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At 4:59 AM -0500 8/28/02, John Lynch wrote:
>Can anyone confirm that trading results are binomially distributed?
I usually find the distribution is a "normal distribution" but it can
be truncated if you use a hard stop for money management.
But the difference between the binomial and normal distribution is
very small. I will use the coin-flip example as an illustration. The
data is in the table below.
#Wins Prob Value ExpValue Normal
0 0.0010 $5.631 $0.005 0.0020
1 0.0098 $9.386 $0.092 0.0115
2 0.0439 $15.643 $0.687 0.0440
3 0.1172 $26.071 $3.055 0.1147
4 0.2051 $43.452 $8.911 0.2039
5 0.2461 $72.420 $17.822 0.2470
6 0.2051 $120.699 $24.753 0.2039
7 0.1172 $201.166 $23.574 0.1147
8 0.0439 $335.276 $14.734 0.0440
9 0.0098 $558.794 $5.457 0.0115
10 0.0010 $931.323 $0.909 0.0020
Sum 1.0000 $100.000 0.9995
Column 1 is the number of wins out of the ten coin flips.
Column 2 is the probability of each occurrence (which is a binomial
distribution).
Column 3 is the value of each combination
Column 4 is the expected value (Column 2 * Column 3) Note that they
sum to $100 as Aaron Schindler pointed out.
Column 5 is a best least-squares-fit normal distribution. Note the
values are extremely close to the value in column 2 except at the
tails of the distribution.
The two are compared in the attached logarithmic plot.
The red curve is the binomial distribution corresponding the the coin
flip problem we have been discussing. The black curve is the best-fit
normal distribution. Note that they are very similar.
Bob Fulks
Attachment:
Description: "Distribution.gif"
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