If $X_1,\ldots,X_n$ is an IID random sample, with $X_i\sim\,\text{Ber}(\theta)$, prove that $Y = \sum_i X_i$ is sufficient using the definition of sufficiency (not the factorization criterion).

Now the definition of sufficiency I'm given is:

A statistic $Y$ is sufficient if the joint distribution of the sample given the statistic does not depend on $\theta$ or

$$ P_\theta(X_1=x_1,\ldots, X_n=x_n | Y_n = y)= f(x_1,\ldots,x_n,y), $$ for some non-negative function $f$ not depending on $\theta$.

My attempt:

Since $Y$ is the sum of $n$ Bernoulli random variables, $Y\sim \text{Bin}(n,\theta)$. Hence

$$ P_\theta(X_1=x_1,\ldots, X_n=x_n | Y_n = y) = \frac{P_\theta(X_1=x_1,\ldots, X_n=x_n, Y_n = y)}{{n \choose y} \theta^y (1-\theta)^{n-y}} $$

but here I'm stuck as I have no idea how to compute the probability in the numerator. Any help or hint is highly appreciated.


1 Answer 1


You are almost there. Note that after $X_1,\ldots, X_n$ has been fixed at $x_1,\ldots, x_n$, $Y$ is automatically fixed. Therefore, your numerator becomes

$$ P_\theta(X_1=x_1,\ldots, X_n=x_n, x_1+\ldots+x_n = y) =\\=P_\theta(X_1=x_1,\ldots, X_n=x_n) \\=\theta^{\sum_i x_i} (1-\theta)^{n-\sum_i x_i}. $$

Now take the ratio and you have the desired result.

  • $\begingroup$ I see, my bad, so trivial! Thank you very much. $\endgroup$
    – laurab
    Nov 22, 2022 at 21:04

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