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Excercise 18.14, page 677

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  • Nima Talebi
    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

    • daashmashty
      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 2 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}}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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