View Full Version : Negative Binomial Question

admin

March 13th 2005, 04:17 PM

Does anyone know what the expected value of the negative binomial distribution is? Originally I thought it was: r(1-p)/p but then I was reading "A first year in prob" by Sheldon and it said that the expected value was r/p. That made more sense because the negative is just like the geometric and the expected value of the geometric is 1/p. B/c the trials are all independent, just r/p made sense. But then I noticed several other books had r(1-p)/p. So I began to think maybe it was a typo in Sheldon. But then I saw there was a proof that it was r/p in sheldon. But then I found a website on the internet that also had a proof that it was equal to r(1-p)/p. They both looked legitimate. How can this be? Which is the real expected value of the negative binomial? Thanks, John

admin

March 13th 2005, 04:17 PM

yah it is....nq/p or u can say r(1-p)/p

krzysio

March 18th 2005, 01:45 AM

Negative binomial distribution either counts how many times you try until you succeed r times, or how many times you fail until you succeed r times. Two possible approaches to basically the same process, but the resulting means are different.

Yours,

Krzys'

leftshadow

March 22nd 2005, 10:25 PM

Negative binomial distribution either counts how many times you try until you succeed r times, or how many times you fail until you succeed r times. Two possible approaches to basically the same process, but the resulting means are different.

Yours,

Krzys'

Yup, I find this rather confusing when I fisrt looked it too.

For the first case you mention :

x is the total times you try until you succeed r times, the expected value will be r/p, (note, here the possible value for x is r, r+1, r+2 ... )

the second case:

x is the times you fail until you succeed r times, the expected value will be r(1-p)/p

(note, here the possible value for x is 0, 1, 2,...)

leerobb1966

October 12th 2008, 12:16 PM

Hi,

I read in two sources that the mean of Geometric Distribution is the reciprocal of the success rate, e.g. 1/p.

But I also read in some solution that the the mean as 1-p/p.

Which one is right?

Thanks for help.

NoMoreExams

October 12th 2008, 02:47 PM

Hi,

I read in two sources that the mean of Geometric Distribution is the reciprocal of the success rate, e.g. 1/p.

But I also read in some solution that the the mean as 1-p/p.

Which one is right?

Thanks for help.

You will benefit from reading this: http://en.wikipedia.org/wiki/Geometric_distribution and also using parantheses where appropriate. "1-p/p" means 1-(p/p) = 1-1 = 0", you obv. meant (1-p)/p :)

leerobb1966

October 12th 2008, 03:07 PM

You will benefit from reading this: http://en.wikipedia.org/wiki/Geometric_distribution and also using parantheses where appropriate. "1-p/p" means 1-(p/p) = 1-1 = 0", you obv. meant (1-p)/p :)

Thank you so much!

and I will be more careful with expressions.:tongue:

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