# Natural parameters of Bernoulli distribution

In Bishop's Pattern Recognition and Machine Learning, he states that exponential family distributions over $\mathbf{x}$ given parameters $\boldsymbol{\eta}$ can be written as $p(\mathbf{x} | \boldsymbol{\eta}) = h(\mathbf{x})g(\boldsymbol{\eta})\exp\{\boldsymbol{\eta}^\top\mathbf{u}(\mathbf{x})\}$.

He gives the natural parameters for the Bernoulli distribution with probability of success $\mu$ as: $$\eta = \ln(\frac{\mu}{1 - \mu})$$ $$u(x) = x$$ $$h(x) = 1$$ $$g(\eta) = \frac{1}{1 + \exp{(\eta)}}$$

I was working this out myself and found $$\boldsymbol{\eta} = [\ln\mu,\: \ln(1 - \mu)]^\top$$ $$\mathbf{u}(x) = [x, \: 1-x]^\top$$ $$h(\mathbf{x}) = 1$$ $$g(\boldsymbol{\eta}) = 1$$ could also work.

How do you know when you've found the natural parameters and is there a rule preventing you from adding arbitrary scaling constants into the expressions for $\boldsymbol{\eta}$, $\mathbf{u}$, $h$, and $g$?

• there's a typo, it must be $u(x)=x$ Jan 14, 2016 at 16:38
• ...and unless I am mistaken, also $g(\eta) = \frac{1}{1 + \exp{(\eta)}}$ Jan 14, 2016 at 16:44

A natural exponential family is one in which the natural parameter $\eta$ and the statistic $u(x)$ are both the identity (see here), which is the case for Bishop's formulation, but not yours.
• Isn't Bishop's $u(x) = x$ though, which is not identity? Jan 14, 2016 at 16:48