# Proving a negative binomial distribution problem

#### StatStudent3

##### New Member
Given that P(k), k=0, 1, etc...denote the probabilities in the negative binomial (r,p) distribution. Show that the consecutive odds ratio are

P(k)/P(k-1) = (r + k -1)*q/k (k=1,2,3, etc) <-- Prove

Isn't P(k) = (r + k -1
k - 1) * p^k * q^r

Isn't P(k-1) = (r + k -2
k -2) * p^k-1 * q^r

I've looked in my textbook, and the examples seem to say the same thing, however, the proof is just not adding up.

#### StatStudent3

##### New Member
What is P(k-1)?

I'm having trouble finding an exact formula for this - I understand P(k), but P(k-1) is tricky.

#### StatStudent3

##### New Member
Something still doesn't add up - I applied P(k-1) exactly how P(k) was applied(as shown in that link), however, the proof in the OP still doesn't add up.

P(k-1) = (k-1+r-1
r-1) * p^k * q^r-1

If I use this,then everything prove's, however, this itself doesn't add up. Shouldn't it be p^k-1?

Edit: if p^k-1 is used, then the proof still doesn't add up.

#### ichbin

##### New Member
Plugging in the formula for the binomial coefficient in terms of factorials,

$$P(k) = \frac{(k+r-1)!}{(r-1)! \, k!} p^k q^r$$

So...

$$P(k) = \frac{(k+r-1)!}{(r-1)! \, k!} \frac{(r-1)! \, (k-1)!}{(k+r-2)!} \frac{p^k q^r}{p^{k-1} q^r} = \frac{(k+r-1)! \, (k-1)!}{(k+r-2)! \, k!} p = \frac{k+r-1}{k} p$$

The only difference between this and your original assertion is $$q \Leftrightarrow p$$, which presumably comes down to your textbook defining p as the probability of failure while the Wikipedia article defines q as the probability of failure.

#### StatStudent3

##### New Member
I have a few more questions on other homework problems, but I'll go ahead and post them in this thread.

The average income(u) in an area is $10K, what is the upper bound for percentage of families with incomes over$50K, and standard deviation of incomes is $8K? The way I set it up was P(|X - u| >=$50K) <= $8K^2 /$50K^2

Where SD = $8K, Var =$8K^2, t = $50K, and t^2 =$50K^2.......From here, with 8K^2/50K^2 = .0256, I got the upper bound of 1/6^2 = 1/36(where k=6). This using Chebychev's.

Is .0256 or 1/36 the upper bound here?