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Let $X_1,X_2,...X_n$ be a random sample from $f(x,\theta)=\frac{1}{2 \theta}e^{\frac{-|x|}{\theta}}$.We know by Factorisation theorem that $\frac{\sum |X_i|}{n}$ is sufficient for $\theta$. But can we show that $\frac{\sum X_i}{n}$ is not sufficient for $\theta$? It is tough to show it by definition as the distribution of $\sum X_i$ cannot be found explicitly. Is there any other way?

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let $S$ is minimal sufficient. $T$ is not sufficient if exist $x,y\in support$

$T(x)=T(y)$ but $S(x)\neq S(y)$

let n=2, $T(x)=x_1+x$ and $S(x)=|x_1|+|x_2|$ (S is minimal sufficient)

$x=(2,-1)$ and $y=(3,-2)$

$T(x)=1=T(y)$ $S(x)=3\neq 5=T(y)$

so $T$ is not sufficient

in general for arbitrary $n$ choose $x=(2,-1,0,\cdots ,0)$ $y=(3,-2,0,\cdots ,0)$

this method is based of this point that minimal sufficient is a function of any sufficient Statistic, and in above we shown that $S$ is not a function of $T$.note $V$ is a function of $U$ if

$\forall x,y \quad U(x)=U(y) \Longrightarrow V(x)=V(y)$ so

$V$ is not a function $U$ if

$\exists x,y \quad V(x)=V(y) \quad but \quad U(x)\neq U(y) $

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Yes, it's quite easy. All you have to do to show that a statistic is not sufficient is to show that there has been some loss of information (about the parameter) as compared to what was present in the original sample.

If you know that one statistic is sufficient, then computing its value for a specific sample will result in no loss of information about $\theta$ compared to the original sample. So if you can find two distinct samples for which the insufficient statistic gives the same value, but the sufficient statistics are unequal, you know you've lost information about $\theta$.

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  • $\begingroup$ Can you enlighten me with an example? $\endgroup$ – user321656 Apr 4 '19 at 4:35

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