How can I show that $\sum X_i$ is not a sufficient statistic for $\theta$?

Question

Let $X_1,\ldots, X_n \sim f(x\mid \theta)=\frac{x}{\theta}e^{-x^2/(2\theta)}, x > 0$ independently. $\theta > 0$ is unknown.

How can I show that $\sum X_i$ is not a sufficient statistic for $\theta$?

I found from the likelihood ratio that

$$\frac{f(\mathbf{x}\mid \theta)}{f(\mathbf{y}\mid \theta)}=\prod^n_{i=1}\left(\frac{x_i}{y_i}\right)\frac{e^{-\frac{1}{2\theta}\sum x_i^2}}{e^{-\frac{1}{2\theta}\sum y_i^2}}$$

Since we need this likelihood ratio to not be a function of $\theta$, $\sum x_i^2=\sum y_i^2$. Thus, $T(\mathbf{X})=\sum x_i^2$ is our minimal sufficient statistic. (Is this logic correct?)

How can I show that $\sum x_i$ cannot be a sufficient statistic since $T(\mathbf{X})=\sum x_i^2$ cannot be written as a function of $\sum x_i$?

Answer 1

It looks like you are trying to figure out how to prove that one thing is not a function of another thing. This is generally quite simple using a proof by counter-example. For any data vector $\mathbf{x}$, let $S(\mathbf{x}) \equiv \sum x_i$ denote its sum. If $T$ can be written as a composite function of $S$ then this would mean that $S(\mathbf{x}) = S(\mathbf{x}')$ implies $T(\mathbf{x}) = T(\mathbf{x}')$. Thus, to show that $T$ cannot be written as a function of $S$, all you need to do is find any two vectors $\mathbf{x}$ and $\mathbf{x}'$ that falsify this implication ---i.e. you need to establish a counter-example with:

$$S(\mathbf{x}) = S(\mathbf{x}') \quad \quad \text{ and } \quad \quad T(\mathbf{x}) \neq T(\mathbf{x}').$$

As whuber points out in the comments, you will actually need to assume that $n>1$ to get this result (since $S$ is sufficient if $n=1$). If you can establish a counter-example of this kind (which should be quite trivial), then you have shown that the minimal sufficient statistic $T$ is not a function of $S$, and so $S$ is not a sufficient statistic.