Let $X_1, X_2 , \dots, X_n$ be i.i.d observations from $N(\theta, \theta)$ (where $\theta$ is the variance). I intend to find the minimal sufficient statistics (M.S.S) for the joint distribution.

My steps ultimately lead me to the following:

$\dfrac{f_\theta(x)}{f_\theta(y)} = \text{exp}{ \left\{ \dfrac{-1}{2\theta} \left[ \left(\sum x_i^2 - \sum y_i^2 \right) + 2\theta \left( \sum y_i - \sum x_i \right) \right] \right) }$.

My M.S.S are therefore $\left( \sum x_i^2 , \sum x_i \right)$

Am I on track?

  • 2
    $\begingroup$ You are on the write track, this is one approach to solve it. But I believe that your solution is incorrect. You have to be careful with appealing to the theorem, which states that T(X) is a minimally sufficient statistic if and only if for T(X) = T(Y), that $\frac{f_{\theta}(x)}{f_{\theta}(y)} = \phi(x,y)$, where $\phi(x,y)$ is free of $\theta \forall \theta \in \Theta$. The hint I'll give is that one of your MSS solutions is not necessary for this to be satisfied. $\endgroup$ Commented Aug 27, 2020 at 3:14
  • $\begingroup$ Since this is a regular exponential family, a minimal sufficient statistic is obtained directly through 'exponential family factorization'. $\endgroup$ Commented Aug 27, 2020 at 7:38
  • $\begingroup$ @Tyrel Stokes: I have realized it. the $\Sigma x_i $ is not necessary. $\endgroup$ Commented Aug 27, 2020 at 12:12
  • $\begingroup$ @Scortchi-ReinstateMonica That term does not depend on $\theta$. Our aim is to find a solution so that the ratio is free of $\theta$. $\endgroup$ Commented Aug 29, 2020 at 3:10
  • $\begingroup$ People use the notation $N(\theta, \theta)$ in two senses, $\theta$ the variance or $\theta$ the standar deviation. You should specify which (but from the rest of the post we can infer first option, so I did add it). $\endgroup$ Commented Apr 20, 2021 at 18:20

2 Answers 2


It turns out that after further expansion, the above becomes

$\dfrac{f_\theta(x)}{f_\theta(y)} = \exp \left\{ - \left[ \dfrac{1}{2\theta} \left( \sum x_i^2 - \sum y_i^2 \right) + \left( \sum y_i - \sum x_i \right) \right] \right\}$.

Since it is only the first term that depends on $\theta$, the M.S.S is $\sum x_i^2$.

  • 1
    $\begingroup$ Perhaps clearer with something like $\exp \left(- \frac{ \Sigma x_i^2 - \Sigma y_i^2}{2\theta}\right)\;\exp\left(- {(\Sigma y_i - \Sigma x_i)}\right)$ $\endgroup$
    – Henry
    Commented Aug 27, 2020 at 12:48

By writing the joint density in exponential family form we find

$$ f(x,\theta) = \frac1{\sqrt{2\pi}} \exp\left\{ -\frac{n}{2}\log\theta - \frac1{2\theta} \sum_i x_i^2 + \sum_i x_i -n\theta^2 \right\} $$ and then the result follows by general exponential family theory, as noted in comments.


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