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f(x) = $\binom{(r+x)-1}{x-1}p^{r}(1-p)^{x}$ noting 0 < p<1 and r>0 $$\sum_{X}^\infty \binom{(r+x)-1}{r-1}p^{r}(1-p)^{x} =1.$$ x=0,1,2,3......... I think it looks like this: $$\sum_{X}^\infty \binom{(r+x)-1}{r-1}p^{r}(1-p)^{x} =$$ I think i put an e^tx inside of the sumation. But do i try to turn all the x's to x+r or the r's to r+x?

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up vote 1 down vote accepted

Your expressions are not equivalent.

Perhaps it would be helpful for you to consider the following identity.

$$ {x+r-1 \choose x-1}={x+r-1 \choose r} $$

In terms of x=failures and r=successes in the first we are counting the number of ways we can choose the x-1 failures (since the xth failure ends the trial and is determined), and in the other, the ways we can choose r successes. They both exist in the same basket that we are counting from and we know that the total number of events in the basket is x+r-1.

To count r successes before x-1 failures doesn't require a change to the pmf. Changing the r to r+x, or x to r+x in the exponents is equivalent to changing the number or success or failures needed to satisfy the problem and would require changes to the combinatoric element of the pmf as well. It would be a whole different probability.

Your understanding of the definition of the moment generation function is correct though.

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i got it thank you! – user43126 Oct 1 '12 at 22:31

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