- #1

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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