我们有N个样本 $X_i$，从均匀分布 $[0,\theta]$ 哪里 $\theta$未知。估计$\theta$ 从数据。

因此，贝叶斯法则...

$f(\theta | {X_i}) = \frac{f({X_i}|\theta)f(\theta)}{f({X_i})}$

可能是：

$f({X_i}|\theta) = \prod_{i=1}^N \frac{1}{\theta}$ （编辑：何时 $0 \le X_i \le \theta$ 对所有人 $i$，否则为0-感谢whuber）

但是没有其他信息 $\theta$，似乎事前应该与 $1$ （即制服）或 $\frac{1}{L}$ （杰弗里斯事前？） $[0,\infty]$但是然后我的积分不收敛，我不确定如何进行。有任何想法吗？

This has generated some interesting debate, but note that it really doesn't make much difference to the question of interest. Personally I think that because $\theta$ is a scale parameter, the transformation group argument is appropriate, leading to a prior of

$$\begin{array}& p(\theta|I)=\frac{\theta^{-1}}{\log\left(\frac{U}{L}\right)}\propto\theta^{-1} & L<\theta<U\end{array}$$

This distribution has the same form under rescaling of the problem (the likelihood also remains "invariant" under rescaling). The kernel of this prior, $f(y)=y^{-1}$ can be derived by solving the functional equation $af(ay)=f(y)$. The values $L,U$ depend on the problem, and really only matter if the sample size is very small (like 1 or 2). The posterior is a truncated pareto, given by:

$$\begin{array}\\ p(\theta|DI)=\frac{N\theta^{-N-1}}{ (L^{*})^{-N}-U^{-N}} & L^{*}<\theta<U & \text{where} & L^{*}=max(L,X_{(N)}) \end{array}$$ Where $X_{(N)}$ is the Nth order statistic, or the maximum value of the sample. We get the posterior mean of $$E(\theta|DI)= \frac{ N((L^{*})^{1-N}-U^{1-N}) }{ (N-1)((L^{*})^{-N}-U^{-N}) }=\frac{N}{N-1}L^{*}\left(\frac{ 1-\left[\frac{L^{*}}{U}\right]^{N-1} }{ 1-\left[\frac{L^{*}}{U}\right]^{N} }\right)$$ If we set $U\to\infty$ and $L\to 0$ the we get the simpler exression $E(\theta|DI)=\frac{N}{N-1}X_{(N)}$.

But now suppose we use a more general prior, given by $p(\theta|cI)\propto\theta^{-c-1}$ (note that we keep the limits $L,U$ to ensure everything is proper - no singular maths then). The posterior is then the same as above, but with $N$ replaced by $c+N$ - provided that $c+N\geq 0$. Repeating the above calculations, we the simplified posterior mean of

$$E(\theta|DI)=\frac{N+c}{N+c-1}X_{(N)}$$

So the uniform prior ($c=-1$) will give an estimate of $\frac{N-1}{N-2}X_{(N)}$ provided that $N\geq 2$ (mean is infinite for $N=2$). This shows that the debate here is a bit like whether or not to use $N$ or $N-1$ as the divisor in the variance estimate.

One argument against the use of the improper uniform prior in this case is that the posterior is improper when $N=1$, as it is proportional to $\theta^{-1}$. But this only matters if $N=1$ or is very small.

Since the purpose here is presumably to obtain some valid and useful estimate of $\theta$, the prior distribution should be consistent with the specification of the distribution of the population from which the sample comes. This does NOT in any way mean that we "calculate" the prior using the sample itself -this would nullify the validity of the whole procedure. We do know that the population from which the sample comes is a population of i.i.d. uniform random variables each ranging in $[0,\theta]$. This is a maintained assumption and is part of the prior information that we possess (and it has nothing to do with the sample, i.e. with a specific realization of a subset of these random variables).

Now assume that this population consists of $m$ random variables, (while our sample consists of $n<m$ realizations of $n$ random variables). The maintained assumption tells us that

$$\max_{i=1,...,n}\{X_i\}\le \max_{j=1,...,m}\{X_j\} \le \theta$$

Denote for compactness $\max_{i=1,...,n}\{X_i\} \equiv X^*$. Then we have $\theta \ge X^*$ which can also be written

$$\theta = cX^*\qquad c\ge 1$$

The density function of the $\max$ of $N$ i.i.d Uniform r.v.'s ranging in $[0,\theta]$ is

$$f_{X^*}(x^*) = N\frac {(x^*)^{N-1}}{\theta^N}$$

for the support $[0,\theta]$, and zero elsewhere. Then by using $\theta = cX^*$ and applying the change-of-variable formula we obtain a prior distribution for $\theta$ that is consistent with the maintained assumption:

$$f_p(\theta) = N\frac {(\frac{\theta}{c})^{N-1}}{\theta^N}\frac 1c = \frac {N}{c^N} \theta^{-1}\qquad \theta \in [x^*, \infty]$$

which may be improper if we don't specify the constant $c$ suitably. But our interest lies in having a proper posterior for $\theta$, and also, we do not want to restrict the possible values of $\theta$ (beyond the restriction implied by the maintained assumption). So we leave $c$ undetermined.
Then writing $\mathbf X = \{x_1,..,x_n\}$ the posterior is

$$f(\theta \mid \mathbf X)\; \propto\; \theta^{-N}\frac {N}{c^N} \theta^{-1} \Rightarrow f(\theta \mid \mathbf X) = A\frac {N}{c^N} \theta^{-(N+1)}$$

for some normalizing constant A. We want

$$\int_{S_{\theta}}f(\theta \mid \mathbf X)d\theta =1 \Rightarrow \int_{x^*}^{\infty}A\frac {N}{c^N} \theta^{-(N+1)}d\theta =1$$

$$\Rightarrow A\frac {N}{c^N}\frac {1}{-N}\theta^{-N}\Big |_{x^*}^{\infty} = 1 \Rightarrow A = (cx^*)^N$$

Inserting into the posterior

$$f(\theta \mid \mathbf X) = (cx^*)^N\frac {N}{c^N} \theta^{-(N+1)} = N(x^*)^N\theta^{-(N+1)}$$

Note that the undetermined constant $c$ of the prior distribution has conveniently cancelled out.

The posterior summarizes all the information that the specific sample can give us regarding the value of $\theta$. If we want to obtain a specific value for $\theta$ we can easily calculate the expected value of the posterior,

$$E(\theta\mid \mathbf X) = \int_{x^*}^{\infty}\theta N(x^*)^N\theta^{-(N+1)}d\theta = -\frac{N}{N-1}(x^*)^N\theta^{-N+1}\Big |_{x^*}^{\infty} = \frac{N}{N-1}x^*$$

Is there any intuition in this result? Well, as the number of $X$'s increases, the more likely is that the maximum realization among them will be closer and closer to their upper bound, $\theta$ - which is exactly what the posterior mean value of $\theta$ reflects: if, say, $N=2 \Rightarrow E(\theta\mid \mathbf X) = 2x^*$, but if $N=10 \Rightarrow E(\theta\mid \mathbf X) = \frac{10}{9}x^*$. This shows that our tactic regarding the selection of the prior was reasonable and consistent with the problem at hand, but not necessarily "optimal" in some sense.

Uniform Prior Distribution Theorem (interval case):

"If the totality of Your information about $\theta$ external to the data $D$ is captured by the single proposition

$$B=\{\{\text{Possible values for } \theta\}=\{\text{the interval } (a,b)\},a<b\}$$ then Your only possible logically-internally-consistent prior specification is $$f(\theta)=\text{Uniform}(a,b)$$

Thus, you prior specification should correspond to the Jeffrey's prior if you truly believe in the above theorem."

Not part of the Uniform Prior Distribution Theorem:

Alternatively you could specify your prior distribution $f(\theta)$ as a Pareto distribution, which is the conjugate distribution for the uniform, knowing that you posterior distribution will have to be another uniform distribution by conjugacy. However, if you use the Pareto distribution, then you will need to specify parameters of the Pareto distribution in some sort of way.