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270Statistical Learning Example Section 20.1

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  • pcomitz
    Dec 3, 2003
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      Has anyone tried to work the example on pages 713/714 (second
      edition)? I cannot got equation 20.1 to match the graph on page 714
      (figure 20.1(a)).

      For example, consider P(h5|d) = alpha * P(d|h5)P(h5). In this case P
      (d = lime|h5) is always 1. For the ten trials,

      P(d1 ...d10|h5) = 1**10 = 1. (20.3)
      P(hi) = .1

      So P(h5|d) = (.1 * alpha) from equation 20.1. Assume alpha = 1/P(d =
      lime) = 1/.5 = 2. So after one candy is unwrapped, P(h5|d) = .2 as
      shown in figure 20.1(a).

      On the next trial (i.e. second candy is also lime) P(d|h5) = 1 (as
      above from eqn 20.3). Then P(h5|d) = (1) * (.1) * alpha. The value on
      figure 20.1 is shown as approximately .3 - implying alpha = 3
      Subsequent calculations for alpha appear to 4, 5 etc.

      I can't seem to see how alpha is calculated ..... Any opinions are
      greatly appreciated.

      Paul Comitz