Terms and Definitions
Firm's / Client's Portfolio
Benchmark Portfolio
Symbol: B B B
Weights: b \pmb b b b
Inner-Product: b ′ 1 = 1 \pmb b' \pmb 1= 1 b b ′ 1 1 = 1
Active Holdings
Symbol: h \pmb h h h
Weights: h \pmb h h h
Inner-Product: h ′ 1 = 0 \pmb h' \pmb 1= 0 h h ′ 1 1 = 0
Active Portfolio
Symbol: A A A
Weights: w = b + h \pmb w = \pmb b + \pmb h w w = b b + h h
Inner-Product: w ′ 1 = 1 \pmb w' \pmb 1= 1 w w ′ 1 1 = 1
I R = E [ r W ] − E [ r B ] T E \begin{align} IR = \frac{E[r_W] - E[r_B]}{TE} \end{align} I R = TE E [ r W ] − E [ r B ]
Tracking error / active risk :
T E = V a r ( r A − r B ) = V a r [ ( w − b ) ′ r ] = V a r ( h ′ r ) = h ′ V h = σ e \begin{align} TE &= \sqrt{Var(r_A - r_B)} \\ &= \sqrt{Var[(\pmb w - \pmb b)' \pmb r]} \\ &= \sqrt{Var(\pmb h'\pmb r)} \\ &= \sqrt{\pmb h'V\pmb h} \\ &= \sigma_e \end{align} TE = Va r ( r A − r B ) = Va r [( w w − b b ) ′ r r ] = Va r ( h h ′ r r ) = h h ′ V h h = σ e
The Problem
Our client holds a portfolio that looks quite a bit like some benchmark. We assume they've optimized their portfolio to have no alpha with respect to to the benchmark (α B F = 0 \alpha_{BF} =0 α BF = 0 ). We also assume, (and will impose by constraint), that the the beta of the active fund with respect to the benchmark is one (β A B = 1 \beta_{AB}=1 β A B = 1 ).
We want to construct an active portfolio which will be constructed as a combination of the benchmark portfolio with active positions .
To provide our client with a higher alpha portfolio, the we can employ one of two distinct methods.
Maximimizing the Information Ratio.
Benchmark Hedging.
Maximizing the Information Ratio
One method to supply alpha to A A A is to maximize the information ratio .
Numerator : It can be shown that when β A B = 1 \beta_{AB} =1 β A B = 1 then h ′ α = E [ r A ] − E [ r B ] \pmb h' \pmb \alpha = E[r_A] - E[r_B] h h ′ α α = E [ r A ] − E [ r B ] .
Denominator : We enforce that h ′ V h ≤ σ ˉ \sqrt{\pmb h' V \pmb h} \leq \bar \sigma h h ′ V h h ≤ σ ˉ . Though we'll do this via a scaling after solving for an unscaled h ~ \pmb{ \tilde h } h ~ h ~ .
Beta Constraint : Turns out that β A B = 1 ⟺ β n B ′ h = 0 \beta_{AB} = 1 \iff \pmb \beta_{\pmb nB}'\pmb h =0 β A B = 1 ⟺ β β n n B ′ h h = 0 . We enforce the latter.
Active Weights Constraint : By definition of h \pmb h h h we want h ′ 1 = 0 \pmb h'\pmb 1=0 h h ′ 1 1 = 0
We can maximize the information ratio and satisfy the constraints by solving:
L = h ′ α − λ h ′ V h − γ ′ ( C h − c ) \begin{align} L = \pmb h' \pmb \alpha - \lambda \pmb h' V \pmb h - \pmb \gamma'(C\pmb h - \pmb c) \end{align} L = h h ′ α α − λ h h ′ V h h − γ γ ′ ( C h h − c c )
Where
C = [ 1 ′ β n B ′ ] , c = [ 0 0 ] \begin{align} C= \begin{bmatrix}\pmb 1' \\\pmb \beta_{\pmb nB}'\end{bmatrix}, \ \ \pmb c=\begin{bmatrix}0 \\ 0\end{bmatrix} \end{align} C = [ 1 1 ′ β β n n B ′ ] , c c = [ 0 0 ]
After solving we find that:
h ~ = ( 2 λ ) − 1 V − 1 [ I − C ′ ( C V − 1 C ′ ) − 1 C V − 1 ] α + V − 1 C ′ ( C V − 1 C ′ ) − 1 c = ( 2 λ ) − 1 V − 1 [ I − C ′ ( C V − 1 C ′ ) − 1 C V − 1 ] α \begin{align} \pmb{\tilde h} &= (2\lambda)^{-1}V^{-1}\big[I - C'(CV^{-1}C')^{-1}CV^{-1}\big]\pmb \alpha + V^{-1}C'(CV^{-1}C')^{-1}\pmb c \\ &= (2\lambda)^{-1}V^{-1}\big[I - C'(CV^{-1}C')^{-1}CV^{-1}\big]\pmb \alpha \end{align} h ~ h ~ = ( 2 λ ) − 1 V − 1 [ I − C ′ ( C V − 1 C ′ ) − 1 C V − 1 ] α α + V − 1 C ′ ( C V − 1 C ′ ) − 1 c c = ( 2 λ ) − 1 V − 1 [ I − C ′ ( C V − 1 C ′ ) − 1 C V − 1 ] α α
Now, we scale our answer to the appropriate tracking error, σ ˉ \bar \sigma σ ˉ . We want to find k k k such that:
( k h ~ ) ′ V ( k h ~ ) = ∣ k ∣ h ~ ′ V h ~ = σ ˉ ⟹ k = σ ˉ h ~ ′ V h ~ = σ ˉ σ e \begin{align} \sqrt{(k\pmb{\tilde h})'V(k\pmb{\tilde h})}= |k|\sqrt{\pmb{\tilde h}'V\pmb{\tilde h}}&= \bar \sigma \\ \implies k &= \frac{\bar \sigma}{\sqrt{\pmb{\tilde h}'V\pmb{\tilde h}}} \\[12pt] &=\frac{\bar \sigma}{\sigma_{e}} \end{align} ( k h ~ h ~ ) ′ V ( k h ~ h ~ ) = ∣ k ∣ h ~ h ~ ′ V h ~ h ~ ⟹ k = σ ˉ = h ~ h ~ ′ V h ~ h ~ σ ˉ = σ e σ ˉ
Finally, h ∗ = k ⋅ h ~ \pmb h^* = k \cdot \pmb {\tilde h} h h ∗ = k ⋅ h ~ h ~ . And the active portfolio will have tracking error equal to σ ˉ \bar \sigma σ ˉ .
We supply w ∗ = h ∗ + b \pmb w^* = \pmb h^* + \pmb b w w ∗ = h h ∗ + b b to the client.
Benchmark Hedging
A difficulty with maximizing the information ratio is that it requires views on every stock in the benchmark. Instead of having N N N predicted alphas, suppose we only have P P P views for constructing the active portfolio. We only invest in the P P P assets for which we have views and a benchmark ETF.
By definition:
V a r ( r E T F ) = b ′ V b ∈ R C o v ( r p , r E T F ) = ( V b ) p ∈ R P × 1 \begin{align} Var(r_{ETF}) &= \pmb b' V \pmb b \in R\\[5pt] Cov(r_p, r_{ETF}) &= (V\pmb b)_p \in R^{P\times 1} \end{align} Va r ( r ETF ) C o v ( r p , r ETF ) = b b ′ V b b ∈ R = ( V b b ) p ∈ R P × 1
We construct:
α A = [ α p 0 ] ∈ R ( P + 1 ) × 1 V A = [ V p ( V b ) p ( V b ) p ′ b ′ V b ] ∈ R ( P + 1 ) × ( P + 1 ) \begin{align} \pmb \alpha_A &=\begin{bmatrix}\pmb \alpha_p \\ 0 \end{bmatrix} \in \mathbb R^{(P+1)\times 1}\\[12pt] V_A &= \begin{bmatrix}V_p & (V\pmb b)_p \\ (V\pmb b)_p' & \pmb b' V \pmb b\end{bmatrix} \in \mathbb R^{(P+1)\times(P+1)} \end{align} α α A V A = [ α α p 0 ] ∈ R ( P + 1 ) × 1 = [ V p ( V b b ) p ′ ( V b b ) p b b ′ V b b ] ∈ R ( P + 1 ) × ( P + 1 )
We want to find:
h A = [ h p h E T F ] ∈ R ( P + 1 ) × 1 \begin{align} \pmb h_A = \begin{bmatrix}\pmb h_p \\ h_{ETF}\end{bmatrix} \in \mathbb R^{(P+1) \times 1} \end{align} h h A = [ h h p h ETF ] ∈ R ( P + 1 ) × 1
The Lagrangian for our problem is:
L = h A ′ α A − λ h A ′ V A h A − γ ′ ( C h A − c ) \begin{align} L = \pmb h_A'\pmb \alpha_A - \lambda \pmb h_A' V_A \pmb h_A - \pmb \gamma'(C\pmb h_A - \pmb c) \end{align} L = h h A ′ α α A − λ h h A ′ V A h h A − γ γ ′ ( C h h A − c c )
Where
C = [ 1 p ′ 1 β p ′ 1 ] , c = [ 0 0 ] \begin{align} C = \begin{bmatrix} \pmb 1_p' & 1 \\ \pmb \beta_p' & 1 \end{bmatrix}, \ \ \pmb c=\begin{bmatrix} 0 \\ 0\end{bmatrix} \end{align} C = [ 1 1 p ′ β β p ′ 1 1 ] , c c = [ 0 0 ]
After solving we find that:
h ~ A = ( 2 λ ) − 1 V A − 1 [ I − C ′ ( C V A − 1 C ′ ) − 1 C V A − 1 ] α + V A − 1 C ′ ( C V A − 1 C ′ ) − 1 c = ( 2 λ ) − 1 V A − 1 [ I − C ′ ( C V A − 1 C ′ ) − 1 C V A − 1 ] α \begin{align} \pmb{\tilde h}_A &= (2\lambda)^{-1}V_A^{-1}\big[I - C'(CV_A^{-1}C')^{-1}CV_A^{-1}\big]\pmb \alpha + V_A^{-1}C'(CV_A^{-1}C')^{-1}\pmb c \\ &= (2\lambda)^{-1}V_A^{-1}\big[I - C'(CV_A^{-1}C')^{-1}CV_A^{-1}\big]\pmb \alpha \end{align} h ~ h ~ A = ( 2 λ ) − 1 V A − 1 [ I − C ′ ( C V A − 1 C ′ ) − 1 C V A − 1 ] α α + V A − 1 C ′ ( C V A − 1 C ′ ) − 1 c c = ( 2 λ ) − 1 V A − 1 [ I − C ′ ( C V A − 1 C ′ ) − 1 C V A − 1 ] α α
Now, we scale our answer to the appropriate tracking error, σ ˉ \bar \sigma σ ˉ . We want to find k k k such that:
( k h ~ ) ′ V A ( k h ~ ) = ∣ k ∣ h ~ ′ V A h ~ = σ ˉ ⟹ k = σ ˉ h ~ ′ V A h ~ = σ ˉ σ e \begin{align} \sqrt{(k\pmb{\tilde h})'V_A(k\pmb{\tilde h})}= |k|\sqrt{\pmb{\tilde h}'V_A\pmb{\tilde h}}&= \bar \sigma \\ \implies k &= \frac{\bar \sigma}{\sqrt{\pmb{\tilde h}'V_A\pmb{\tilde h}}} \\[12pt] &=\frac{\bar \sigma}{\sigma_{e}} \end{align} ( k h ~ h ~ ) ′ V A ( k h ~ h ~ ) = ∣ k ∣ h ~ h ~ ′ V A h ~ h ~ ⟹ k = σ ˉ = h ~ h ~ ′ V A h ~ h ~ σ ˉ = σ e σ ˉ
Scaling, h A ∗ = k ⋅ h ~ A \pmb h^*_A = k \cdot \pmb {\tilde h}_A h h A ∗ = k ⋅ h ~ h ~ A . And the active portfolio will have tracking error equal to σ ˉ \bar \sigma σ ˉ .
We need to unpack h A ∗ = [ h p ∗ ′ h E T F ∗ ] ′ \pmb h^*_A= \begin{bmatrix} \pmb h_p^{*\prime} & h_{ETF}^* \end{bmatrix}' h h A ∗ = [ h h p ∗ ′ h ETF ∗ ] ′ to recover the weights we offer the client:
w A ∗ = ( h A ∗ + b ) + h E T F ∗ ⋅ b = h A ∗ + ( 1 + h E T F ∗ ) ⋅ b \begin{align} \pmb w^*_A &= (\pmb h^*_A + \pmb b) + h_{ETF}^* \cdot \pmb b \\[5pt] &= \pmb h^*_A + (1+h_{ETF}^*)\cdot \pmb b \end{align} w w A ∗ = ( h h A ∗ + b b ) + h ETF ∗ ⋅ b b = h h A ∗ + ( 1 + h ETF ∗ ) ⋅ b b