Let $X_1$ and $X_2$ be iid random variables from a $Bernoulli(p)$ distribution. Verify if the statistic $X_1+2X_2$ is sufficient for $p$.

I calculated and found out $X_1+X_2$ as a sufficient statistic for $p$. Is this enough to rule out the possibility of $X1+2X2$ as a sufficient statistic? Is there a better way to show that explicitly?

Addendum: I want to check for $X_1+2X_2$, sorry for the mistake. It is now edited.

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    $\begingroup$ Calculate the mle of $p$ using $X_1$ and $X_2$. Then, calculate the MLE of $p$ using $X_1$ + $X_2$ and see if it gives you the same MLE. If it does, then the sum is sufficient. Technically speaking, I don't think this is the perfectly correct definition of sufficiency ( I forget it at the moment ) but it will lead to the correct result. $\endgroup$
    – mlofton
    Commented Jun 26, 2020 at 17:01
  • $\begingroup$ I edited the question. Sorry for the mistake. See if you can help me with this estimator $X1+2X2$ being sufficient or not? $\endgroup$
    – Nisha
    Commented Jun 28, 2020 at 17:05
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    $\begingroup$ Just check definition of sufficiency, i.e. whether the distribution of $(X_1,X_2)$ given $T=X_1+2X_2$ depends on $p$ or not. The answer is obvious once you note the possible values of $T$ and how they occur. $\endgroup$ Commented Jun 28, 2020 at 18:03
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    $\begingroup$ Hint: given $X_1+2X_2,$ you can recover the values of both $X_1$ and $X_2,$ making this statistic the equivalent of $(X_1,X_2).$ $\endgroup$
    – whuber
    Commented Jun 28, 2020 at 18:20

2 Answers 2


Let $T=X_1+2X_2$ , $S=X_1+X_2$. We know $S$ is a minimal sufficient statistics.

$\{T=0\}=\{ (0,0)\}$

$\{T=1\}=\{ (1,0)\}$

$\{T=2\}=\{ (0,1)\}$

$\{T=3\}=\{ (1,1)\}$

$\sigma(T)=\sigma\bigg( \color{red}\{(0,0)\color{red}\} ,\color{red}\{(1,0)\color{red}\} , \color{red}\{(0,1)\color{red}\},\color{red}\{(1,1)\color{red}\} \bigg)$

$\sigma(S)=\sigma\bigg( \color{red}\{(0,0)\color{red}\} ,\color{red}\{(1,0), (0,1)\color{red}\} , \color{red}\{(1,1)\color{red}\} \bigg)$ where $\sigma(T)$ denotes the sigma generated by T and $\sigma(S)$ denotes the sigma generated by S.

Since $\sigma(S)\subset \sigma(T)$ (the information in $T$ is more than $S$) ,$S$ is a minimal sufficient statistic and $S$ is a function of $T$ ,hence $T$ is a sufficient statistic(But not a minimal one). We can also compare it with $\sigma(X_1,X_2)$ and find $\sigma(X_1,X_2)=\sigma(T)$ ($T$ and $(X_1,X_2)$ have a same information) and obtain that $T$ is a sufficient statistics.

We can also use the definition of a sufficient statistics as follows:

\begin{eqnarray} P(X_1=x_1,X_2=x_2|T=t)= \left\{ \begin{array}{cc} *1 & t=0 \\ *2 & t=1 \\ *3 & t=2 \\ *4 & t=3 \end{array} \right. \end{eqnarray} and find *1,*2,*3 and *4. For example(*1)

\begin{eqnarray} P(X_1=x_1,X_2=x_2|T=0)= \left\{ \begin{array}{cc} 1 & x_1=0,x_2=0 \\ 0 & O.W. \end{array} \right. \end{eqnarray} and in all cases it does not depend of the parameter.


Since $T \equiv X_1+X_2$ is a sufficient statistic, the question boils down to whether or not you can recover the value of this sufficient statistic from the alternative statistic $T_* \equiv X_1 + 2 X_2$. Formally, is there any function that maps $T_*$ to $T$? The other answer by Masoud gives you the information you need to construct such a mapping, so use this to have a go constructing a function of this kind. You can then appeal directly to the Fisher-Neyman factorisation to show that the latter statistic is also sufficient.


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