1. ## factorization criterion help

Q: If $Y_{1},Y_{2},...,Y_{n}$ denote a random sample from a geometric distribution with paramter $p$, show that $\bar{Y}$ is sufficient for $p$.

A:

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

So, I have that $h(y)=1$ and $g(\sum_{i}^{n-1}y_{i=1}-1,p)=(1-p)^{\sum_{i=1}^{n-1}y_{i}-1}p^{n}$ . Since the unkown parameter $p$ only interacts with the data $y$ through $U=\sum_{i}^{n-1}y_{i}-1$, $U=\sum_{i=1}^{n-1}y_{i}-1$ is sufficient for $p$.

Now, can I just rewrite the sum as follows to show that $\bar{Y}$ is sufficient for $p$,

$\sum_{i=1}^{n-1}y_{i}=(n-1)\bar{Y}$?

Also, would it be alright if I started the index from zero or do I have to start from one, since the collection of random variables in the problem statement are started at $i=1,2,...n$? I think that would help clean things up a bit.

Thanks

2. Hello,

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

and hence $\bar Y$ is sufficient.

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

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

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

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

3. Originally Posted by Moo
Hello,

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

and hence $\bar Y$ is sufficient.

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

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

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

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

$g(\bar{Y},p)=(1-p)^{(n-1)\bar{Y}-1}p^{n}$ works as my g function, as does $g'(\bar{Y},p)=(1-p)^{(n)\bar{Y}}p^{n}$?

I guess I was getting mixed up, because I the RV's can be expressed in two ways

for k = 1, 2, 3, ....
Equivalently, if the probability of success on each trial is p, then the probability that there are k failures before the first success is
for k = 0, 1, 2, 3, ....

4. Originally Posted by Danneedshelp
So,

$g(\bar{Y},p)=(1-p)^{(n-1)\bar{Y}-1}p^{n}$ works as my g function, as does $g'(\bar{Y},p)=(1-p)^{(n)\bar{Y}}p^{n}$?
No, your g is not correct. There are always n variables, it doesn't depend on the type of geometric distribution you choose.
I was just using (n-1) etc... to show how your formula would work, but it was wrong.

I guess I was getting mixed up, because I the RV's can be expressed in two ways

for k = 1, 2, 3, ....
Equivalently, if the probability of success on each trial is p, then the probability that there are k failures before the first success is
for k = 0, 1, 2, 3, ....
Yup, I would've guessed...

For the first one, the likelihood function is $p(1-p)^{y_1-1}\times \dots \times p(1-p)^{y_n-1}=p^n (1-p)^{\sum_{i=1}^n \{y_i-1\}}=p^n(1-p)^{\sum_{i=1}^n y_i-n}$

and we'd have $g(\bar Y,p)=p^n(1-p)^{n\bar Y-n}$

For the second one, the likelihood function is, in a similar way, $p^n(1-p)^{\sum_{i=1}^n y_i}$

which gives $g(\bar Y,p)=p^n (1-p)^{n\bar Y}$

Sorry if I confused you when I tried to pursue your reasoning and only correcting the mistake at the end...

5. Originally Posted by Moo

Sorry if I confused you when I tried to pursue your reasoning and only correcting the mistake at the end...
No worries, I got it now. Thank you ver much.