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 $\sum x_i$?

  • 2
    $\begingroup$ Hint: You will need to assume $n \ne 1$ for this to be true. $\endgroup$
    – whuber
    Feb 6, 2020 at 4:57
  • $\begingroup$ @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! $\endgroup$
    – Ron Snow
    Feb 6, 2020 at 17:23

1 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.

  • 2
    $\begingroup$ 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? $\endgroup$
    – Ron Snow
    Feb 6, 2020 at 17:28
  • 1
    $\begingroup$ Yes, that is a valid counter-example. $\endgroup$
    – Ben
    Feb 6, 2020 at 20:41
  • $\begingroup$ Awesome- thank you for the clarification! $\endgroup$
    – Ron Snow
    Feb 7, 2020 at 1:36

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