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Let $X_1,X_2, X_3$ be a random sample from Bernoulli distribution $B(p)$. Which of the following is sufficient statistic for $p$ ?

$(A) X_{1}^{2}+X_{2}^{2}+X_{3}^{2}$

$(B) X_1+2X_{2}+X_{3}$

$(C) 2X_1-X_{2}-X_{3}$

$(D) X_1+X_{2}$

$(E) 3X_1+2X_{2}-4X_{3}$

We know $T=sum_i^3 X_i$ is sufficient statistic for $p$ via Neymann Factorization theorem.

The reasoning for different options.

$(A)$ It's not a one-one function of sufficient statistic $T$ (If it was $T^2$ , was it sufficient statistic then? My answer is yes because it is one to one function of sufficient statistic our random variable is positive am I right?).

$(B)X_1+X_{2}+X_{3}+X_{2}$ It contains original statistic therefor it is suffient statistic for $p$.

$(D)$ It doesn't include $X_3$ So it's not sufficient statistic for $p$

All other options include subtraction in it so I ruled out all of them.

I think my reasoning is not very good I lack some intuition behind finding sufficient statistic. Correct me here, please.

Since $T=X_1+X_2+X_3$ is a minimal sufficient statistic for $p$, implying that it is a function of every other sufficient statistic, option (D) is eliminated.

Now $X_i^2$ is a one-to-one function function of $X_i$ because $X_iin{0,1}$ for each $i$.

So $X_1^2+X_2^2+X_3^2$ is also a one-to-one function of $T$, implying that the former is sufficient for $p$. (Thanks to @Alex for pointing this out)

For the remaining options I create a table of the possible values that the statistics can take:

begin{array}{|c|c|c|}
hline (X_1,X_2,X_3)&T_1=X_1+2X_2+X_3&T_2=2X_1-X_2-X_3&T_3=3X_1+2X_2-4X_3\ hline(0,0,0)&0&0&0\
hline(0,0,1)&1&-1&-4\
hline(0,1,0)&2&-1&2\
hline (0,1,1)&3&-2&-2\
hline (1,0,0)&1&2&3\
hline (1,0,1)&2&1&-1\
hline (1,1,0)&3&1&5\
hline (1,1,1)&4&0&1\
hline
end{array}

I am looking whether $T_1,T_2,T_3$ are one-to-one functions of the sample $(X_1,X_2,X_3)$ or not. If they are, then they are sufficient statistics. If you can see right away that the only bijection is $$T_3:{0,1}^3to{-4,-2,ldots,3,5}$$, then you are done. Because a bijection with the sample implies that observing $T_3(X_1,X_2,X_3)$ and observing $(X_1,X_2,X_3)$ are equivalent. In other words, $T_3$ is a sufficient statistic for $p$.

Alternatively, if you consider the case $T_1=2$, you will find that the conditional distribution $P({X_1,X_2,X_3}mid T_1)$ depends on $p$. Similar argument holds for $T_2$ if you consider the case $T_2=0$. Also it is apparent from the table above that $P({X_1,X_2,X_3}mid T_3)$ is independent of $p$ for all possible values of $T_3$, because $T_3$ takes 8 distinct values corresponding to 8 different tuples.

I used the following threads for reference:

• Verification of sufficiency of a linear combination of the sample $(X_i)_{ige1}$ where $X_istackrel{text{i.i.d}}simtext{Ber}(theta)$




• Sufficiency or Insufficiency