Optimal Blend of Two Correlated Alpha Signals
You have two daily alpha signals $s_1$ and $s_2$ on the same equity universe. Both have IC $= 0.05$ (sample, assumed stable) and pairwise correlation $\rho = 0.4$. Volatilities are normalized so both signals are unit-variance cross-sectionally. (a) Find the linear combination $w_1 s_1 + w_2 s_2$ that maximizes the combined IC subject to unit norm. (b) What is the combined IC? (c) A risk manager insists on $w_1 = w_2$ (equal-weight) because it is 'robust.' What is the cost of that simplification here?
Hints
- Two-asset Markowitz in IC space. Optimal weight is $\Sigma^{-1} i$ up to normalization.
- Under symmetric IC and correlation, equal-weight is the optimal weight.
- Equal-weight is robust to estimation error — the risk manager is making a real point even though in this particular symmetric problem the optimal weight IS equal-weight.
Worked Solution
How to Think About It: Two symmetric signals, equal IC, partial correlation — this is the textbook two-asset Markowitz problem in IC space. The IC-maximizing weight is symmetric (equal-weight is optimal here because the symmetry forces it). The interesting part is the COST when symmetry is broken or when correlation moves.
Quick Estimate: For two equal-IC equally-correlated signals, the combined IC is $\text{IC} \cdot \sqrt{2/(1+\rho)}$. Plug in: $0.05 \cdot \sqrt{2/1.4} = 0.05 \cdot 1.195 \approx 0.0598$. So combining the two gets you about a 20% IC bump.
Approach: Maximize $w^\top i / \sqrt{w^\top \Sigma w}$ subject to $\|w\|=1$. With $i = (0.05, 0.05)^\top$ and $\Sigma = \begin{pmatrix} 1 & \rho \\ \rho & 1 \end{pmatrix}$, the symmetric structure forces $w_1 = w_2 = 1/\sqrt{2}$.
Derivation: Optimal weight is $w^{*} \propto \Sigma^{-1} i$. Computing $\Sigma^{-1}$: $ \Sigma^{-1} = \frac{1}{1-\rho^2} \begin{pmatrix} 1 & -\rho \\ -\rho & 1 \end{pmatrix}. $ So $\Sigma^{-1} i = \frac{0.05}{1-\rho^2} (1 - \rho, 1 - \rho)^\top$ — equal weights. After normalizing to unit-norm, $w^{*} = (1/\sqrt{2}, 1/\sqrt{2})$. Plug in to get combined IC: $ \text{IC}_{\text{combined}} = \frac{w^\top i}{\sqrt{w^\top \Sigma w}} = \frac{\sqrt{2} \cdot 0.05}{\sqrt{1+\rho}} = 0.05 \cdot \sqrt{2/1.4} \approx 0.0598. $
(c) Cost of equal-weight under symmetry: Zero. Equal-weight IS optimal here. The risk manager is right that equal-weight is robust to estimation error in IC and $\rho$, and in this symmetric problem there is no extra IC to be had. The cost only appears when symmetry breaks — different ICs, or different correlations. For example if $\text{IC}_1 = 0.07$ and $\text{IC}_2 = 0.03$ (sum still 0.10), optimal weight tilts toward signal 1 and the combined IC is meaningfully higher than equal-weight. Under estimation noise though, equal-weight often beats the math-optimal weight out of sample (a standard 'shrinkage beats MLE' result).
Answer: $w^{*} = (1/\sqrt{2}, 1/\sqrt{2})$, combined IC $\approx 0.0598$ (~20% bump). Equal-weight is optimal here due to symmetry; the cost of forcing equal-weight only arises when the underlying IC or correlation is asymmetric.
Intuition
The Markowitz lesson: when two signals are correlated, the cost of redundancy shows up as a $\sqrt{1+\rho}$ in the denominator of the combined IC. Higher $\rho$ = more redundancy = worse blend.
The broader practical lesson: math-optimal weights computed from sample ICs and sample correlations are usually noisy. Shrinkage toward equal-weight, or Bayesian priors that pull the weights toward simple structures, beat the MLE solution out of sample most of the time. The risk manager who insists on equal-weight is often empirically right even when she is theoretically wrong.