# Homework Help: Probability problem

1. Aug 16, 2004

### jimbo007

hi
i am trying to show that for 2 probabilities $$p_{0}$$ and $$p_{1}$$ that $$X=\frac{p_{1}}{p_{0}}$$ cannot be further from 1 than $$Y=\frac{p_{1}(1-p_{0})}{p_{0}(1-p_{1})}$$

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

for case 1 we have $$X=Y=1$$ so X is not further from 1 than Y

for case 2 we require that $$\frac{1-p_{0}}{1-p_{1}}<1$$ for Y to be less than X
this is true if $$1-p_{0}<1-p_{1}$$
<=> $$p_{0}>p_{1}$$
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.

2. Aug 16, 2004

### HallsofIvy

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.

3. Aug 16, 2004

### jimbo007

$$\frac{1-p_{0}}{1-p_{1}}<1$$ (*)
(*) is true iff $$1-p_{0}<1-p_{1}$$
iff $$p_{0}>p_{1}$$ (**)