Beta分布是否具有共轭先验?


Answers:


25

看来您已经放弃了共轭。仅作记录,我见过人们正在做的一件事(但不记得确切的位置,对不起)是这样的重新参数化。如果是条件IID,给定的α β,使得X | α β Ë 一个α β ,记住 È [ X | α β ] = αX1,,Xnα,βXiα,βBeta(α,β)V- [R[X|αβ]=αβ

E[Xiα,β]=αα+β=:μ
因此,可能会重新参数化的可能性在以下方面 μ σ 2和使用作为现有 σ 2 | μ ü [ 0 μ 1 - μ ]
Var[Xiα,β]=αβ(α+β)2(α+β+1)=:σ2.
μσ2 现在,您可以计算后验,并通过自己喜欢的计算方法进行探索。
σ2μU[0,μ(1μ)]μU[0,1].

4
不,不是MCMC这东西!正交这个东西!仅2个参数-正交是小尺寸后代的“黄金标准”,无论是时间还是准确性。
概率

3
另一种选择是将视为精度的度量,然后再次使用μ = αψ=α+β为平均值。这始终通过Dirichlet流程完成,并且beta分发是一种特殊情况。所以也许在ψ上扔一个伽玛或对数正态μ=αα+βψ抛弃μ
家伙

2
可以肯定的是,这不是共轭的,对吗?
2013年

3
当然不!
2013年

嗨,@ Zen,我现在正在处理这个问题,但是我是贝叶斯人的新手,我不确定我是否理解这个想法。我发现您打算找到011μ(1μdμ,然后使用重新参数,但当然这不是想法,你可以帮我明白了。?
红噪声

23

是的,它在指数族中有一个共轭先验。考虑三个参数族 为的某些值一个bp,这是积的,虽然我还没有完全想出其中(I相信p0一个<0b<0应该工作-p=0对应于独立指数分布,从而绝对有效,共轭更新涉及增量

π(α,βa,b,p){Γ(α+β)Γ(α)Γ(β)}pexp(aα+bβ).
(a,b,p)p0a<0,b<0p=0所以这表明 p > 0pp>0也可以)。

该问题,并且在的原因至少一部分没有人使用它,是 即归一化常数没有封闭形式。

00{Γ(α+β)Γ(α)Γ(β)}pexp(aα+bβ)=?

啊。那是有问题的。无论如何,我一直在寻找共轭先验的非信息性版本,所以看起来我还是最好从两个参数的统一先验开始。谢谢。
垃圾平衡

如果您只是在比较可能性,则无需对其进行归一化处理
Neil G

我认为您也可能在exp期内缺少的作用。它可能应该是p 一个αpexppaα
尼尔摹

@NeilG p is in the exp, you just have to express things in terms of logΓ() instead of Γ(). Doing paα is just a reparmetrization, it changes nothing. Not sure what you mean "just comparing likelihoods". You can't implement a Gibbs sampler with this prior without using something like Metropolis, which kills the advantage of conditional conjugacy, the normalizing constant depends on a and b which kills putting a prior on them or estimating them by likelihood methods, etc...
guy

2
@NeilG integral is over α and β since those are the random variables.
guy

9

In theory there should be a conjugate prior for the beta distribution. This is because

However the derivation looks difficult, and to quote A Bouchard-Cote's Exponential Families and Conjugate Priors

An important observation to make is that this recipe does not always yields a conjugate prior that is computationally tractable.

Consistent with this, there is no prior for the Beta distribution in D Fink's A Compendium of Conjugate Priors.


3
The derivation is not difficult — See my answer: mathoverflow.net/questions/63496/…
Neil G

3

我不相信有一个“标准”(即指数族)分布是β分布之前的共轭物。但是,如果确实存在,则必须是双变量分布。


我不知道这个问题,但是我确实找到了这个方便的共轭先验图,似乎支持了您的答案:johndcook.com/conjugate_prior_diagram.html
贾斯汀·博佐尼尔

共轭先验在指数族中,具有三个参数-不为两个。
Neil G

1
@尼尔,你绝对正确。我想我应该说它必须至少有两个参数。

-1: this answer is clearly wrong in the claim that "conjugate prior does not exist in the exponential family", as is demonstrated in the answer above...
Jan Kukacka

3

Robert and Casella (RC) happen to describe the family of conjugate priors of the beta distribution in Example 3.6 (p 71 - 75) of their book, Introducing Monte Carlo Methods in R, Springer, 2010. However, they quote the result without citing a source.

Added in response to gung's request for details. RC state that for distribution B(α,β), the conjugate prior is "... of the form

π(α,β){Γ(α+β)Γ(α)Γ(β)}λx0αy0β

where {λ,x0,y0} are hyperparameters, since the posterior is then equal to

π(α,β|x){Γ(α+β)Γ(α)Γ(β)}λ(xx0)α((1x)y0)β."

The remainder of the example concerns importance sampling from π(α,β|x) in order to compute the marginal likelihood of x.


2
I don't have Robert's book available but the posterior is π(α,β)(Γ(α+β)Γ(α)Γ(β))λ+1(xx0)α1(y0(1x))β1. Robert also posted on this topic here mathoverflow.net/questions/20399/…
Fred Schoen

1
I humbly recommend that the original poster updates the post to indicate that the posterior given in the textbook is incorrect, per Fred Schoen's comment (which is easily verified).
RMurphy
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