Negative Binomial distribution function - a question on range of X

**Volga** · January 23rd 2011, 08:38 PM

Consider the mean formula of the Negative Binomial distribution:

$\Sigma^{\infty}_{x=r}{{x-1}\choose{r-1}}p^rq^{x-r}$

I don't understand why the range for X in the summation formula starts from r. I've looked up several textbooks and they seem to be treating this as obvious - but this is not obvious to me.

(The mean formula is just an example of one of the functions of NegBin: I have the same question about any other (variance, moment generating function) formulas for NegBin which include the same range of X.)

Here is the most relevant comment on the range of R I found (internet):

"Let X be the random variable which is the number of trials up to and including r-th success. This means that the range of X is the set {r, r+1, r+2,...}."

This sounds couter-intuitive to me, if X is the number of trials UP to r, then why do we start counting from r? why not start from x=1 (first trial) until x=r (the last successful outcome we need to 'complete' the r)?

Appreciate your help.

**emakarov** · January 24th 2011, 12:41 AM

"Let X be the random variable which is the number of trials up to and including r-th success. This means that the range of X is the set {r, r+1, r+2,...}."

As this quote says, the total number of trials X is greater than or equal to the number of successes r. If you are told to wait for r successes, then you know that you need at least r trials. The actual number of trials can be r with some probability, r + 1 with some other probability and so on. The mean, therefore, is $r\cdot P(X=r) + (r+1)\cdot P(X=r+1)+\dots=\sum_{x=r}^\infty xP(X=x)$ . Since $P(X=x)={x-1\choose r-1}p^rq^{x-r}$ , the mean is $\sum_{x=r}^\infty x{x-1\choose r-1}p^rq^{x-r}$ , so I think there is a factor $x$ missing in your formula.

**mr fantastic** · January 24th 2011, 12:59 AM

Originally Posted by Volga

Consider the mean formula of the Negative Binomial distribution:

$\Sigma^{\infty}_{x=r}{{x-1}\choose{r-1}}p^rq^{x-r}$

I don't understand why the range for X in the summation formula starts from r. I've looked up several textbooks and they seem to be treating this as obvious - but this is not obvious to me.

(The mean formula is just an example of one of the functions of NegBin: I have the same question about any other (variance, moment generating function) formulas for NegBin which include the same range of X.)

Here is the most relevant comment on the range of R I found (internet):

"Let X be the random variable which is the number of trials up to and including r-th success. This means that the range of X is the set {r, r+1, r+2,...}."

This sounds couter-intuitive to me, if X is the number of trials UP to r, then why do we start counting from r? why not start from x=1 (first trial) until x=r (the last successful outcome we need to 'complete' the r)?

Appreciate your help.

Why are you summing?

The minimum number of trials is always going to be equal to the number of successes.

Read this: Negative binomial distribution - Wikipedia, the free encyclopedia

From what I can see, what you're calling a success Wikipedia is calling a failure and x = k + r. Since $0 \leq k < +\infty$ it follows that $r \leq x < +\infty$ .

**Volga** · January 24th 2011, 01:20 AM

Originally Posted by emakarov

If you are told to wait for r successes, then you know that you need at least r trials.

thanks! now THAT makes sense to me )))

yes, you are right, x is missing from the mean formula (I was deliberating which 'summation' formula to include, and ultimately made a mistake)

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