Timeline for What the dimension of an exponential family tell us about that family?
Current License: CC BY-SA 3.0
8 events
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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 |