# Bayesian Statistics - Factorization Theorem

• Feb 24th 2011, 09:54 AM
sirellwood
Bayesian Statistics - Factorization Theorem
Hi all, having real problems doing this question. Can anyone explain the steps of how its done (leaving the process of actually carrying out the steps to me! :-))

If X1,X2, . . . ,Xn are independent random variables from the Binomial distribution
Bin(k, $\theta$), show that S = $\displaystyle\sum\limits_{i=1}^n$ Xi is sufficient for $\theta$
(i) by using the Factorization Theorem;
(ii) by finding the distribution of X1,X2, . . . ,Xn conditional on S = s.
• Feb 24th 2011, 10:41 AM
theodds
Quote:

Originally Posted by sirellwood
Hi all, having real problems doing this question. Can anyone explain the steps of how its done (leaving the process of actually carrying out the steps to me! :-))

If X1,X2, . . . ,Xn are independent random variables from the Binomial distribution
Bin(k, $\theta$), show that S = $\displaystyle\sum\limits_{i=0}^n i^3$ Xi is sufficient for $\theta$
(i) by using the Factorization Theorem;
(ii) by finding the distribution of X1,X2, . . . ,Xn conditional on S = s.

Did you write S correctly? $T = \sum X_i$ is minimal sufficient for theta in this problem, and it doesn't look to me like the S you gave is a function of T.
• Feb 24th 2011, 11:07 AM
sirellwood
Argh! My apologies.... original post is now edited. Thank you theodds
• Feb 24th 2011, 11:29 AM
harish21
Note that $X_i$s are independent , and for binomial RVs, the sum $\displaystyle S=\sum_{i=1}^n X_i \sim Binomial(kn,\theta)$
• Feb 24th 2011, 01:41 PM
sirellwood
ok so im guessing that i begin like that for part (ii) of this question?
• Feb 24th 2011, 03:43 PM
sirellwood
Ok i can do part (ii), the direct method now, but still having problems with the factorization method. I know that f $\theta$(x) = g(T(x), $\theta$)h(x)

but im not able to express this for the binomial distribution :-(
• Feb 24th 2011, 05:22 PM
harish21
the RVs $X_1, \cdots, X_n$ form Binomial $(k, \theta)$, and the pmf is

$f(x|\theta)= \dbinom{k}{x} {\theta}^x (1-\theta)^{k-x}\;I_{0,\cdots,k}(x)$

Now,

$\displaystyle f(x_1,\cdots,x_n|\theta)= \prod_{i=1}^n f(x_i|\theta)=\prod_{i=1}^n \bigg[\dbinom{k}{x_i}\theta^{x_i} (1-\theta)^{k-x_i}\;I_{0,\cdots,k}(x_i)\bigg]$

$\displaystyle = {\theta}^{\sum_{i=1}^n x_i} (1-\theta)^{\sum_{i=1}^n 1-x_i}\;\prod_{i=1}^n\bigg[\dbinom{k}{x_i} I_{0,\cdots,k}(x_i)\bigg]$

simplify further and remember the factorization theorem to come to your conclusion.