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I've attached my formula as a png, hope it gets through, in case it doesn't...
Text:
E=|_sum_{{i=0};{i=M-m};{|_pmatrix1x2_{{M};{m+i}}ε^{m+i}}}
LaTeX:
\mbox{E}=\sum_{i=0}^{i=M-m}{\left(\begin{array}{c} M \\ m+i \end{array}\right)\epsilon ^{m+i}}
...where m is the bigger half - that is the `half' that has just enough power to out-weight the other half, for example where M = 20, m = 11, where M = 5, m = 3.
Again, I don't think I have the right, or at least most elegant solution to the problem, I feel it has to be much simpler - otherwise the question wouldn't seriously expect us to calculate the above formula for M = 20 as that sum would expand into 9 terms!
Nima
On Fri, Oct 24, 2008 at 12:37 AM, Nima Talebi <nima.talebi@...> wrote:
Hi,
I'm interested to see if my solution is even remotely correct, the question is to derive the error formula given the number of experts/hypotheses (M) and also the error of each hypotheses (e) - assuming all to be the same, and that all are independent of one another...
I treated this as a probability problem (and unsure if I should have done so), and hence derived a formula which yields the following results...
M:5, e:0.10, 0.0100
M:5, e:0.20, 0.0800
M:5, e:0.40, 0.6400
M:10, e:0.10, 0.0002
M:10, e:0.20, 0.0134
M:10, e:0.40, 0.8602
M:20, e:0.10, 0.0000
M:20, e:0.20, 0.0034
M:20, e:0.40, 7.0448
The third column is the total error for the network I calculated; it certainly looks wrong from looking at the last value, but I'd like to know how to go about solving this as it's rather intriguing.
My `formula' in Python form is as follows...
def fact(k):
return reduce(lambda i, j : i*j, range(1, k+1))
for M in 5, 10, 20:
for e in 0.1, 0.2, 0.4:
m = M/2+1
_m = M-m
print "M:%d, e:%0.2f, %0.4f"%(M, e, (fact(M) * pow(e, m))/(fact(_m) * fact(m)))
Thanks to anyone who has a crack at solving this =)
Nima
--
Nima Talebiw: http://ai.autonomy.net.au/
p: +61-4-0667-7607 m: nima@...
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