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Jun 12, 2014 at 15:44 vote accept beuhbbb
Jun 12, 2014 at 15:29 comment added beuhbbb I recasted my question that now fits very well with your answer and my example. Thanks again
Jun 12, 2014 at 15:28 comment added beuhbbb @Neil G. You right. I realise I must be tired because my question makes no sense. I ask by the negative exactly what I state preliminarly to my question. Thanks.
Jun 12, 2014 at 15:24 comment added Neil G @peuhp: Isn't your example exactly the one defined to be a curved exponential family? The dimension of $\theta$ is 1; the dimension of $\eta$ is 2. It's a parabolic slice through the cartesian plane.
Jun 12, 2014 at 15:13 comment added beuhbbb @whuber. Thanks. From the book (and roughly speaking), the dimension of the exponential family is the one of $\nu(\theta)$, so I am a bit lost. It seems to me still contradictory: $\theta$ is a scalar ($d=1$) but $\nu(\theta)$ is cleary a 2d vector ($s=2$). So $d>s$. Thanks for helping me to clarify my mind.
Jun 12, 2014 at 14:27 comment added whuber @peuhp There seem to be two senses of the word "dimension" in operation here: one is the dimension of a manifold (coordinatized by the parameters $\theta$, of dimension $d$) and the other is a dimension of a space in which it is embedded (with coordinates $\eta$, of dimension $s$). The latter was likely meant in the C. Robert quotation--but whether that is the correct interpretation depends on exactly how Robert defines "dimension."
Jun 12, 2014 at 12:42 comment added beuhbbb Thanks. From "The Bayesian choice", C. Robert p 138: $x \sim N(\theta, \theta^2)$ induces an exponential family of dimension 2. Is not it contradictory with our answer ? Thanks in advance.
Jun 12, 2014 at 9:52 history answered Neil G CC BY-SA 3.0