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If $X_1$ follows Binomial distribution with parameter $m$ and $p$ where $m$ is the number of trials and $p$ is the probability of success , that is , $X_1\sim B(m,p)$ and $X_2\sim B(n,p)$ then how can i prove that $$\hat p=\frac{X_1+X_2}{m+n}$$ is a sufficient statistic.

I know the way to find sufficient statistic by Factorization theorem. Also I have seen the link.

But i couldn't proceed to prove $\hat p$ is a sufficient statistic.

Community
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user 31466
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2 Answers2

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You have that $X_1+X_2 \sim B(m+n,p)$ (assuming that $X_1, X_2$ are independent). Then the result follows from the proof that the mean (or the sample proportion $\left(\hat{p}=\frac{X}{n}\right)$ is a sufficient statistic for the binomial distribution. This proof can be found as example in many books or online internet sources. For example see here.

Jimmy R.
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    $X_1+X_2$ is not a $B(m+n,p)$ random variable unless $X_1$ and $X_2$ are independent random variables. – Dilip Sarwate Feb 24 '14 at 14:24
  • @DilipSarwate. I thought it was obvious, but actually it was not mentioned in the question. Thanks for the comment, I edited the answer to include it. – Jimmy R. Feb 24 '14 at 15:58
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By theorem:

If $p(x|\theta)$ is the joint pdf or pmf of X and $q(t|\theta)$is the pdf or pmf of T(X), then T(X) is a sufficient statistic for $\theta\space$if, for every x in the sample space, the ratio $p(x|\theta)/q(T(x)|\theta)$is constant as a function of $\theta.$ $\frac{p^{x_1+x_2}*(1-p)^{m+n-x_1-x_2}}{\binom{m+n}{t}p^t(1-p)^{m+n-t}}$ $=\frac{1}{\binom{m+n}{t}}$which is constant as a function of $\theta$. So T(X)=$\frac{X1+X2}{m+n}$is a sufficient statistic

Dong Xu
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