The attachment was stripped as suspected, so here's the readable version...

http://ai.autonomy.net.au/chrome/site/AIMA-18.4.png

Nima

--- In aima-talk@yahoogroups.com, "Nima Talebi" <nima.talebi@...> wrote:
>
> Okay, I can already see a flaw in this, here's my revised version...
>
> 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@...
>