of the return model. Its Sharpe ratio for that period is

SR⁡=E[R]Var⁡(R)=μσ.\operatorname{SR} = \frac{\mathbb{E}[R]}{\sqrt{\operatorname{Var}(R)}} = \frac{\mu}{\sigma} .

It is dimensionless, depends on the length of the period, and does not change when every position is scaled by a constant c>0c > 0, because the excess return becomes cRcR. From observations R1,…,RnR_1, \dots, R_n the estimate is SR⁡^=Rˉ/σ^\widehat{\operatorname{SR}} = \bar R / \hat\sigma, with the sample mean Rˉ\bar R and the sample standard deviation σ^\hat\sigma, and it has a sampling distribution. The following result is stated here, not derived. If R1,…,RnR_1, \dots, R_n are independent and identically distributed with E[R4]<∞\mathbb{E}[R^4] < \infty and σ>0\sigma > 0, a joint central limit theorem for (Rˉ,σ^2)(\bar R, \hat\sigma^2) and the delta method give, in distribution as n→∞n \to \infty,

n (SR⁡^−SR⁡)⟶N(0,v),v=1−κ3SR⁡+14(κ4−1)SR⁡2,\begin{aligned} &\sqrt{n}\,\bigl(\widehat{\operatorname{SR}} - \operatorname{SR}\bigr) \\[0.4em] &\quad \longrightarrow \mathcal{N}(0, v), \\[0.4em] &v = 1 - \kappa_3 \operatorname{SR} \\[0.4em] &\quad + \tfrac{1}{4}(\kappa_4 - 1)\operatorname{SR}^2, \end{aligned}

where κ3=E[(R−μ)3]/σ3\kappa_3 = \mathbb{E}[(R - \mu)^3]/\sigma^3 is the skewness and κ4=E[(R−μ)4]/σ4\kappa_4 = \mathbb{E}[(R - \mu)^4]/\sigma^4 the kurtosis. For normal returns κ3=0\kappa_3 = 0 and κ4=3\kappa_4 = 3, so v=1+SR⁡2/2v = 1 + \operatorname{SR}^2/2. The condition σ>0\sigma > 0 is needed because the ratio μ/σ\mu/\sigma is not smooth at σ=0\sigma = 0. With serial dependence the variance vv involves long-run covariances of the moment terms and needs suitable dependence conditions. When many strategies are compared and the best one is reported, its estimate is biased upward, and these formulas for a single preselected strategy do not account for the search.

In plain words The Sharpe ratio is the mean excess return per unit of standard deviation over one period. Doubling every position doubles the mean and the standard deviation of the excess return and leaves the ratio unchanged. An estimated Sharpe ratio is a noisy number, and the decimals a program prints say nothing about how noisy it is.

Example A constructed monthly excess return with μ=0.01\mu = 0.01 and σ=0.1\sigma = 0.1 has SR⁡=0.1\operatorname{SR} = 0.1. With n=120n = 120 independent normal months, the asymptotic standard error of the estimate is 1.005/120≈0.0915\sqrt{1.005/120} \approx 0.0915, and the approximate 95 percent interval

0.1±1.96⋅0.0915≈[−0.079, 0.279]\begin{aligned} &0.1 \pm 1.96 \cdot 0.0915 \\[0.4em] &\quad \approx [-0.079, \, 0.279] \end{aligned}

contains zero. In a simulation of 200000200000 such samples the estimates had standard deviation about 0.0920.092. In a second simulation, twenty independent strategies with true Sharpe ratio 00 were each tested on 120120 normal months and the best estimate was kept. Its average was about 0.170.17, well above the true 0.10.1 of the first strategy.

Serial dependence

Definition A return series (Rt)(R_t) has serial dependence when returns at different dates are dependent. Suppose E[Rt2]<∞\mathbb{E}[R_t^2] < \infty for every tt and the series is covariance stationary, with E[Rt]=μ\mathbb{E}[R_t] = \mu and

γk=Cov⁡(Rt,Rt+k)\gamma_k = \operatorname{Cov}(R_t, R_{t+k})

the same for every tt and every integer kk. Then γ−k=γk\gamma_{-k} = \gamma_k, the lag-zero value γ0\gamma_0 is the one-period variance, and when γ0>0\gamma_0 > 0 the lag-kk autocorrelation is ρk=γk/γ0\rho_k = \gamma_k/\gamma_0. The part of the dependence that enters the variance of a sum is serial correlation, a nonzero γk\gamma_k at some lag k≥1k \ge 1. Uncorrelated returns can still be dependent, for example in a GARCH model with mean-zero shocks, whose volatility clustering leaves every γk\gamma_k with k≠0k \ne 0 at zero but changes the sampling distribution of estimates. Not every sequence of numbers is an autocovariance function. For every qq, the q×qq \times q matrix with entries γ∣i−j∣\gamma_{\lvert i - j \rvert} is the covariance matrix of (R1,…,Rq)(R_1, \dots, R_q) and must be positive semidefinite. The family γk=σ2ρ∣k∣\gamma_k = \sigma^2 \rho^{\lvert k \rvert} with ∣ρ∣<1\lvert \rho \rvert < 1 is valid, since it is the autocovariance of the stationary AR(1) process Rt−μ=ρ(Rt−1−μ)+etR_t - \mu = \rho(R_{t-1} - \mu) + e_t driven by white noise of variance σ2(1−ρ2)\sigma^2(1 - \rho^2). If ∑k∣γk∣<∞\sum_k \lvert \gamma_k \rvert < \infty, the long-run variance is

σLR2=∑k=−∞∞γk=γ0+2∑k=1∞γk,\begin{aligned} \sigma^2_{\mathrm{LR}} &= \sum_{k=-\infty}^{\infty} \gamma_k \\[0.4em] &= \gamma_0 + 2\sum_{k=1}^{\infty} \gamma_k , \end{aligned}

and the identity of the next card shows that Var⁡(R1+⋯+Rq)/q→σLR2\operatorname{Var}(R_1 + \dots + R_q)/q \to \sigma^2_{\mathrm{LR}} as q→∞q \to \infty.

In plain words Positive serial correlation means that periods with above-average returns tend to be followed by more of them, so a sum of returns spreads out more than it would for uncorrelated periods. Negative serial correlation makes nearby periods partly offset each other.

Example Overlapping returns create serial correlation mechanically, as in Time series validation and overlapping labels. Let monthly returns be uncorrelated with constant variance σ2\sigma^2, and record each month the return over the last three months. Consecutive three-month returns share two months and returns two months apart share one, so γ0=3σ2\gamma_0 = 3\sigma^2, γ1=2σ2\gamma_1 = 2\sigma^2, γ2=σ2\gamma_2 = \sigma^2 and γk=0\gamma_k = 0 for k≥3k \ge 3, giving ρ1=2/3\rho_1 = 2/3 and ρ2=1/3\rho_2 = 1/3. For the constructed AR(1) series of the next card, σ2=0.01\sigma^2 = 0.01 and ρ=0.5\rho = 0.5 give γ1=0.005\gamma_1 = 0.005, γ2=0.0025\gamma_2 = 0.0025 and the long-run variance 0.01⋅1.5/0.5=0.030.01 \cdot 1.5/0.5 = 0.03, three times the one-period variance.

Sharpe ratio aggregation

Definition Let (Rt)(R_t) be a covariance-stationary excess-return series with mean μ\mu and autocovariances γk\gamma_k, as in the previous card, and let the qq-period return be the additive sum Sq=R1+⋯+RqS_q = R_1 + \dots + R_q. Then E[Sq]=qμ\mathbb{E}[S_q] = q\mu and

Vq=Var⁡(Sq)=qγ0+2∑k=1q−1(q−k)γk.\begin{aligned} V_q &= \operatorname{Var}(S_q) \\[0.4em] &= q\gamma_0 \\[0.4em] &\quad + 2\sum_{k=1}^{q-1} (q - k)\gamma_k . \end{aligned}

The proof expands the variance of the sum into the q2q^2 covariances Cov⁡(Rs,Rt)\operatorname{Cov}(R_s, R_t), each equal to γ∣s−t∣\gamma_{\lvert s - t \rvert}. The qq diagonal terms give qγ0q\gamma_0. A lag k≥1k \ge 1 occurs for the q−kq - k pairs of dates (s,s+k)(s, s + k) inside the window and once more in reverse order, which gives 2(q−k)γk2(q - k)\gamma_k. Only finite second moments and covariance stationarity are used, not independence. When Vq>0V_q > 0, the Sharpe ratio of the additive qq-period aggregate is

SR⁡q=qμVq.\operatorname{SR}_q = \frac{q\mu}{\sqrt{V_q}} .

If γk=0\gamma_k = 0 for 1≤k≤q−11 \le k \le q - 1, then Vq=qγ0V_q = q\gamma_0 and SR⁡q=q SR⁡\operatorname{SR}_q = \sqrt{q}\,\operatorname{SR}. This square-root rule, used for example to annualize a monthly ratio with 12\sqrt{12} or a daily one with 252\sqrt{252}, is therefore exact for uncorrelated returns with constant mean and variance, in particular for independent, identically distributed returns, and it fails whenever the weighted covariance sum ∑k=1q−1(q−k)γk\sum_{k=1}^{q-1}(q - k)\gamma_k is not zero. With μ>0\mu > 0, a positive weighted sum pushes the ratio below q SR⁡\sqrt{q}\,\operatorname{SR} and a negative one pushes it above. With μ<0\mu < 0, a positive weighted sum moves the ratio toward zero. The variance VqV_q can be zero, for example when Rt=μ+(−1)tZR_t = \mu + (-1)^t Z for a random variable ZZ with mean 00 and positive variance and qq is even, and then the ratio is undefined. The identity is exact for sums of returns, and for log returns the sum is exactly the multi-period log return. It is not an exact formula for the compounded simple return ∏t(1+Rt)−1\prod_t (1 + R_t) - 1, and it is a population identity that says nothing about the sampling error of an estimated ratio.

In plain words Adding qq periods multiplies the mean by qq, but the variance grows by the factor qq only when the periods are uncorrelated. Every covariance between two dates inside the window adds to or subtracts from the variance, and the lag kk appears q−kq - k times in each direction.

Example Take constructed AR(1) excess returns with μ=0.01\mu = 0.01, γ0=0.01\gamma_0 = 0.01 and γk=0.01⋅0.5k\gamma_k = 0.01 \cdot 0.5^{k}, so the one-period Sharpe ratio is 0.10.1. For q=12q = 12 the weighted correlation terms (12−k) 0.5k(12 - k)\,0.5^{k} are 11⋅0.5=5.511 \cdot 0.5 = 5.5, then 10⋅0.25=2.510 \cdot 0.25 = 2.5, and so on down to 1⋅0.5111 \cdot 0.5^{11}, and they add to 10+2−1110 + 2^{-11}. The variance multiplier V12/γ0V_{12}/\gamma_0 is

12+2(10+2−11)=32+2−10≈32.000977,\begin{aligned} &12 + 2\bigl(10 + 2^{-11}\bigr) \\[0.4em] &\quad = 32 + 2^{-10} \\[0.4em] &\quad \approx 32.000977, \end{aligned}

so V12≈0.3200098V_{12} \approx 0.3200098 with square root about 0.5656940.565694, and

SR⁡12=0.120.565694≈0.212129.\begin{aligned} \operatorname{SR}_{12} &= \frac{0.12}{0.565694} \\[0.4em] &\approx 0.212129 . \end{aligned}

The square-root rule would give 12⋅0.1≈0.346410\sqrt{12} \cdot 0.1 \approx 0.346410, about 6363 percent too high. Setting ρ=0\rho = 0 with the same mean and variance removes every covariance term, brings the multiplier back to 1212 and restores 0.3464100.346410. As qq grows, the ratio SR⁡q/(q SR⁡)\operatorname{SR}_q / (\sqrt{q}\,\operatorname{SR}) tends to (1−ρ)/(1+ρ)=1/3≈0.577\sqrt{(1 - \rho)/(1 + \rho)} = 1/\sqrt{3} \approx 0.577 for this series.

Factor attribution

Definition A factor attribution decomposes a return relative to a declared linear factor model

Rt=α+β⊤ft+εt,R_t = \alpha + \beta^{\top} f_t + \varepsilon_t ,

where RtR_t is the strategy’s excess return, ft∈Rmf_t \in \mathbb{R}^m the vector of factor returns over the same period, β∈Rm\beta \in \mathbb{R}^m a fixed vector of loadings, α\alpha a constant intercept and εt\varepsilon_t a residual, with finite second moments and expectations under P\mathbb{P}. If E[εt]=0\mathbb{E}[\varepsilon_t] = 0, taking expectations gives the exact mean attribution

E[Rt]=α+∑j=1mβj E[fj,t],\mathbb{E}[R_t] = \alpha + \sum_{j=1}^{m} \beta_j\, \mathbb{E}[f_{j,t}],

in which βjE[fj,t]\beta_j \mathbb{E}[f_{j,t}] is the contribution of factor jj and the intercept α\alpha is the part of the mean that the declared exposures do not account for. A variance decomposition needs a further condition. With Σf=Var⁡(ft)\Sigma_f = \operatorname{Var}(f_t),

Var⁡(Rt)=β⊤Σf β+Var⁡(εt)+2β⊤Cov⁡(ft,εt),\begin{aligned} \operatorname{Var}(R_t) &= \beta^{\top} \Sigma_f\, \beta \\[0.4em] &\quad + \operatorname{Var}(\varepsilon_t) \\[0.4em] &\quad + 2\beta^{\top} \operatorname{Cov}(f_t, \varepsilon_t), \end{aligned}

and the cross term vanishes when Cov⁡(ft,εt)=0\operatorname{Cov}(f_t, \varepsilon_t) = 0. If Σf\Sigma_f is invertible, the population linear projection β=Σf−1Cov⁡(ft,Rt)\beta = \Sigma_f^{-1}\operatorname{Cov}(f_t, R_t) and α=E[Rt]−β⊤E[ft]\alpha = \mathbb{E}[R_t] - \beta^{\top}\mathbb{E}[f_t] makes both conditions hold by construction. In an ordinary least squares fit with an intercept, the normal equations make the fitted residuals average to zero and have zero sample covariance with every factor in the fitting sample. These are identities of the fit. They do not show that the loadings or the orthogonality persist outside the sample, and an attribution describes association with the declared factors, not cause. The intercept changes with the list of declared factors, and an estimated intercept carries sampling error, so it is not by itself evidence of skill.

In plain words Attribution splits the strategy’s mean return into the parts that come with its exposures to the declared factors and a remainder, the intercept. Calling the whole mean alpha ignores the exposures, and a different list of factors can give a different alpha for the same returns.

Example Take the constructed values α=0.002\alpha = 0.002, β=(1.2,−0.5)\beta = (1.2, -0.5) and factor means 0.010.01 and 0.0040.004 per period. The contributions are 1.2⋅0.01=0.0121.2 \cdot 0.01 = 0.012 and −0.5⋅0.004=−0.002-0.5 \cdot 0.004 = -0.002, so

E[Rt]=0.002+0.012−0.002=0.012,\begin{aligned} \mathbb{E}[R_t] &= 0.002 + 0.012 \\[0.4em] &\quad - 0.002 \\[0.4em] &= 0.012, \end{aligned}

a mean of 1.21.2 percent of which only 0.20.2 percent is intercept. Suppose further that the two factors are uncorrelated and that εt\varepsilon_t is uncorrelated with both. The projection of RtR_t on the first factor alone then keeps the loading 1.21.2, because the omitted term −0.5f2,t-0.5 f_{2,t} is uncorrelated with f1,tf_{1,t}, and its intercept is 0.012−1.2⋅0.01=00.012 - 1.2 \cdot 0.01 = 0. The same returns have an alpha of 0.20.2 percent under the two-factor model and 00 under the one-factor model, whose intercept absorbs the second factor’s contribution of −0.002-0.002 together with the two-factor alpha. Related factor models appear in Factor models and covariance estimation.

Check yourself

  1. Three assets have the signals −1-1, 00 and 11 and the next-period excess returns −0.01-0.01, 0.0020.002 and 0.0080.008. Compute the sample information coefficient and the excess return of the rule with weights (−0.5,0,0.5)(-0.5, 0, 0.5). What happens to each if every return is ten times smaller?
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Both samples have mean 00, so the deviations are the values themselves. The sum of products is (−1)(−0.01)+0+1⋅0.008=0.018(-1)(-0.01) + 0 + 1 \cdot 0.008 = 0.018, and the sums of squares are 22 and 0.0001680.000168, so

IC⁡^=0.0182⋅0.000168≈0.982.\begin{aligned} \widehat{\operatorname{IC}} &= \frac{0.018}{\sqrt{2 \cdot 0.000168}} \\[0.4em] &\approx 0.982 . \end{aligned}

The rule earns (−0.5)(−0.01)+0.5⋅0.008=0.009(-0.5)(-0.01) + 0.5 \cdot 0.008 = 0.009, or 0.90.9 percent before costs. With every return ten times smaller, the correlation is unchanged at about 0.9820.982, because it ignores a positive rescaling, while the excess return falls to 0.00090.0009. The coefficient alone does not fix the profit, which also depends on the dispersion of returns, the position sizes and the costs.

  1. A strategy’s mean monthly excess return is 0.0080.008. A declared two-factor model has the loading 0.90.9 on a market factor with mean 0.0050.005, the loading 0.40.4 on a value factor with mean 0.0030.003, and a residual with mean zero. Find each factor contribution and the intercept. A report calls the whole 0.80.8 percent the strategy’s alpha. What is wrong with that, and what changes if the residual turns out to be correlated with the market factor?
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The contributions are 0.9⋅0.005=0.00450.9 \cdot 0.005 = 0.0045 and 0.4⋅0.003=0.00120.4 \cdot 0.003 = 0.0012, together 0.00570.0057, so the intercept is 0.008−0.0057=0.00230.008 - 0.0057 = 0.0023, or 0.230.23 percent. The report counts the 0.570.57 percentage points associated with the declared exposures as alpha. Even the 0.230.23 percent is specific to this model, could change with other factors, and carries sampling error when the loadings and means are estimated. A residual correlated with the market factor leaves the mean attribution intact, since it needs only a residual with mean zero. The variance can then no longer be split into a factor part and a residual part, because the cross term 2β⊤Cov⁡(ft,εt)2\beta^{\top}\operatorname{Cov}(f_t, \varepsilon_t) is not zero. The stated loadings are then also not the projection coefficients, so a regression on these factors would, in large samples, estimate other loadings and another intercept.

  1. Monthly excess returns are covariance stationary with mean 0.0050.005, standard deviation 0.040.04 and lag-one autocorrelation ρ1\rho_1, and have zero autocorrelation at every lag of two or more. Find the twelve-month Sharpe ratio of the additive aggregate for ρ1=0.2\rho_1 = 0.2 and for ρ1=−0.3\rho_1 = -0.3, and compare both with the square-root rule. Why can ρ1\rho_1 not be −0.6-0.6 in such a series?
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The monthly ratio is 0.005/0.04=0.1250.005/0.04 = 0.125, and the square-root rule gives 0.12512≈0.4330.125\sqrt{12} \approx 0.433. Only the lag-one term survives in the identity, with weight 1111, so the variance multiplier is 12+22ρ112 + 22\rho_1 and

SR⁡12=0.060.0412+22ρ1.\begin{aligned} &\operatorname{SR}_{12} \\[0.4em] &= \frac{0.06}{0.04\sqrt{12 + 22\rho_1}} . \end{aligned}

For ρ1=0.2\rho_1 = 0.2 the multiplier is 16.416.4 and the ratio is about 0.3700.370, below the rule. For ρ1=−0.3\rho_1 = -0.3 it is 5.45.4 and the ratio is about 0.6450.645, above the rule, because the positive mean is unchanged while consecutive months partly offset each other. For ρ1=−0.6\rho_1 = -0.6 the multiplier would be 12−13.2=−1.212 - 13.2 = -1.2, a negative variance, so no covariance-stationary series has these autocorrelations. The 12×1212 \times 12 correlation matrix with 11 on the diagonal and −0.6-0.6 beside it has smallest eigenvalue 1−1.2cos⁡(π/13)≈−0.1651 - 1.2\cos(\pi/13) \approx -0.165, so it is not positive semidefinite. A covariance-stationary series with zero autocorrelation beyond lag one exists exactly when ∣ρ1∣≤1/2\lvert \rho_1 \rvert \le 1/2, the range of the lag-one autocorrelation of a moving average of order one.

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