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If $X_1$ and $X_2$ are random samples from an exponential$(\frac{1}{\theta})$ distribution, how can I show that $T = X_1 + 2X_2$ is not sufficient for $\theta$ (mean of the distribution)?

I know that a statistic can be considered sufficient if I can write the probability function as:

$p(X) = h(X)g(\theta,T(X))$

Which is the factorization theorem. But how can I show that it is not possible to write it this way? Or is there any other way?

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  • $\begingroup$ Although it doesn't affect the solution you may want to specify if $\theta$ is a mean or scale parameter. $\endgroup$
    – dsaxton
    Commented Oct 18, 2016 at 19:50
  • $\begingroup$ $\theta$ is the mean. I've edited the question a bit. $\endgroup$ Commented Oct 18, 2016 at 19:55

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The idea is to find a way to change the data that does not alter $T$ but does alter the likelihood $P$ in an important way: namely, by an amount that truly depends on the parameter $\theta$. That would show that $T$ does not tell us all we need to know about $\theta$, insofar as it might be revealed through $P$. One elementary (yet fully rigorous) approach is sketched below.


Because the PDF of the underlying probability law is

$$f_\theta(x) = \frac{1}{\theta}\exp\left(x/\theta\right),$$

the assumption that $X_1$ and $X_2$ are independent implies

$$P(\mathbf{x}) = f_\theta(x_1)f_\theta(x_2) = \frac{1}{\theta^2}\exp((x_1+x_2)/\theta).$$

Suppose (in order to derive a contradiction) that this likelihood can be factored as

$$P(\mathbf{x}) = h(\mathbf{x})g(\theta, T(\mathbf{x})) = h(x_1,x_2)g(\theta, x_1+2x_2).$$

Taking logarithms will simplify things:

$$-2\log\theta + \frac{x_1+x_2}{\theta}=\log P(\mathbf{x}) = \log h(x_1,x_2) + \log g(\theta, x_1+2x_2).\tag{1}$$

Since both $x_1$ and $x_2$ are almost surely positive, for sufficiently small but nonzero $\epsilon \lt x_2$ both $x_1 + 2\epsilon$ and $x_2-\epsilon$ will still be positive, so it makes sense to plug them into both sides. Notice how this combination of changes in the $x_i$ was chosen to leave $g$ unchanged, because $$T(x_1,x_2) = T(x_1+2\epsilon, x_2-\epsilon).$$

At this juncture, compute the change in the right hand side of $(1)$ and the change in the left hand side when $(x_1,x_2)$ is changed to $(x_1+2\epsilon, x_2-\epsilon)$. (This requires only simple algebra.) Observe that the change in the RHS depends only on $x_1, x_2,$ and $\epsilon$ (by construction), but that the change in the LHS depends ineluctably on $\theta$ (because $\epsilon$ is nonzero). Draw your conclusions from this contradiction.

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  • $\begingroup$ So, after computing the change on both sides I have: $-2log (\theta) + \frac{x_1 + \epsilon + x_2}{\theta} = log [h(x_1,x_2)] + log[ g(\theta,x_1+2x_2)]$ As there was change on a side that depended on $\theta$ even tough there was no change on the other side, I can conclude that there is more information to be obtained than $x_1 + 2x_2$ gives us, is that correct? $\endgroup$ Commented Oct 20, 2016 at 0:03
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Here's an alternative approach. It is straight forward that $S=X_1+X_2$ is the minimal sufficient statistic ($\frac{f_\theta(x)}{f_\theta(y)}$ is free of $\theta$ if and only if $S(x)=S(y)$). So if $T$ is also a sufficient statistic, there exists a function, say $g$, maps $T$ to $S$. Hence $X_1+X_2=S(X)=g(T(X))=g(X_1+2X_2)$. This is clearly not possible because if we let $(X_2, X_2)=(1,2)$ and $(3,1)$, we get $g(5)=3=4$.

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If that's the case, the corresponding unbiased estimator would be $T/3$. But what do we know of the variance bounds of this unbiased estimator? Will it hit them? If it doesn't hit the variance bounds, we know better statistics must be out there, so we'd make a conclusion about whether T is sufficient or not.

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    $\begingroup$ So, I get the unbiased estimator $T/3$ then check its variance bounds. If it doesnt hit the variance bounds, we can be sure it is not sufficient? $\endgroup$ Commented Oct 18, 2016 at 19:39

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