## Excercise 18.14, page 677

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• 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
Message 1 of 3 , Oct 23, 2008
• 0 Attachment
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

• 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:*
Message 2 of 3 , Oct 23, 2008
• 0 Attachment
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 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@...
• The attachment was stripped as suspected, so here s the readable version... http://ai.autonomy.net.au/chrome/site/AIMA-18.4.png Nima
Message 3 of 3 , Oct 23, 2008
• 0 Attachment
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}}Îµ^{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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