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Let $T$ be a sufficient statistic. Suppose $f(T)$ is not a one-to-one function of $T$. Show $f(T)$ is not a sufficient statistic.  

I think this should be proved by contradiction. Since $f$ is not one-to-one, $\exists t_1 \ne t_2 \ni g(t_1)=g(t_2)$. I sough a contradiction in the factorization theorem by writing $f_\theta(x)=h(x)g_\theta(f^{-1}\circ f(T(X)))$. But, I didn't succeed. I tried from the definition of a sufficient statistic, but I don't see a way there too.

I think, the proof can be done without specifying a distribution for the sample. If that's not the case, we may assume, the sample comes from a $\operatorname{Bernouilli}(\theta)$.

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The intuition is that if $\mathbb P(X \mid T,\theta) = \mathbb P(X \mid T)$, then $\mathbb P(X \mid f(T),\theta) \neq \mathbb P(X \mid f(T))$ because you've lost information since $f$ is not injective. –  cardinal Oct 15 '11 at 22:24

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up vote 5 down vote accepted

The statement as it stands is not true.

If, as you request, we look at the sample $\mathbf{X} =(X_1, X_2,\ldots, X_n)$ representing $n$ i.i.d. $\operatorname{Bernouilli}(\theta)$ random variables taking the values $0$ or $1$, then $\mathbf{X}$ is a sufficient statistic for itself in the sense that $$\Pr(\mathbf{X}=\mathbf{x}|\mathbf{X}=\mathbf{x},\theta)=\Pr(\mathbf{X}=\mathbf{x}|\mathbf{X}=\mathbf{x})=1$$ but letting $f(\mathbf{x}) = \sum x_i$, then $f(\mathbf{X})$ is also a sufficient statistic for $\mathbf{X}$ as $$\Pr(\mathbf{X}=\mathbf{x}|f(\mathbf{X})=f(\mathbf{x}),\theta)=\Pr(\mathbf{X}=\mathbf{x}|f(\mathbf{X})=f(\mathbf{x}))=\frac{1}{n \choose f(\mathbf{x})}.$$

Clearly if $f(\mathbf{X})$ is not always either $0$ or $n$, then $f$ is not injective, i.e. is not a one-to-one function.

Perhaps the statement should have been

Let T be a minimal sufficient statistic. Suppose f(T) is not a one-to-one function of T. Show f(T) is not a sufficient statistic.

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You have a very good point Henry. I picked this question during a class as is. I'm surprised it's ill-posed. I think your amended statement is the right one. But, I will not correct my question. Your answer made me learn a lot. Maybe it will benefits others. Thanks. –  Nicolas Essis-Breton Oct 16 '11 at 1:32
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@NicolasEssis-Breton: I wouldn't call the statement of the question ill-posed. I'd just call it false. :) (+1 Henry for the nice answer). –  cardinal Oct 16 '11 at 1:53

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