Now we'll consider the following Density Distribution function: (1)       f(x) = A e-k |x-m| where m is the mean and |x-m| is the absolute value of x-m and A and k are as-yet-unknown constants. This has more "tail" than a function which decays like the Normal distribution. A sample of such a function looks like this Remember (from Part III) a magic requirement for the function f(x), namely

This requires that A = k/2, so that our function now looks ...
>I assume you can prove that?
Want to see the proof?
>Definitely not.
Our function now looks like:

(2)       f(x) = (k/2) e^{-k |x-m|}

Further, we'll borrow from the Log-normal scheme and assume that the logarithms of the Gain Factors have this distribution.
That gives us:

(3)       f(x) = (k/2x) e^{-k |log(x/G)|}     where x > 0   since x is now a Gain Factor

>Don't tell me! m is now the average of the logs, so m=log(G) where ...
Yes, as in Part III, G is the geometric Mean of the numbers {g} whose distribution we're discussing.
>And you're doing that funny dz = (1/x) dx because of the logarithm so that f gets divided by x and ...
Of course.

>So if the numbers we're discussing are g1, g2, etc. then ... ... then G = { g1g2...gN }(1/N) >Why do you always interrupt? Can't I finish ...? Pay attention. We're forging on. Here's a picture of our distribution with k = 25 ... chosen to be the best match to the S&P 500. Notice that our Mystery Distribution has a tail greater than the actual S&P distribution, meaning that there is a greater opportunity for returns far from the average.

>What's the Cumulative distribution look like, compared to ...? Here they are, where we're now back to the monthly gains (rather than the Gain Factors). To do this we just subtract "1" from the Gain Factors which slides the graph to the left by an amount "1". >This Mystery Distribution is your invention? No, not really. It's called the exponential distribution. >I asume there's a magic formula for the mystery F(x)? Yes, and it's pretty simple: F(x) = (1/2) (x/G)k when 0 < x < G = 1 - (1/2)(x/G)-k when x > G

>Where x is the Gain Factor, namely 1+Gain Fraction so it's always positive, and G is ...
G is the Geometric Mean of the Gain Factors which are always greater than zero.
>So what's the magic formula for f(x)? It's the slope of F(x), right?
Yes. Although we've given it above, the simpler version is:

f(x) = (k/2x) (x/G)k when 0 < x < G = (k/2x)(x/G)-k when x > G or f(x) = (k/2G)(x/G)p where p = k-1   when 0 < x < G and p = -k-1   when x > G and x is a Gain Factor and G is the Geometric Mean of the Gain Factors

>Is that your Mystery Distribution? Well, I can give you another ... >Please don't! Here's another:     (1)       f(x) = A e -k(x-m)2/(b+|x-m|) where we can choose the numbers A, k and b so as satisfy the magic distribution requirement (noted above) and so as to match, for example, the historical S&P 500 returns. The figure on the right has such a choice and it seems a pretty good match, eh?

Undoubtedly, you'll have noticed that, when x is far from its mean (which is denoted by m), this distribution function has an exponential decrease similar to the first Mystery Distribution (in equation (1), above). In fact, for large values of |x-m|, our new distribution looks like: A e^-k|x-m|, which means ...
>Undoubtedly?
... which means we've still got those bigger tails.
>So, what do you intend to do with this stuff?
Hmmm ... give me a minute to think about it.

for next Part

    See also Distribution Stuff