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Suppose we have a random variable $X$ having p.d.f of the form $$f(x|\theta)=\exp\{c(\theta)'T(x)-B(\theta)\}h(x),$$ then we say $X$ is from exponential family. Further, we say that an exponential family is minimal if the functions $c(\theta)$ and the statistics $T(X)$, are linearly independent respectively.

I am trying to determine the minimal exponential family form of the following case.

Here, $X$ has p.d.f $$f(x|\alpha,\beta)=\exp\left\{\sum_{i=1}^nx_{i}(\alpha+\beta z_i)-\sum_{i=1}^n\log(1+\exp(\alpha+\beta z_i))\right\},$$

which is equal to

$$f(x|\alpha,\beta)=\exp\left\{\sum_{i=1}^nx_{i}\alpha + \sum_{i=1}^n\beta z_ix_i-\sum_{i=1}^n\log(1+\exp(\alpha+\beta z_i))\right\}$$

For the first equation the natural sufficient statistic $T$ is , $T=(X_1, \dots,X_n)$, while the second is $T=(\sum_{i=1}^nX_i,\sum_{i=1}^nX_iz_i)$.

I wonder which one is the minimal exponential family form?

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How can you have a sufficient statistic with more dimensions than the number of parameters? If that were true, then the expectation parametrization would have more dimensions than the natural parametrization.

I think if $n \le 1$, it's the first form (and the parameter is just $\alpha + z_1 \beta$), otherwise it's the second form.

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  • $\begingroup$ Look into curved exponential families $\endgroup$ – kjetil b halvorsen Feb 23 at 7:54
  • $\begingroup$ @kjetilbhalvorsen I don't think that the sufficient statistic of a curved exponential family can have more dimensions than the number of parameters. Can you provide an example of such a sufficient statistic? It's certainly not true for, for example, the normal distribution with known variance (one sufficient statistic, one parameter). $\endgroup$ – Neil G Feb 23 at 7:59

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