Timeline for Is there any strong argument about objective/non-informative improper prior?
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8 events
| when toggle format | what | by | license | comment | |
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| Apr 22, 2022 at 16:11 | comment | added | kjetil b halvorsen♦ | If you found this answer helpful, then please consider upvoting and/or accepting it. | |
| Apr 21, 2022 at 15:04 | comment | added | kjetil b halvorsen♦ | Please see stats.meta.stackexchange.com/questions/6304/my-upvoting-policy, when you find a question sufficiently clear to write an answer, consider to upvote the question! | |
| Apr 16, 2022 at 17:11 | comment | added | Celi | In fact I was perhaps wrong when saying the probability is $0$. Maybe I can reformulate as : the data is made of $x_1, ..., x_n$ and $n$, the sample size. If I suddenly tell you that $x_1, ..., x_n$ were negative, then when evaluating your posterior, you should do it with $n = 0$, i.e. my data was fake data. So your posterior probability equals your prior probability. Isn't it more convincing like that ? In any case from my point of view, even in your answer, it seems that the posterior cannot depend on $z$ only. | |
| Apr 16, 2022 at 17:01 | comment | added | Xi'an | You cannot calculate the posteriori with negative values of the $x_i$'s just as you cannot calculate the posterior with complex or matrix values of the $x_i$'s. Bayesian inference functions in M-closed mode, ie under the assumption that the data is potentially generated from the model. Once data clashes with this assumption, everything stops (and the posterior is not zero). | |
| Apr 16, 2022 at 16:58 | comment | added | Celi | Say you have calculated that posterior, which is supposed to depend on z only. Then I come to you with a vector z, and ask you what's the value of the posterior for that z, at e.g. xi = 1. Then you tell me a positive value. And then I tell you that in fact I calculated z with only negative values for x1, ..., xn. : ) So the postive value should have been a 0 ! And hence knowing z only seems not suffcient for evaluating that posterior. ... and hence in fact we need all the indicators, not just the one for x1 | |
| Apr 16, 2022 at 16:42 | comment | added | Xi'an | Why should the posterior be multiplied by this indicator as a function of the parameter? The indicators all equal one, given the observations. | |
| Apr 16, 2022 at 16:38 | comment | added | Celi | I think there is an error in point a. : the posterior does depend on x1, in addition to z, via an indicator function that ensures x1 is positive (the pdf of an exponential random variable being defined for positive values only) ... that is, the posterior must be written as g(z) * 1{x1 > 0} ... or perhaps 1{x1 >= 0}, depending on how the exponential distribution is defined | |
| Apr 16, 2022 at 15:31 | history | answered | Xi'an | CC BY-SA 4.0 |