Bayesian Statistics

**sirellwood** · February 24th 2011, 09:54 AM

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.

**theodds** · February 24th 2011, 10:41 AM

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.

**sirellwood** · February 24th 2011, 11:07 AM

Argh! My apologies.... original post is now edited. Thank you theodds

**harish21** · February 24th 2011, 11:29 AM

Note that $X_i$ s are independent , and for binomial RVs, the sum $\displaystyle S=\sum_{i=1}^n X_i \sim Binomial(kn,\theta)$

**sirellwood** · February 24th 2011, 01:41 PM

ok so im guessing that i begin like that for part (ii) of this question?

**sirellwood** · February 24th 2011, 03:43 PM

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 :-(

**harish21** · February 24th 2011, 05:22 PM

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.

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