Bayesian Estimation: Posterior Mean Estimator

[economist13]

Homework Statement

$$X_1 , \dots ,X_n \sim U[0, \theta]$$ iid. $$\theta \sim U[0,1]$$

derive the Bayesian posterior mean estimator

Homework Equations

$$f(\theta |\vec{X}) = \frac{f( \vec{X}|\theta)f(\theta )}{f( \vec{X})}$$

The Attempt at a Solution

My line of thinking...

First, the marginal for X, $$f( \vec{X}) = \int f( \vec{X}|\theta)f(\theta )d\theta$$

Let $$x_M \equiv max_i \{X_i\}$$. Since it must be that $$\theta \geq x_M$$

$$f( \vec{X}) = \int_{x_M}^1 \frac{1}{\theta^n} d\theta = \frac{x_M^{-(n-1)}-1}{n-1}$$

Then $$f( \theta | \vec{X}) = \frac{1/\theta^n}{\frac{x_M^{-(n-1)}-1}{n-1}} = \frac{n-1}{\theta^n(x_M^{-(n-1)} -1)}$$

Then $$E( \theta | \vec{X}) = \int_0^1 \theta f(\theta |X) d \theta = \frac{n-1}{n-2} \frac{1}{1-x_M^{-(n-1)}$$I think my bounds of integration are wrong, or something to that effect...where did I go wrong?

 

[mighty2000]

Your approach is correct, but there are a few small mistakes in your calculations. Here is the correct solution:

First, the marginal for X is given by:

f(\vec{X}) = \int_0^1 \frac{1}{\theta^n} d\theta = \frac{1}{(n-1)x_M^{n-1}}

Next, the posterior distribution for \theta is given by:

f(\theta | \vec{X}) = \frac{f(\vec{X} | \theta) f(\theta)}{f(\vec{X})} = \frac{1}{\theta^n} \frac{1}{(n-1)x_M^{n-1}} = \frac{1}{(n-1)\theta^n x_M^{n-1}}

Finally, the posterior mean estimator is given by:

E(\theta | \vec{X}) = \int_0^1 \theta f(\theta | \vec{X}) d\theta = \frac{n-1}{n-2} \frac{1}{1-x_M^{-(n-1)}}

Note that the bounds of integration are from 0 to 1, since \theta is uniformly distributed on [0,1]. Also, the denominator in the final expression should be (n-2), not (n-1).

Hope this helps!