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Homework Help: Probability problem

  1. Aug 16, 2004 #1
    hi
    i am trying to show that for 2 probabilities [tex]p_{0}[/tex] and [tex]p_{1}[/tex] that [tex]X=\frac{p_{1}}{p_{0}}[/tex] cannot be further from 1 than [tex]Y=\frac{p_{1}(1-p_{0})}{p_{0}(1-p_{1})} [/tex]

    how i went about the problem was as follows:
    i split it into 3 cases, case 1 is where [tex]p_{0}=p_{1}[/tex], case 2 is where [tex]p_{0}>p_{1}[/tex], and case 3 is [tex]p_{0}<p_{1}[/tex]

    for case 1 we have [tex]X=Y=1[/tex] so X is not further from 1 than Y

    for case 2 we require that [tex]\frac{1-p_{0}}{1-p_{1}}<1[/tex] for Y to be less than X
    this is true if [tex]1-p_{0}<1-p_{1}[/tex]
    <=> [tex]p_{0}>p_{1}[/tex]
    which is what we were assuming in first place

    case 3 is a similar argument but with the inequality sign changed from less than to greater than.

    i dont think the reasoning is that great, for example can i assume that multiplying a number by another number that isless than 1 but greater than 0 will result in a number smaller than the original. also unsure about the implications in case 2.

    i need to be shown how to improve the reasoning as i'm not satisfied with the current way it is but i cant think of any way to improve it.

    thanks in advance
     
  2. jcsd
  3. Aug 16, 2004 #2

    HallsofIvy

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    Science Advisor

    A note on the way you expressed this: you said "X cannot be further from 1 than Y" and I at first interpreted this to mean |1-X|< Y. But you clearly mean |1-X|< |1-Y|.

    In tiny steps- If p0> p1 then p0/p1= X > 1 so |1-X|= X-1. Also, in this case, -p0> -p1 so 1-p0> 1-p1 and, finally, (1-p0)/(1-p1)> 0. Then
    Y> X> 1 so |1-Y|= Y-1. You need to show X-1< Y-1 or X< Y (which we said above).
    If p0 < p1 then the inequalities work the other way: Y< X< 1 so |1-X|= 1-X and |1-Y|= 1-Y. Now you need to show 1-X< 1-Y or X>Y (again, we just said that).

    By the way, since probabilites CAN be equal to 0 or 1 you should state explicitely that your inequality does not work in those cases.
     
  4. Aug 16, 2004 #3
    thanks for the reply!
    i'm not sure i completely understand. for starters i cant see how you can say that -p0> -p1 and p0>p1. doesnt mutliplying by a negative change the inequality. and in your tiny step guide i see it boils down to showing whether X<Y, which was the the part i wasnt sure how to do.
    case 2 for example (p0>p1):
    do i show X<Y by first saying (along with inlcuding your "tiny steps") that we require
    [tex]\frac{1-p_{0}}{1-p_{1}}<1[/tex] (*)
    (*) is true iff [tex]1-p_{0}<1-p_{1}[/tex]
    iff [tex]p_{0}>p_{1}[/tex] (**)
    but we know (**) by assumption so (*) must also be true hence X<Y?

    i think there were some subtleness in your proof but it kind of went *whoosh* straight over my head
     
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