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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

$$m_tequivfrac{1}{B(t, t)}sum_{i<t} B(i, t)r_i,$$

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 $color{red}{mathbf{I}[0][0]}$…
$$
begin{align}
m_t & = -0.000394790246734left( frac{1}{B(t,t)} right) \
& = -0.000394790246734left( color{red}{mathbf{I}[0][0]} right) \
& = -0.000394790246734left( 678.1834365 right) \
& = -0.267740206251 \
end{align}
$$
Did I solve for $m_t$ correctly?

Quantitative Finance Asked by joshwa on November 17, 2021

2 Answers

2 Answers

Why you don't just do a least square regression ? It is likely not stable no ?

Answered by EntrustName on November 17, 2021

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.

Answered by Klaus on November 17, 2021

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