Scaling of Returns

Module 3: Modelling with Randomness

If log returns are independent draws from one stable law, their distribution at every time horizon is a rescaled copy of the one-step distribution. (mantegna1995scaling?) tested this idea on the S&P 500 index. This page states the prediction and their result.

The scaling prediction

Let the one-step log returns be i.i.d. symmetric stable with index $\alpha$ and density $p_1$. By Stable Distributions, the return over $\Delta t$ steps, $r(\Delta t)$, is stable with the same $\alpha$ and a scale multiplied by $\Delta t^{1/\alpha}$. Its density is

\[ p_{\Delta t}(x) = \Delta t^{-1/\alpha}\, p_1\!\left( x\, \Delta t^{-1/\alpha} \right). \tag{1}\]

At $x = 0$ this gives the probability of return to the origin

\[ p_{\Delta t}(0) = p_1(0)\, \Delta t^{-1/\alpha} . \tag{2}\]

On log-log axes $p_{\Delta t}(0)$ against $\Delta t$ is a straight line with slope $-1/\alpha$. For Gaussian returns $\alpha = 2$ and the slope is $-1/2$. The peak of the distribution is well sampled, so Equation 2 is easier to measure than the tail.

What Mantegna and Stanley found

(mantegna1995scaling?) used S&P 500 index values sampled every minute from 1984 to 1989. They report that:

$p_{\Delta t}(0)$ follows Equation 2 for $\Delta t$ from 1 to 1000 minutes, three orders of magnitude.
The slope is about $-0.71$, which gives $\alpha \approx 1.4$ ($1.40 \pm 0.05$ in the paper). The exponent is stable over the six years of data.
After rescaling with Equation 1, the central part of the distributions at different $\Delta t$ matches a Lévy stable law.
The far tails fall off faster than a Lévy law, approximately exponentially, so the variance is finite.

The last point led to the truncated Lévy flight: a Lévy flight whose largest steps are cut off. Sums of such steps look stable for a long range of $n$ and only converge to a Gaussian after very many steps (mantegna1994stochastic?). Later work on the tails of individual US stocks found a power-law tail exponent close to 3, outside the stable range $0 < \alpha \le 2$ (gopikrishnan1998inverse?). This is consistent with the heavy-tail and aggregational Gaussianity facts of Stylized Facts of Returns.

A simulated check

The figure applies the $p_{\Delta t}(0)$ method to two synthetic series of $10^6$ i.i.d. returns: symmetric stable with $\alpha = 1.4$ and Gaussian. The density at zero is estimated as the fraction of aggregated returns in a window $|r| < \varepsilon/2$ divided by $\varepsilon$, with $\varepsilon$ equal to one tenth of the interquartile range at that horizon.

Figure 1: Estimated probability of return to the origin against horizon for i.i.d. stable ($\alpha = 1.4$) and Gaussian returns. Lines have slopes $-1/1.4$ and $-1/2$.

Exercise

Fit the slope of $\log p_{\Delta t}(0)$ against $\log \Delta t$ for both series and convert it to an estimate of $\alpha$. Then repeat the fit for the stable series using only $\Delta t \ge 100$. Why does the estimate get noisier?

Solution

for label, r in series.items():
    p0 = np.array([p_zero(aggregate_returns(r, h)) for h in horizons])
    s_all = np.polyfit(np.log(horizons), np.log(p0), 1)[0]
    big = horizons >= 100
    s_big = np.polyfit(np.log(horizons[big]), np.log(p0[big]), 1)[0]
    name = label.replace("$", "").replace("\\alpha", "alpha")
    print(f"{name:18s} alpha (all) = {-1 / s_all:.2f}, alpha (dt >= 100) = {-1 / s_big:.2f}")

stable, alpha = 1.4 alpha (all) = 1.37, alpha (dt >= 100) = 1.15
Gaussian           alpha (all) = 2.08, alpha (dt >= 100) = 2.52

With $10^6$ returns there are only 2000 non-overlapping blocks at $\Delta t = 500$, and few of them fall inside the small window around zero. Long horizons cost data, which is why (mantegna1995scaling?) needed minute-by-minute prices to span three decades of $\Delta t$.

--- title: "Scaling of Returns" subtitle: "Module 3: Modelling with Randomness" format: html --- If log returns are independent draws from one stable law, their distribution at every time horizon is a rescaled copy of the one-step distribution. @mantegna1995scaling tested this idea on the S&P 500 index. This page states the prediction and their result. ## The scaling prediction Let the one-step log returns be i.i.d. symmetric stable with index $\alpha$ and density $p_1$. By [Stable Distributions](stable-distributions.qmd), the return over $\Delta t$ steps, $r(\Delta t)$, is stable with the same $\alpha$ and a scale multiplied by $\Delta t^{1/\alpha}$. Its density is $$ p_{\Delta t}(x) = \Delta t^{-1/\alpha}\, p_1\!\left( x\, \Delta t^{-1/\alpha} \right). $$ {#eq-scaling-density} At $x = 0$ this gives the probability of return to the origin $$ p_{\Delta t}(0) = p_1(0)\, \Delta t^{-1/\alpha} . $$ {#eq-scaling-p0} On log-log axes $p_{\Delta t}(0)$ against $\Delta t$ is a straight line with slope $-1/\alpha$. For Gaussian returns $\alpha = 2$ and the slope is $-1/2$. The peak of the distribution is well sampled, so @eq-scaling-p0 is easier to measure than the tail. ## What Mantegna and Stanley found @mantegna1995scaling used S&P 500 index values sampled every minute from 1984 to 1989. They report that: - $p_{\Delta t}(0)$ follows @eq-scaling-p0 for $\Delta t$ from 1 to 1000 minutes, three orders of magnitude. - The slope is about $-0.71$, which gives $\alpha \approx 1.4$ ($1.40 \pm 0.05$ in the paper). The exponent is stable over the six years of data. - After rescaling with @eq-scaling-density, the central part of the distributions at different $\Delta t$ matches a Lévy stable law. - The far tails fall off faster than a Lévy law, approximately exponentially, so the variance is finite. The last point led to the truncated Lévy flight: a Lévy flight whose largest steps are cut off. Sums of such steps look stable for a long range of $n$ and only converge to a Gaussian after very many steps [@mantegna1994stochastic]. Later work on the tails of individual US stocks found a power-law tail exponent close to 3, outside the stable range $0 < \alpha \le 2$ [@gopikrishnan1998inverse]. This is consistent with the heavy-tail and aggregational Gaussianity facts of [Stylized Facts of Returns](stylized-facts.qmd). ## A simulated check The figure applies the $p_{\Delta t}(0)$ method to two synthetic series of $10^6$ i.i.d. returns: symmetric stable with $\alpha = 1.4$ and Gaussian. The density at zero is estimated as the fraction of aggregated returns in a window $|r| < \varepsilon/2$ divided by $\varepsilon$, with $\varepsilon$ equal to one tenth of the interquartile range at that horizon. ```{python} #| label: fig-p0-scaling #| fig-cap: "Estimated probability of return to the origin against horizon for i.i.d. stable ($\\alpha = 1.4$) and Gaussian returns. Lines have slopes $-1/1.4$ and $-1/2$." #| echo: false import matplotlib.pyplot as plt import numpy as np from amlab.finance import aggregate_returns from amlab.random_walks import sample_symmetric_stable def p_zero(x): """Density at 0 estimated in a window of one tenth of the IQR.""" q75, q25 = np.percentile(x, [75, 25]) eps = 0.1 * (q75 - q25) return np.mean(np.abs(x) < eps / 2) / eps rng = np.random.default_rng(24) series = { "stable, $\\alpha = 1.4$": sample_symmetric_stable(1.4, 10**6, rng=rng), "Gaussian": rng.standard_normal(10**6), } horizons = np.array([1, 2, 5, 10, 20, 50, 100, 200, 500]) fig, ax = plt.subplots(figsize=(6, 3.8)) for (label, r), slope, marker in zip(series.items(), [-1 / 1.4, -0.5], ["o", "s"]): p0 = np.array([p_zero(aggregate_returns(r, h)) for h in horizons]) ax.loglog(horizons, p0, marker, label=label) ax.loglog(horizons, p0[0] * horizons**slope, "k-", lw=0.8) ax.set_xlabel(r"horizon $\Delta t$") ax.set_ylabel(r"$p_{\Delta t}(0)$") ax.legend() plt.tight_layout() plt.show() ``` ::: {.callout-note} ## Exercise Fit the slope of $\log p_{\Delta t}(0)$ against $\log \Delta t$ for both series and convert it to an estimate of $\alpha$. Then repeat the fit for the stable series using only $\Delta t \ge 100$. Why does the estimate get noisier? ::: ::: {.callout-tip collapse="true"} ## Solution ```{python} for label, r in series.items(): p0 = np.array([p_zero(aggregate_returns(r, h)) for h in horizons]) s_all = np.polyfit(np.log(horizons), np.log(p0), 1)[0] big = horizons >= 100 s_big = np.polyfit(np.log(horizons[big]), np.log(p0[big]), 1)[0] name = label.replace("$", "").replace("\\alpha", "alpha") print(f"{name:18s} alpha (all) = {-1 / s_all:.2f}, alpha (dt >= 100) = {-1 / s_big:.2f}") ``` With $10^6$ returns there are only 2000 non-overlapping blocks at $\Delta t = 500$, and few of them fall inside the small window around zero. Long horizons cost data, which is why @mantegna1995scaling needed minute-by-minute prices to span three decades of $\Delta t$. :::