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

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 $$\sum x_i$$?

• Hint: You will need to assume $n \ne 1$ for this to be true. – whuber Feb 6 at 4:57
• @whuber I think I see what you're implying: to use $s^2$. I'll try playing around with this a bit more. Thank you! – Ron Snow Feb 6 at 17:23

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.
• Such as letting $x_1=-2, x_2=2$ and $x'_1=-3, x'_2=3$? It appears that $S(\mathbf{x})=S(\mathbf{x'})=0$, while $T(\mathbf{x})=4, T(\mathbf{x'})=9$. Is this the right idea? – Ron Snow Feb 6 at 17:28