panic09

03-06-2008, 11:28 AM

Help needed. Here is the problem....

If Y is a negative binomial random variable (i.e. Y~nb(p,m)) then EY = mq/p and VarY = mq/p^2. Show that [Y - EY] / VarY converges in distribution to a standard normal distribution.

Here is my answer so far....

By Central Limit Theorem, I know sqrt(n)*(Ybar - EY)/sqrt(VarY) is distributed N(0,1). Ybar is the sum of n i.i.d. random variables divided by n so the above can be rewritten sqrt(n)*(Y - EY)/sqrt(VarY) but that is not quite what I need. I need that sqrt(n) to go away but i'm not sure where i've gone wrong.

Any help would be much appreciated...Thanks

If Y is a negative binomial random variable (i.e. Y~nb(p,m)) then EY = mq/p and VarY = mq/p^2. Show that [Y - EY] / VarY converges in distribution to a standard normal distribution.

Here is my answer so far....

By Central Limit Theorem, I know sqrt(n)*(Ybar - EY)/sqrt(VarY) is distributed N(0,1). Ybar is the sum of n i.i.d. random variables divided by n so the above can be rewritten sqrt(n)*(Y - EY)/sqrt(VarY) but that is not quite what I need. I need that sqrt(n) to go away but i'm not sure where i've gone wrong.

Any help would be much appreciated...Thanks