- 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:**

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!

NimaOn 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@... - 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@...

>