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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
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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 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 Talebi
      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
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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:
        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@...
      • 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 3 of 3 , Oct 23, 2008
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          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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