Re: Is there a probability distribution that looks similar to the McDonald's 'M'?

On May 3, 7:09 pm, Robert Moulton <robertjohnmoul...@xxxxxxxxx> wrote:

My main question now is: Is this the right approach to collect annual
return data (ie. these overlapping time periods)? Maybe the McDonalds
shape arises because of the general nature of stock market
performance: slow climbs and quick retreats?

So here is one possible explanation for the phenomenon you are
observing, although I am sure it is not the only one.

Let g(d) be the *daily* return on date d. The annual return on date D
is then

A(D) = [prod_{d = D - 364}^{D} g(d)] - 1 .

If g(d) is very close to 1, which will usually be the case, then
defining g(d) = 1 + e(d), we can to a good approximation (we do not
need this approximation, but it simplifies things) write the annual
return as

A(D) = sum_{d} Y(D - d) e(d) ,

where Y(x) = 1 if x is between 0 and 364, and 0 otherwise, and the sum
is over all dates, not just the previous year. So the annual return is
just a linear filtering of e.

The intuition behind the explanation is then this. The reason you find
the double-peaked histogram is that you are filtering the "signal", e,
at a frequency that corresponds to one of its main periodicities, in
this case, a frequency of about one year. Your filter is 'looking' for
this periodicity, and since it is present everywhere in the signal,
the output of the filter is almost never zero. Thus you get peaks at
finite values rather than at zero. The rest of this post is the
detail.

Since A(D) is a linear filtering, it can be expressed very easily in
terms of the Fourier components of the filter Y and the signal e:

A(D) = sum_{k} e^{i k D} Y(k) e(k) ,

where I have used the same symbol for the Fourier components as for
the original functions. Now suppose that e has a particularly strong
periodicity. This means that one frequency, call it k', and its
negative, -k', have very strong amplitudes e(k') and e(-k') (= e(k')*
because e is real). These two terms will dominate the sum, and we can
write the annual return as

A(D) = B cos(k'D) ,

where B = |Y(k ') e(k')|.

Now to the histogram. The histogram can be expressed like this:

H(a) = sum_{D} delta(a, A(D)) ,

where delta(x, y), gives 1 if y lies in a bucket centred on x, and
zero otherwise ,and the sum over D is over all values for which we
have data (think of this as infinite if the boundaries bother you). So

H(a) = sum_{D} delta(a, B cos(k' D)) .

It is easiest to see what this means when D is not a discrete variable
but continuous, and we shrink the bucket size to zero. The equation
for the histogram then becomes

H(a) = \int dD delta(a, B cos(k' D)) .

Setting k' = 1 (this is just a question of units), changing variables
and integrating, we find that

H(a) = 1 / sqrt(B^{2} - a^{2}) .

This is a double peaked histogram, going to infinity at a = B. This
extreme result comes from assuming that there is only one frequency in
e and that the dates are continuous. The existence of several large
amplitude frequencies, clustered together, coupled with the discrete
nature of dates, plus inevitable noise, combine to mean that in
practice the histogram will be blurred, and the peaks broader. The
result can easily come to resemble the histogram you showed.

If you average over much longer time periods, you should see this
structure disappear, and the histogram head towards Gaussian.

More generally, one can predict the histogram based on a model, i.e. a
probability distribution for the signal. It would be interesting to
see what is needed to reproduce what you observe. Certainly no
stationary Gaussian process can produce it, since the histogram is
then the same as the probability distribution of a marginal, and this
will also be Gaussian. Thus a Brownian process is ruled out, for
example.

illywhacker;

.

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