13A.9 Hedge ratios: OLS, total least squares and the Kalman filter

13A.9.1Three static estimators#

With centred sums $S_{xx}, S_{yy}, S_{xy}$ of log prices:

• OLS of $y$ on $x$: $\hat\beta_{y|x} = S_{xy}/S_{xx}$ (minimises vertical distances).
• Reverse OLS, inverted: $1/\hat\beta_{x|y} = S_{yy}/S_{xy}$ (minimises horizontal distances).
• Total least squares (orthogonal regression): minimises perpendicular distances.

Proof

The squared perpendicular distance from $(x_i,y_i)$ to the line $y = \alpha+\beta x$ is $(y_i-\alpha-\beta x_i)^2/(1+\beta^2)$. Minimising over $\alpha$ gives $\alpha = \bar y - \beta\bar x$, leaving $f(\beta) = (S_{yy}-2\beta S_{xy}+\beta^2S_{xx})/(1+\beta^2)$. Then $f'(\beta) = 0$ reduces (multiply out and cancel the $\beta^3$ terms) to

$$S_{xy}\beta^2 + (S_{xx}-S_{yy})\beta - S_{xy} = 0.$$

The two roots have product $-1$ (perpendicular directions); the "+" root has the sign of $S_{xy}$ and is the minimum, the other the maximum. This is (13A.30). $f(\beta)$ is the Rayleigh quotient of the covariance matrix in direction $(-\beta, 1)$, so the minimiser is orthogonal to the smallest eigenvector, i.e. along the first principal component. Swapping $x$ and $y$ swaps $S_{xx}$ and $S_{yy}$; substituting shows the new root is the reciprocal. For the ordering, differentiate: $f'(\beta) = 2q(\beta)/(1+\beta^2)^2$ with $q(\beta) = S_{xy}\beta^2+(S_{xx}-S_{yy})\beta-S_{xy}$. Direct substitution gives $q(S_{xy}/S_{xx}) = S_{xy}(S_{xy}^2 - S_{xx}S_{yy})/S_{xx}^2 \le 0$ and $q(S_{yy}/S_{xy}) = (S_{xx}S_{yy}-S_{xy}^2)/S_{xy}\ge0$, both by Cauchy–Schwarz. With $S_{xy}>0$, $q$ is a convex parabola with $q(0) = -S_{xy}<0$, so on the positive axis it is negative to the left of its positive root and positive to the right. Hence the positive root lies between the two OLS slopes.∎

Also, $\hat\beta_{y|x}\cdot\hat\beta_{x|y} = S_{xy}^2/(S_{xx}S_{yy}) = \hat\rho^2$: the two OLS slopes agree only if the fit is perfect. TLS is not invariant to rescaling one variable, so apply it to log prices (or standardised series), never to rupee prices of very different magnitudes.

If the pair is truly cointegrated, all three estimators are superconsistent for the same $\beta$; their differences are a finite-sample matter. They are large precisely when the residual spread is wide relative to the common trend, which is the case worth trading. Problem 13A.10 has numbers.

13A.9.2A time-varying hedge ratio: the state-space model#

with $\varepsilon,\eta$ independent of each other and over time, and prior $\beta_0\sim N(b_0, P_0)$. For a pair, $x_t = p^2_t$ (scalar) or $x_t = (1, p^2_t)'$ with a time-varying intercept. Write $\mathcal F_t$ for the information in $y_1,\dots,y_t$ (the $x_t$ are treated as known).

Proof

By induction. Suppose $\beta_{t-1}\mid\mathcal F_{t-1}\sim N(b_{t-1|t-1},P_{t-1|t-1})$ (true at $t=1$ by the prior).

Predict. $\beta_t = \beta_{t-1}+\eta_t$ with $\eta_t$ independent of $\mathcal F_{t-1}$ and $\beta_{t-1}$, so $\beta_t\mid\mathcal F_{t-1}$ is normal with mean $b_{t-1|t-1}$ and variance $P_{t-1|t-1}+Q$: (13A.32).

Joint law. Given $\mathcal F_{t-1}$, $y_t = x_t'\beta_t+\varepsilon_t$ is a linear function of the Gaussian pair $(\beta_t,\varepsilon_t)$, so $(\beta_t, y_t)$ is jointly normal with

$$\begin{aligned} \mathbb E[y_t\mid\mathcal F_{t-1}] &= x_t'b_{t|t-1}, \\ \operatorname{Var}(y_t\mid\mathcal F_{t-1}) &= x_t'P_{t|t-1}x_t+R = F_t, \\ \operatorname{Cov}(\beta_t,y_t\mid\mathcal F_{t-1}) &= P_{t|t-1}x_t. \end{aligned}$$

Update. For jointly normal $(U,V)$, $U\mid V\sim N\big(\mathbb EU + \Sigma_{UV}\Sigma_{VV}^{-1}(V-\mathbb EV),\ \Sigma_{UU}-\Sigma_{UV}\Sigma_{VV}^{-1}\Sigma_{VU}\big)$. With $U = \beta_t$, $V = y_t$ this is (13A.34). Conditioning on $y_t$ in addition to $\mathcal F_{t-1}$ is conditioning on $\mathcal F_t$, which completes the induction.∎

Likelihood The innovations $e_t$ are independent $N(0,F_t)$ given the past, so by the prediction-error decomposition

Maximise over $(Q,R)$ numerically. Two facts help. First, scaling $Q$, $R$ and $P_0$ by the same constant leaves $K_t$ and $b_{t|t}$ unchanged (every $P$ and $F$ scales by it), so the filtered hedge ratio depends only on the signal-to-noise ratio $\delta = Q/R$. Second, with a diffuse prior the first few $F_t$ are huge; drop them from (13A.35) (kalman_loglik(burn=...)).

Proof

Combining (13A.32) and (13A.34), $P_{t+1|t} = P_{t|t-1} - \frac{x^2P_{t|t-1}^2}{x^2P_{t|t-1}+R} + Q = \frac{P_{t|t-1}R}{x^2P_{t|t-1}+R} + Q$. A fixed point satisfies $\bar P(x^2\bar P + R) = \bar PR + Q(x^2\bar P+R)$, i.e. $x^2\bar P^2 - Qx^2\bar P - QR = 0$, whose positive root is (13A.36). The map $P\mapsto PR/(x^2P+R)+Q$ is increasing and concave with derivative $R^2/(x^2P+R)^2<1$ on $[0,\infty)$, so iteration converges to $\bar P$ from any $P\ge0$. Then $b_{t|t} = b_{t-1|t-1} + \bar K(y_t - xb_{t-1|t-1}) = (1-\bar Kx)b_{t-1|t-1} + \bar Kx\,(y_t/x)$.∎

So the "Kalman hedge ratio" with a random-walk state is exponential smoothing with a smoothing constant chosen by maximum likelihood, the observation Muth (1960) made for forecasting.

13A.9.3Using the filter for a pairs trade, and its trap#

The natural signal is the standardised innovation $z_t = e_t/\sqrt{F_t}$, the surprise in $y_t$ given the hedge ratio predicted from the past (Chan 2013, ch. 3). Trade when $|z_t|$ is large.

• Use $b_{t|t-1}$, not $b_{t|t}$, to form the spread you trade at $t$. $b_{t|t}$ already uses $y_t$. The smoothed estimates $b_{t|T}$ use the whole sample and are look-ahead (the same trap as M13 · 13.5 for regime probabilities).
• The trap. A large $\delta$ lets $\beta_t$ chase every move in $y_t/x_t$. The innovation then shrinks, the spread $y_t - x_tb_{t|t}$ is close to zero by construction, and the mean reversion you wanted to trade has been absorbed into the "hedge ratio". A drifting $\beta_t$ is also an admission that the relationship is not cointegrated with a fixed vector. Choose $\delta$ by likelihood or out-of-sample, and check the economic plausibility of the path of $\beta_t$.

Exercise link m13a_kalman.py checks the one-step update by hand, the steady state (13A.36), that the filter tracks a simulated random-walk $\beta_t$ with lower RMSE than a 120-day rolling OLS, that the standardised innovations have unit variance, and that the likelihood (13A.35) picks the true $Q$ on a grid. With $Q = 0$ the filter is recursive least squares and its final state equals full-sample OLS.