# Calculate the Bayes estimate for the improper prior $\pi (\theta)=1$ under uniform distribution

Let $$X_1, X_2,...,X_n$$ be iid uniform $$U(0, \theta), 0<\theta <\infty$$.

Assume a quadratic loss function. Calculate the Bayes estimate for the improper prior $$\pi (\theta)=1, 0<\theta <\infty$$

Verify whether the Bayes estimate is consistent for $$\theta$$

My attempt:

Given the quadratic loss function, the Bayes estimator is the posterior mean. I first try to get the posterior distribution of $$\theta$$. I use capital $$X$$ to denote the n observations.

$$\pi (\theta | X) = f(X |\theta) \pi (\theta)= \theta^{-n} 1_{X_{n}< \theta}$$ What this distribution is? I know if I just focus on $$\theta^{-n}$$, then this is Beta(1-n, 1).

The solution is:

To get the posterior distribution of $$\theta$$, you use Bayes' Theorem: $$\pi(\theta \mid X) = \frac{f(X \mid \theta) \pi(\theta)}{\int_{0}^{\infty} f(X \mid \theta) \pi(\theta) d\theta}.$$ Note that we need the integral in the denominator to make the posterior a proper distribution.

Since we know that $$f(X \mid \theta) = \theta^{-n} I_{(X_{(n)} < \theta)} \textrm{ and } \pi(\theta) = 1$$ we get $$\pi(\theta \mid X) = \frac{\theta^{-n} I_{(X_{(n)} < \theta)}}{\int_{0}^{\infty} \theta^{-n} I_{(X_{(n)} < \theta)} d\theta} = \frac{\theta^{-n} I_{(X_{(n)} < \theta)}}{\int_{X_{(n)}}^{\infty} \theta^{-n} d\theta}.$$

Therefore $$E(\theta \mid X) = \int_{0}^{\infty}\theta \pi(\theta \mid X) d\theta = \frac{\int_{0}^{\infty}\theta \theta^{-n} I_{(X_{(n)} < \theta)}d\theta}{\int_{X_{(n)}}^{\infty} \theta^{-n} d\theta} = \frac{\int_{X_{(n)}}^{\infty}\theta \theta^{-n}d\theta}{\int_{X_{(n)}}^{\infty} \theta^{-n} d\theta}.$$

This gives you the formula that you put a question mark next to in your question.

Then we have $$\int_{X_{(n)}}^{\infty}\theta \theta^{-n}d\theta = \int_{X_{(n)}}^{\infty}\theta^{-n+1}d\theta = \left[\frac{\theta^{-n+2}}{-n+2}\right]_{X_{(n)}}^\infty = \frac{X_{(n)}^{-n+2}}{n-2}$$ and also $$\int_{X_{(n)}}^{\infty} \theta^{-n} d\theta = \left[\frac{\theta^{-n+1}}{-n+1}\right]_{X_{(n)}}^\infty = \frac{X_{(n)}^{-n+1}}{n-1}.$$

So the fraction is $$\frac{n-1}{n-2}X_{(n)},$$ and since $$(n-1)/(n-2) \rightarrow 1$$ as $$n \rightarrow \infty$$, and $$X_{(n)} \xrightarrow{P} \theta$$ we are done.

Proof that $$X_{(n)} \xrightarrow{P} \theta$$:

To show convergence in probability, we need to show that for any $$\epsilon > 0$$, $$\lim_{n\rightarrow \infty} P(\mid X_{(n)} - \theta \mid > \epsilon) = 0.$$

The random variable $$X_{(n)} = \max(\{X_1, \dots, X_n\})$$ is guaranteed to be less than $$\theta$$, so $$\mid X_{(n)} - \theta \mid = \theta - X_{(n)}$$. Therefore $$P(\mid X_{(n)} - \theta \mid > \epsilon) = P(\theta - X_{(n)} > \epsilon) = P(\theta - \epsilon > X_{(n)}).$$

But $$\theta - \epsilon > X_{(n)}$$ if and only if $$\theta - \epsilon > X_i$$ for every $$i$$, since $$X_{(n)}$$ is the maximum. Therefore $$P(\theta - \epsilon > X_{(n)}) = P(\theta - \epsilon > X_1, \theta - \epsilon > X_2, \dots, \theta - \epsilon > X_n).$$ The $$X_i$$ are independent and identically distributed so $$P(\theta - \epsilon > X_1, \dots, \theta - \epsilon > X_n) = P(\theta - \epsilon > X_1) \times \cdots \times P(\theta - \epsilon > X_n) = P(\theta - \epsilon > X_i)^n.$$ Since $$X_i \sim U(0, \theta)$$ then as long as $$0 < \epsilon < \theta$$ we have $$P(\theta - \epsilon > X_i) = \int_{\theta-\epsilon}^\theta \frac{1}{\theta} dx = \frac{\epsilon}{\theta}.$$ Putting this together we have that if $$0 < \epsilon < \theta$$ then $$P(\mid X_{(n)} - \theta \mid > \epsilon) = \left(\frac{\epsilon}{\theta}\right)^n \rightarrow 0 \textrm{ as } n \rightarrow \infty,$$ and if $$\epsilon \geq \theta$$ then $$P(\mid X_{(n)} - \theta \mid > \epsilon) = 0$$ since $$0 < X_{(n)} < \theta$$.

In either case we have $$\lim_{n\rightarrow \infty} P(\mid X_{(n)} - \theta \mid > \epsilon) = 0$$.

• How to see $X_{(n)} \xrightarrow{P} \theta$? Commented May 20, 2023 at 12:42
• @Jackie I've edited my answer to include the proof of convergence in probability
– Alex
Commented May 20, 2023 at 16:15