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This tutorial written and reproduced with permission from Peter Ponzo

Once upon a time, a British government bureaucrat named Harold Edwin Hurst studied 800 years of records of the Nile’s flooding. He noticed that there was a tendency for a high flood year to be followed by another high flood year, and for a low flood year to be followed by another low flood year.

Was that accidental … or was there really some correlation between levels?
Did the height at year 5 have an effect on the height in year 6?

#### Are you talking about river levels … or financial stuff?

Patience. To analyze, we might do something like this:

Note the heights of the n flood levels:

h(1), h(2), … h(n)
Let m be the Mean of these levels:
M = (1/n) [ h(1)+h(2)+…+h(n) ]

Calculate the deviations from the mean:

x(1) = h(1) – M
x(2) = h(2) – M
…
x(n) = h(n) – M

Note that the set of xs have zero mean. Positive x’s indicate that the Nile level was above the average.

Now calculate the Sums:

Y(1) = x(1)
Y(2) = x(1) + x(2)
…
Y(n) = x(1) + x(2) + …+ x(n)

Note that the set of partial sums, the Y’s, are sums of zero-mean variables. They will be positive if there’s a preponderance of positive x’s. Note, too, that Y(k) = Y(k-1) + x(k).

Let R(n) = MAX[Y(k)] – MIN[Y(k)]

This difference between the maximum and minimum of the n values is called the Range. Let s(n) be the standard deviation of the set of n h-values.

As it turns out, the probability theorist William Feller proved that if a series of random variables (like the x’s) had finite standard deviation and were independent, then the so-called R/s statistic (formed over n observations) would increase in proportion to n1/2 (for large values of n).

#### Huh? The so-called R/s statistic?

Yes. Apparently lots of people are interested in this animal. This guy, R/s, is called the rescaled range

Anyway, we now have:
R(n) / s(n) ∼ kn1/2 … where k is some constant

If that were true, then we’d expect that: log(R/s ) ∼ log(k) + (1/2) log(n)

So, if we were to plot log(R/s ) vs log(n), we’d expect it to be approximately a straight line with slope (1/2).

#### A logarithm to what base?

It doesn’t matter. Anyway, what Hurst apparently found, was that the plot had a slope closer to 0.7 (rather than 0.5).

#### So, what’s that mean?

I guess it means that the annual Nile levels weren’t independent, but this year’s level might be expected to affect next year’s level. Indeed, if the slope of the log(R / s ) vs log(n) “best fit line” is H, then we’d expect:  R / s ∼ knH

You got it.

#### So what’s it got to do with financial stuff?

Patience.

Patience. The interesting thing is that many things seem to exhibit this long term patterns or dependence … seven years of plenty followed by seven years of plenty.

#### Sounds like a biblical reference.

Yes. It’s called the Joseph Effect

#### Do you realize that don’t have a single picture?

A picture is worth a thousand …

### Hurst Examples

Okay, let’s look at 300 daily returns for Exxon stock.

We’ll call them h(1), h(2), … h(300).

We calculate the Mean of these 300 returns. We’ll call it M.
M = (1/300) [ h(1) + h(2) + … + h(300) ]

Then we calculate x(1), x(2), … x(300), the 300 deviations from the Mean:
x(1) = h(1) – M, x(2) = h(2) – M, … x(300) = h(300) – M.

These devations are shown in green, in Figure 1.
(The average of these deviations is zero!)

Now we calculate the Y’s:
Y(1) = x(1), Y(2) = x(1)+x(2), … Y(300) = x(1)+x(2)+…x(300).

The Y’s are shown in red, in Figure 1. Figure 1

Now we find the maximum Y and the minimum Y and subtract them. That’s the Range, R = Max[Y] – Min[Y] … in blue. Finally we calculate the Standard Deviation of the h’s:

s = STDEV[h(k)]

#### And you get a Hurst exponent … somehow?

Okay, from the above scheme we note two magic numbers:
n = 300, R / s = 0.225 / 0.0462 = 4.87.

That’ll give us one point on our log(R/s) vs log(n) chart, namely
log(300) = 5.70 and log(4.87) = 1.58.

Now we repeat the above scheme for 310 points, then 320 points etc. etc., each time generating a point on our chart, and … Figure 2

#### Just give us the chart, okay?

Okay, see Figure 2. You see our first point? We calculated points up to n = 550.

#### And the Hurst exponent is … uh, the slope?

Yes. At least it’s an estimate of the Hurst Exponent H = 0.478.

#### Pretty close to 1/2, eh?

Yes, and that’d imply that daily returns for XOM are random, uncorrelated, a Brownian motion, independent …

Patience …