Using Return Correlations for Linear Prediction and Trading Signals
Article Quant Q&A · Author: joshwa
Summary
The document presents a strategy attributed to Didier Sornette that predicts a return from past returns using their correlation structure. It describes the best linear predictor under a variance-minimizing criterion: weight past observations using entries associated with the inverse correlation matrix, then use the predicted return's sign as a buy or sell signal. The question applies this idea to a short return series by building lagged observations, calculating a correlation matrix, and inverting it.
The exchange challenges the calculation's indexing and scaling, noting confusion about which inverse-matrix entries represent the coefficients and whether the diagonal term should be multiplied or used as a divisor. Another response suggests least-squares regression and raises stability concerns. The material does not resolve the worked example or establish profitability. Its brief sample also leaves the estimation sensitive to how the correlation matrix is constructed, so the formula and signal would require careful validation before practical use.
Key ideas
- A linear predictor can combine past returns using weights derived from the inverse correlation matrix.
- The stated predictor minimizes prediction error variance under its linear setup.
- The sign of the predicted return is proposed as a directional trading signal.
- Correct matrix indexing and scaling are essential to calculating the predictor.
- The exchange raises concerns about stability and does not demonstrate trading performance.
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Full text
# Didier Sornette's Strategy to Exploit Return Correlations
# Didier Sornette's Strategy to Exploit Return Correlations
In his book, "Why Stock Markets Crash", Didier Sornette discusses a trading strategy that exploits return correlations.
> Consider a return $r$ that occurred at time $t$ and a return $r'$ that occurred at a later time $t'$, where $t$ and $t'$, are multiples of some time unit (say 5 minutes). $r$ and $r'$ can be decomposed into an average contribution and a varying part. We are interested in quantifying the correlation $C(t, t')$ between the uncertain varying part, which is defined as the average of the product of the varying part $r$ and of $r'$ normalized by the variance (volatility) of the returns, so that $C(t, t' = t) = 1$ (perfect correlation between $r$ and itself). A simple mathematical calculation shows that the best linear predictor $m_t$ for the return at time $t$, knowing the past history $r_{t-1}, \> r_{t-2}, \ldots ,r_i, \ldots,$ is given by $$m_t\equiv\frac{1}{B(t, t)}\sum_{i<t} B(i, t)r_i,$$ where each $B(i, t)$ is a factor that can be expressed in terms of the correlation coefficient $C(t', t)$ and is usually called the coefficient $(i, t)$ of the inverse correlation matrix. This formula expresses that each past return $r_i$ impacts on the future return $r_t$ in proportion to its value with a coefficient $B(i, t)/B(t, t)$ which is nonzero only if there is nonzero correlation between time $i$ and time $t$. With this formula, you have the best linear predictor in the sense that it will minimize the errors in variance. Armed with this prediction, you have a powerful trading strategy: buy if $m_t \> > 0$ (expected future price increase) and sell if $m_t < 0$ (expected future price decrease).
I'm trying to solve for $m_t$ given the following dataset...
```
+----------+-----------------+
| t | r |
+----------+-----------------+
| 15:50:00 | 0.003705090715 |
| 15:51:00 | 0.003873999746 |
| 15:52:00 | 0.002158853672 |
| 15:53:00 | 0.001246754886 |
| 15:54:00 | 0.005646756563 |
| 15:55:00 | -0.001073638262 |
| 15:56:00 | -0.001804395665 |
| 15:57:00 | 0.002322446782 |
| 15:58:00 | 0.001803468933 |
| 15:59:00 | -0.001686730014 |
| 16:00:00 | 0.0008781111203 |
+----------+-----------------+
```
First I create a matrix for time lags $r_{t}, r_{t-1}, r_{t-2}, r_{t-3}, r_{t-4}$, called $\color{blue}{\mathbf M}$... $$ \begin{bmatrix} \color{blue}{0.00087811} & \color{blue}{-0.00168673} & \color{blue}{0.00180347} & \color{blue}{0.00232245} & -0.0018044 \\ -0.00168673 & 0.00180347 & 0.00232245 & -0.0018044 & -0.00107364 \\ 0.00180347 & 0.00232245 & -0.0018044 & -0.00107364 & 0.00564676 \\ 0.00232245 & -0.0018044 & -0.00107364 & 0.00564676 & 0.00124675 \\ -0.0018044 & -0.00107364 & 0.00564676 & 0.00124675 & 0.00215885 \\ \end{bmatrix} $$ With $\color{blue}{\mathbf M}$ I am able to create a correlation matrix, called $\color{green}{\mathbf C}$... $$ \begin{bmatrix} 1. & \color{green}{-0.15885375} & -0.88120533 & 0.52518141 & 0.32904361 \\ -0.15885375 & 1. & \color{green}{-0.27689212} & -0.87324435 & 0.43345156 \\ -0.88120533 & -0.27689212 & 1. & \color{green}{-0.16496963} & -0.38678436 \\ 0.52518141 & -0.87324435 & -0.16496963 & 1. & \color{green}{-0.18292072} \\ 0.32904361 & 0.43345156 & -0.38678436 & -0.18292072 & 1. \\ \end{bmatrix} $$ and its inverse, called $\color{red}{\mathbf I}$... $$ \begin{bmatrix} \color{red}{678.1834365} & -47.5405215 & 305.9166473 & -205.4095813 & -138.3472675 \\ -47.5405215 & 557.4221492 & 89.6736779 & 245.3640449 & -138.7710611 \\ 305.9166473 & 89.6736779 & 111.2290498 & -89.4926475 & 61.0341470 \\ -205.4095813 & 245.3640449 & -89.4926475 & 323.4950056 & -2.4011426 \\ -138.3472675 & -138.7710611 & 61.0341470 & -2.4011426 & 120.3071511 \\ \end{bmatrix} $$ Next I solve for... $$ \sum_{i<t} \color{green}{B(i, t)}\color{blue}{r_i}, $$ as follows... $$ \color{green}{\mathbf{C}[0][1]} \left( \color{blue}{\mathbf{M}[0][0]} \right) + \color{green}{\mathbf{C}[1][2]} \left( \color{blue}{\mathbf{M}[0][1]} \right) + \color{green}{\mathbf{C}[2][3]} \left( \color{blue}{\mathbf{M}[0][2]} \right) + \color{green}{\mathbf{C}[3][4]} \left( \color{blue}{\mathbf{M}[0][3]} \right) = -0.000394790246734 $$ Finally I multiply by $\frac{1}{B(t,t)}$ which I think is equal to $\color{red}{\mathbf{I}[0][0]}$... $$ \begin{align} m_t & = -0.000394790246734\left( \frac{1}{B(t,t)} \right) \\ & = -0.000394790246734\left( \color{red}{\mathbf{I}[0][0]} \right) \\ & = -0.000394790246734\left( 678.1834365 \right) \\ & = -0.267740206251 \\ \end{align} $$ Did I solve for $m_t$ correctly?
## Answer by Klaus (score 3, accepted)
https://quant.stackexchange.com/a/36461
I do not understand why t is zero in B(t,t) and 1,2,3,4 in B(i,t). And why did you interprete the value B(i,t) as the inverse of C, but not B(t,t)? I think B(i,t) is the first column of I. An inverse is defined by Matrix^-1. But, B(*) is a value. So, you should divide by B(t,t). Here, you multiplicate it.
## Answer by EntrustName (score 1)
https://quant.stackexchange.com/a/36495
Why you don't just do a least square regression ? It is likely not stable no ?Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.