# How to identify one-one correspondance in Sufficient Statistics?

The correct answer to the given question is (1),(3) and (4). I understood how 3 and 4 are correct but I could not understand how (1) is also a correct answer.

I know that here $$\sum_i X_i$$ is a sufficient statistic and any one-one function of this statistic is also a sufficient statistic but according to my calculations $$X_1 + 2X_2$$ is not a one-one function of $$X_1 + X_2$$.

My argument being that when $$X_1 = 1$$ and $$X_2= 0$$ or $$X_1 =0$$ and $$X_2=1$$ then $$X_1 + X_2 =1$$ but in case of $$X_1 + 2X_2$$, it takes value 1 or 2 respectively. 1 is matched with two values 1 and 2. Therefore, it is not a one-one function.

• Think about the possible values of $(X_1,X_2)$ and of the corresponding values of $X_1+2X_2$. – Xi'an Dec 15 '19 at 17:59
• I did, but i found that X1 + 2X2 is not a one one function. Can you confirm that? – napoleon Dec 16 '19 at 1:59
• Okay, got you. So it is actually one-one. Thanks! – napoleon Dec 16 '19 at 9:33

On a general basis, the function \begin{align*} h: &\mathbb N^2 \longmapsto \mathbb N\\ &(x,y) \longmapsto x+2y\\ \end{align*} is neither bijective nor in one-to-one relation with the function \begin{align*} h: &\mathbb N^2 \longmapsto \mathbb N\\ &(x,y) \longmapsto x+y\\ \end{align*} However, the function \begin{align*} h: &\{0,1\}^2 \longmapsto \{0,1,2,3\}\\ &(x,y) \longmapsto x+2y\\ \end{align*} happens to be bijective, that is, for each different value of $$x+2y$$ there exists a single value of $$(x,y)$$. Therefore $$X_1+2X_2$$ contains the same amount of information as $$(X_1,X_2)$$ and is therefore trivially sufficient. It is correct however that it is not in bijection with $$X_1+X_2$$.
Addendum: I reproduced the multiple choice question in an exam last month and none of my students were able to select $$X_1+2X_2$$ with a proper explanation.