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Re: Excercise 18.14, page 677   Message List  
Reply Message #846 of 868 |
Excercise 18.14, page 677

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



Thu Oct 23, 2008 1:37 pm

daashmashty
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Message #846 of 868 |
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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:* ...
Nima Talebi
daashmashty
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Oct 23, 2008
4:23 pm

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...
Nima Talebi
daashmashty
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Oct 23, 2008
4:23 pm

The attachment was stripped as suspected, so here's the readable version... http://ai.autonomy.net.au/chrome/site/AIMA-18.4.png Nima...
daashmashty
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Oct 26, 2008
10:46 pm
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