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Math Help - factorization criterion help

  1. #1
    Senior Member Danneedshelp's Avatar
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    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)<br />
=(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
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  2. #2
    Moo
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    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)
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  3. #3
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by Moo View Post
    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, ....
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  4. #4
    Moo
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    Quote Originally Posted by Danneedshelp View Post
    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...
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  5. #5
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by Moo View Post


    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.
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