# Sufficient statistics for bernoulli distribution

Let $$Y_1, \ldots, Y_n$$ be a random sample of size $$n$$ where each $$Y_i \sim \textrm{Bernoulli}(p),$$ and let $$Y = \sum Y_i$$ for $$i = 1, \ldots, n.$$

The estimator is $$W= (Y+1)/(n+2).$$

Is the estimator a sufficient statistics for parameter p?

I wanted to use the factorization theorem for this problem and wrote out the joint pdf, but I was stuck on rearranging the joint pdf

$$f(y_1,\ldots, y_n;p)= \prod p^{y_i}(1-p)^{1-y_i}.$$

• Do you mean to ask whether the estimator is a function of the sufficient statistic? You know what the sufficient statistic is. Why not just show that W is not a function of that? To factorize the bernoulli likelihood, write it in exponential form, i.e. f(...) = exp(a + bT) where a, the ancillary parameter, does not depend on p. May 3, 2022 at 16:45
• $Y$ is sufficient. If you can find a formula to compute $Y$ from $W$--an easy task--then $W$ must be sufficient, too.
– whuber
May 3, 2022 at 16:56
• Please add the self-study tag & read its wiki. May 7, 2022 at 18:39

$$f(y_1...y_n;p)= \prod^n_{i=1} p^{y_i}(1-p)^{1-y_i}=$$
$$= p^{y_1+...+y_n}(1-p)^{n-(y_1+...+y_n)}=$$
$$=p^{\sum y_i}(1-p)^{n-\sum y_i}$$
By Neyman-Fisher factorization theorem the statistic $$Y$$ is sufficient for parameter $$p$$.
$$Y = (n+2)W-1$$, i.e. $$W$$ is a one-to-one function of a sufficient statistic $$Y$$. Hence, $$W$$ is also a sufficient statistic.