# 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.

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.
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.