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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.

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Edit 13/12/2018: both are sufficient. See: Verification of sufficiency of a linear combination of the sample $(X_i)_{i\ge1}$ where $X_i\stackrel{\text{i.i.d}}\sim\text{Ber}(\theta)$

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.

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