# Mixed parameterization of sample from normal distribution

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.

• 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. – Xi'an Apr 9 '19 at 12:16
• 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. – whuber Apr 9 '19 at 13:24
• thanks, @whuber, but the intention is to have variantion independent parameter components, so I need $\theta_1$ to be free of $\sigma^2$. – Johan Larsson Apr 9 '19 at 13:50
• What do you mean by free? Both $\mu$ and $\mu/\sigma^2$ are free to vary in $\Bbb R$. – Xi'an Apr 9 '19 at 13:54
• 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. – whuber Apr 9 '19 at 13:59