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I want to show that the family of normal distributions is not a single-parameter exponential family, i.e. there aren't functions $h,g,\eta,T$ such that

$$h(x)g(\mu,\sigma^2)\exp(\eta(\mu,\sigma^2) T(x)) = \frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{1}{2\sigma^2}(x-\mu)^2\right)$$

for all $x\in\mathbb R$ and $(\mu,\sigma^2)\in\mathbb R\times\mathbb R_{>0}$.

I tried fixing one argument, in order to get some information about the other functions. For example $\mu=0$ looks much simpler but it didn't lead me anywhere.

Can anybody give me a hint?

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  • $\begingroup$ Freezing $\mu$ won't work, because at fixed $\mu$ (say $\mu=0$ for this purpose but it doesn't matter) you can take $T(x)=x^2,\eta(\mu,\sigma^2)=\frac{1}{2\sigma^2},h \equiv 1$ and choose $g$ for normalization. (Indeed $g$ is always determined by normalization.) Presumably the point is that $\frac{1}{2\sigma^2}(x-\mu^2)$ is not of the form $\eta(\mu,\sigma^2) T(x)$ (and that it can't be fudged by multiplication by a function of only $x$ and another function of only the parameters). $\endgroup$ – Ian Nov 9 '17 at 19:30
  • $\begingroup$ @Ian I thought if I can show that if $\mu=0$ then $T,\eta,h$ must be of the above form it would lead to a contradiction when unfreezing $\mu$, but this didn't work. $\endgroup$ – Tim B. Nov 9 '17 at 19:33
  • $\begingroup$ There's too much freedom if you have fixed $\mu$, because you can shuffle terms back and forth between the exponent and the functions $h,g$. $\endgroup$ – Ian Nov 9 '17 at 19:34

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