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

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

  • $\begingroup$ 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. $\endgroup$
    – AdamO
    May 3, 2022 at 16:45
  • 2
    $\begingroup$ $Y$ is sufficient. If you can find a formula to compute $Y$ from $W$--an easy task--then $W$ must be sufficient, too. $\endgroup$
    – whuber
    May 3, 2022 at 16:56
  • $\begingroup$ Please add the self-study tag & read its wiki. $\endgroup$ May 7, 2022 at 18:39

1 Answer 1


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

$$ = p^{y_1+...+y_n}(1-p)^{n-(y_1+...+y_n)}= $$

$$ =p^{\sum y_i}(1-p)^{n-\sum y_i} $$

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.


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