# Sufficient statistic of Bernoulli Trial

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.

• 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. Commented Jun 26, 2020 at 17:01
• I edited the question. Sorry for the mistake. See if you can help me with this estimator $X1+2X2$ being sufficient or not? Commented Jun 28, 2020 at 17:05
• 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. Commented Jun 28, 2020 at 18:03
• Hint: given $X_1+2X_2,$ you can recover the values of both $X_1$ and $X_2,$ making this statistic the equivalent of $(X_1,X_2).$
– whuber
Commented Jun 28, 2020 at 18:20

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.