Binomial Distribution Related Proof

**bzgeb** · March 10th 2010, 09:14 AM

I'm having a tough time with this problem:

I managed to do the proof (part a) no problem. However, I don't know where to start when it comes to part b.

If somebody could help me get started that'd be great.

**qmech** · March 10th 2010, 09:41 AM

All you need to do is check if P(k+1) is bigger or less than P(k).

This is determined by the multiplier:
$\frac{p}{1-p}\frac{n-k}{k+1}$

and you need to check when
$\frac{p}{1-p}\frac{n-k}{k+1} < 1$

because that's when the P() terms start to decrease.

**bzgeb** · March 10th 2010, 06:12 PM

I still don't see how to do it. Lets say I want to determine if P(k=0) is less than or equal to P(k=1). This would allow me to show that the function starts by increasing. However, if you plug the values in, you get $\frac{p^2(n^2-n)}{2(1-p)^2}$ and $\frac{pn}{1-p}$ . I don't see how to determine which one is greater without knowing the values.

That being sad, I also don't know how to determine when $\frac{p}{1-p}\frac{n-k}{k+1} < 1$ for the same reason. It's too abstract, I don't see how you can determine it without knowing some of the values.

**qmech** · March 11th 2010, 09:15 AM

It's algebra:
$\frac{p}{1-p}\frac{n-k}{k+1} < 1$

multiply out the denominators:

$p(n-k) < (1+k)(1-p)$

rearrange:

$pn-pk < 1+k -p-pk$
$pn+p-1 < k$
$k > pn+p-1$

So the multiplier is less than 1 when k > pn+p-1. That means the largest P(k) happened before this...

**bzgeb** · March 12th 2010, 06:36 AM

Ah it makes sense now. I understand the problem more clearly. Thanks for you help.

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