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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.

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


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