# Check for sufficiency

Suppose $X_1,X_2\ \text{and}\ X_3$ are three independent and identically distributed Bernoulli random variables with parameter p, 0 < p < 1. Verify if the following statistics are suﬃcient for p

$(a) X_1 + 2X_2 + X_3 \\ (b) 2X_1 + 3X_2 + 4X_3$

I cannot find any one-one correspondence between the given statistics and $T(X)=X_1+X_2+X_3$,so that I can use the invariant property of sufficient statistics.

## 1 Answer

Your intuition is correct. The fact that you can't write $$T(X)=X_1+X_2+X_3$$ as a function of statistics $$(a)$$ and $$(b)$$ makes $$(a)$$ and $$(b)$$ not sufficient.

You can prove this in various ways. I'll write two here:

1. (a) and (b) don't hold the Fisher–Neyman factorization theorem. In fact, the joint distribution of the random sample is (since the random variables are iid, aka independent and identically distributed): $$\prod_{i=1}^3 p^{x_i} (1-p)^{1-x_i}=p^{\sum_{i=1}^3 x_i}(1-p)^{3-\sum_{i=1}^3 x_i}=p^{x_1+x_2+x_3}(1-p)^{3-(x_1+x_2+x_3)}$$ You can see that it is not possible to write this expression as a function of neither $$(a)$$ nor $$(b)$$.

2. You can calculate a minimal sufficient statistic for your distribution and check that it cannot be represented as a function of statistics $$(a)$$ or $$(b)$$. You can easily calculate a minimal sufficient statistic using that: $$\frac{f_p(x)}{f_p(y)} \mathrm{\ is\ independent\ of\ }p \iff T(x)=T(y)$$ In this case, a minimal sufficient statistic is $$T(X)=X_1+X_2+X_3$$ and statistics $$(a)$$ and $$(b)$$ cannot be represented as a function of T(X), hence, they are not sufficient.