= Bernoulli sample sum as a sufficient statistic
{c}
{title2=$T=\sum_{i=1}^nX_i$}
Let $n$ be fixed and known, and let $X_1,\ldots,X_n$ be <independent and identically distributed> <Bernoulli random variables> with common success <probability> $p$. Their joint <probability mass function>, on the binary <sample space> $\{0,1\}^n$, is
$$
P_p(X=x)=\prod_{i=1}^n p^{x_i}(1-p)^{1-x_i}=p^{t}(1-p)^{n-t},\qquad t=T(x)=\sum_{i=1}^nx_i.
$$
There are $\binom nt$ arrangements with $t$ successes, so $T$ has the <binomial distribution>. For $0<p<1$ and $t\in\{0,\ldots,n\}$, its exact conditional full-data <probability mass function> is
$$
P_p(X=x\mid T=t)=\frac{p^t(1-p)^{n-t}\mathbf1_{\{T(x)=t\}}}{\binom nt p^t(1-p)^{n-t}}=\frac{\mathbf1_{\{T(x)=t\}}}{\binom nt}.
$$
This <conditional distribution> is uniform over arrangements, with no dependence on $p$: the sum is a <sufficient statistic>. <Sufficiency (Stat 210a, Fall 2025) by Will Fithian>[Sufficiency, Will Fithian, “Independent Bernoulli sequence”: conditional pmf]{ref}
\Image[/bernoulli-sufficient-statistic-grouped-sequences.png]
{title=Four Bernoulli observations grouped by their sufficient sum}
{description=All sixteen binary sequences are grouped by $T$. Within each group the <conditional distribution> is uniform; changing $p$ changes the group's total <probability>, not the conditional weights within it.}
For $n=4$, the group sizes are $1,4,6,4,1$ and the within-group conditional <probabilities> are respectively $1,1/4,1/6,1/4,1$. For example, $T=2$ leaves exactly $0011,0101,0110,1001,1010,1100$, each with conditional <probability> $1/6$. At $p=1/4$, each such sequence has unconditional <probability> $9/256$ and the group has <probability> $54/256$. At $p=1/2$, these values are $1/16$ and $6/16$; at $p=3/4$ they return to $9/256$ and $54/256$. The conditional weights stay $1/6$ in all three cases.
At the boundary $p=0$, only $T=0$ has positive <probability>; at $p=1$, only $T=n$ does. Dividing by $P_p(T=t)$ on the other groups would divide by zero. Nevertheless, the same uniform-on-arrangements <Markov kernel> is a valid common version for the entire family $0\leq p\leq1$: it gives the correct point mass on the one observed group at each boundary and assigns permissible versions on its unobserved groups. Define the joint <probability mass function> there by the original product of the one-trial <Bernoulli> <probability mass functions>, or use the limiting convention $0^0=1$ in the compact formula.
The <sample mean> $T/n$ is an <unbiased estimator> of $p$ and retains the <sufficient statistic> because $n$ is known. By contrast, $X_1$ is also an <unbiased estimator> of $p$ but, for $n\geq2$, is not sufficient: the conditional success <probability> of the <independent> $X_2$ given $X_1$ is still $p$. Thus unbiasedness does not imply sufficiency.
Sufficiency also does not imply <completeness of a statistic>. For $n\geq2$, the full-data <sufficient statistic> $X$ admits the nonzero function $g(X)=X_1-X_2$ with <expectation> zero for every $p$. Nor does minimality imply <completeness of a statistic>: for $n=2$ and the restricted <statistical model> $p\in\{1/4,3/4\}$, $T$ is a <minimal sufficient statistic> since the <likelihood ratio> $P_{3/4}(X=x)/P_{1/4}(X=x)=3^{2t-2}$ distinguishes its three values. Yet $g(0)=3,g(1)=-5,g(2)=3$ has <expectation> $(27-30+3)/16=0$ at both parameter values and is never zero. Changing the parameter family can change <completeness of a statistic> even while preserving sufficiency.
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