Originally Posted by

**Moo** Hello,

You can define $\displaystyle g(x,y)=(1-y)^{(n-1)x-1}y^n$ so that $\displaystyle g(\bar Y,p)=\text{ what you got}$

and hence $\displaystyle \bar Y$ is sufficient.

you can define g in any way that is more suitable to you, as long as you don't write false things ^^

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as for the sum, I don't understand why you sum from 1 to n-1...

we should have :

$\displaystyle L(y_{1},y_{2},...,y_{n}|p)=P(y_{1},y_{2},...,y_{n} |p)=(1-p)^{y_{1}}p\times\\...\times\\(1-p)^{y_{n}}p=(1-p)^{\sum_{i}^n y_i}p^{n}$

(and we should then have $\displaystyle g(x,y)=(1-y)^{nx}y^n$)