I am studying exponential families and mixed parameterizations. Now, I am told that $$ \mathbf{\theta} = \begin{bmatrix}\mu\\ -\frac{1}{2\sigma^2}\end{bmatrix} $$ is the parameter in a variation-independent mixed parameterization for a sample of $i=1,\dots,n$ observations, sampled from $X \sim \mathcal{N}(\mu,\sigma^2)$ independently, with $\mu$ and $\sigma$ unknown but I have not been able to derive this for myself.

The joint distribution of the samples is given by $$ f(\mathbf{x}|\mu,\sigma^2) = (2\pi \sigma^2)^{-n/2} \exp \left[ -\frac{1}{2\sigma^2} \sum_{i=1}^n \left( x_i^2 - 2x_i\mu + \mu^2\right) \right]. $$ I want to rewrite this density function so that it takes the form $$ a(\mathbf{\theta})h(\mathbf{x})\exp[\mathbf{\theta}^\intercal t\mathbf({x})] $$ and this is where I am stuck. I can see that $$ \begin{aligned} f(\mathbf{x}|\mu,\sigma^2) &= (2\pi \sigma^2)^{-n/2} \exp \left[ -\frac{1}{2\sigma^2} \sum_{i=1}^n \left( x_i^2 - 2x_i\mu + \mu^2\right) \right] \\ &= (2\pi \sigma^2)^{-n/2} \exp \left[ -\frac{\sum_{i=1}^n x_i^2}{2\sigma^2} + \frac{\mu\sum_{i=1}^n x_i}{\sigma^2} \right] \exp \left[ \frac{\mu^2}{\sigma^2} \right] \\ \end{aligned}, $$ but this only leads me to the normal canonical parameterization with $$ \mathbf{\theta} = \begin{bmatrix}\frac{\mu}{\sigma^2} \\ -\frac{1}{2\sigma^2}\end{bmatrix}\quad t(\mathbf{x}) = \begin{bmatrix}\sum_{i=1}^n x_i \\ \sum_{i=1}^n x_i^2\end{bmatrix} $$ which is not what I need.

Can anybody give me a hint as to how to proceed to uncover the mixed parameterization?

Note: This is related to a course assignment, so please avoid complete solutions for the time being, thank you.

  • 1
    $\begingroup$ Please add self-study as a tag to ensure someone does not provide a full resolution. And please recall the definition of a mixed parameterisation. I have never heard the term before. $\endgroup$
    – Xi'an
    Apr 9 '19 at 12:16
  • 1
    $\begingroup$ It seems to me at the last line you have derived exactly what you need. Simply give $\mu/\sigma^2$ a new name--let's call it $\theta_1$--and observe that now $$\theta = \pmatrix{\theta_1 \\ -\frac{1}{2\sigma^2}}$$ is in precisely the form you begin your question with. $\endgroup$
    – whuber
    Apr 9 '19 at 13:24
  • $\begingroup$ thanks, @whuber, but the intention is to have variantion independent parameter components, so I need $\theta_1$ to be free of $\sigma^2$. $\endgroup$ Apr 9 '19 at 13:50
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    $\begingroup$ What do you mean by free? Both $\mu$ and $\mu/\sigma^2$ are free to vary in $\Bbb R$. $\endgroup$
    – Xi'an
    Apr 9 '19 at 13:54
  • $\begingroup$ Could you explain what it might mean for $\theta_1$ "to be free of" $\sigma^2$? It explicitly has nothing to do with $\sigma^2.$ I'm guessing here, but it sounds like you are aiming for two incompatible results: in some unstated way, you want the components of $\theta$ to have some specified relationship with the first two moments of the distribution. $\endgroup$
    – whuber
    Apr 9 '19 at 13:59

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