Beta分布是否具有共轭先验？

34

我知道beta分布与二项式共轭。但是beta的共轭先验是什么？谢谢。

beta-distribution conjugate-prior

— 碰撞平衡
source

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，使得，记住 $X_1,\dots,X_n$ $\alpha,\beta$ $X_i\mid\alpha,\beta\sim\mathrm{Beta}(\alpha,\beta)$ 和

E [X_{i} ∣ α, β] = \frac{α}{α + β} =: μ

$\mathbb{E}[X_i\mid\alpha,\beta]=\frac{\alpha}{\alpha+\beta} =: \mu$

因此，可能会重新参数化的可能性在以下方面

和

和使用作为现有

V a r [X_{i} ∣ α, β] = \frac{α β}{(α + β)^{2} (α + β + 1)} =: σ^{2} .

$\mathbb{Var}[X_i\mid\alpha,\beta] = \frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)} =: \sigma^2 \, .$

μ

$\mu$

σ^{2}

$\sigma^2$

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。

σ^{2} ∣ μ \sim U [0, μ (1 - μ)] μ \sim U [0, 1] .

$\sigma^2\mid\mu \sim \mathrm{U}[0,\mu(1-\mu)] \qquad \qquad \mu\sim\mathrm{U}[0,1] \, .$

— 禅
source

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

ψ = α + β

$\psi = \alpha + \beta$

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

上扔一个伽玛或对数正态

μ = \frac{α}{α + β}

$\mu = \frac{\alpha}{\alpha + \beta}$

ψ

$\psi$ 在

上

。

μ

$\mu$

— 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

\int_{0}^{1} \frac{1}{μ (1 - μ} d μ

$\int_{0}^{1}{\frac{1}{\mu(1-\mu}}d\mu$ ，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，这是积的，虽然我还没有完全想出其中（I相信和应该工作-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, β ∣ a, b, p) \propto {\frac{Γ (α + β)}{Γ (α) Γ (β)}}^{p} \exp (a α + b β) .

$\pi(\alpha, \beta \mid a, b, p) \propto \left\{\frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha)\Gamma(\beta)}\right\}^p \exp\left(a\alpha + b\beta \right).$

(a, b, p)

$(a, b, p)$

p \geq 0

$p \ge 0$

a < 0, b < 0

$a < 0, b < 0$

p = 0

$p = 0$

所以这表明

p

$p$

p > 0

$p > 0$ 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。

\int_{0}^{\infty} \int_{0}^{\infty} {\frac{Γ (α + β)}{Γ (α) Γ (β)}}^{p} \exp (a α + b β) = ?

$\int_0^\infty \int_0^\infty \left\{\frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha)\Gamma(\beta)}\right\}^p \exp\left(a\alpha + b\beta \right) = ?$

— 家伙
source

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

期内缺少

的作用。它可能应该是

等

p

$p$

\exp

$\exp$

p a α

$pa\alpha$

— 尼尔摹

@NeilG

p

$p$ is in the

\exp

$\exp$ , you just have to express things in terms of

\log Γ (\dots)

$\log \Gamma(\cdots)$ instead of

Γ (\dots)

$\Gamma(\cdots)$ . Doing

p a α

$pa\alpha$ 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

$a$ and

b

$b$ which kills putting a prior on them or estimating them by likelihood methods, etc...

— guy

2

@NeilG integral is over

α

$\alpha$ and

β

$\beta$ since those are the random variables.

— guy

9

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

the beta distribution is one of the exponential family distributions, and
in theory it should be possible to derive a prior. See, e.g., wikipedia, D Blei's lecture on exponential families.

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.

— TooTone
source

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 $B(\alpha, \beta)$ , the conjugate prior is "... of the form

π (α, β) \propto {\frac{Γ (α + β)}{Γ (α) Γ (β)}}^{λ} x_{0}^{α} y_{0}^{β}

$\pi(\alpha,\beta) \propto \Big\{ \frac{\Gamma(\alpha+\beta)} {\Gamma(\alpha)\Gamma(\beta)} \Big\} ^{\lambda} x_0^{\alpha} y_0^{\beta}$

where $\{\lambda, x_0, y_0\}$ are hyperparameters, since the posterior is then equal to

π (α, β | x) \propto {\frac{Γ (α + β)}{Γ (α) Γ (β)}}^{λ} (x x_{0})^{α} ((1 - x) y_{0})^{β} . "

$\pi(\alpha,\beta \vert x) \propto \Big\{ \frac{\Gamma(\alpha+\beta)} {\Gamma(\alpha)\Gamma(\beta)} \Big\} ^{\lambda} (xx_0)^{\alpha} ((1-x)y_0)^{\beta}."$

The remainder of the example concerns importance sampling from $\pi(\alpha,\beta \vert x)$ in order to compute the marginal likelihood of $x$ .

— user37239
source

2

I don't have Robert's book available but the posterior is

π (α, β) \propto {(\frac{Γ (α + β)}{Γ (α) Γ (β)})}^{λ + 1} (x x_{0})^{α - 1} {(y_{0} (1 - x))}^{β - 1}

$\pi(\alpha, \beta) \propto \left( \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha) \Gamma(\beta)} \right)^{\lambda +1} (x x_0)^{\alpha-1} \left(y_0 (1-x) \right)^{\beta-1}$ . 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