# Find a sufficient statistic $Y$ for $\theta$ then find the Bayes estimator $w(Y)$

Let $$X_1,...,X_n$$ be a iid random sample having pdf $$\theta x^{\theta-1}1(0 < x \le 1)$$

Find a sufficient statistic $$Y$$ for $$\theta$$ then find the Bayes estimator $$w(Y)$$ based on this statistic using the loss function $$L(\theta,a) = (a-\theta)^2$$ where the prior distribution is exponential with mean $$\frac{1}{\beta}$$.

First sufficiency:

The likelihood function is $$\displaystyle L(\theta) = \Pi_{i = 1}^n\theta x_i^{\theta -1} = \theta^n(x_1\cdots x_n)^\theta(x_1\cdots x_n)^{-1}$$ thus by the factorization theorem we can take $$Y = (x_1\cdots x_n)^{-1}$$.

Bayes Estimator:

For square-error loss the estimator $$w(Y) = \hat{\theta} = E[\theta \mid Y\,]$$ i.e. the mean of the posterior.

For the posterior need to first solve $$m(y) = \displaystyle \int_0^\infty \beta e^{-\beta \theta}y^{1-\theta}d\theta$$ Is this a well known integral? I was trying to solve by u-substitution but I am making a mistake somewhere. I am trying $$u = y^{-\theta}, du = -y^{-\theta}\log(y)d\theta$$ but for some reason I can't see how to take care of $$e^{-\beta\theta}$$.

Before continuing would appreciate to know if this is correct:

EDIT: $$y^{-\theta} = e^{-\theta \log(y)}$$ so re-write as $$\displaystyle \beta y \int_{0^\infty}e^{-\theta(\beta + \log(y))}d\theta$$ and set $$u = -\theta(\beta + \log(y))$$

Then we will have $$\displaystyle -\frac{\beta y}{\beta + y}e^{-\theta(\beta + \log(y))} \bigg \vert_{\theta = 0}^\infty= \frac{\beta y}{\beta + y}$$

Would still like to know if this a well known integral.

Now the next step is to solve $$\displaystyle E[\theta \mid Y] = \int_0^\infty \theta \frac{y+\beta}{\beta y}\beta e^{-\beta \theta}\theta^ny^{1-\theta}d\theta$$ correct? and this will give use the estimator we seek.

Using factorization theorem the sufficient statistic for $$\theta$$ is $$y=\prod_i X_i$$. This because the function $$g(\theta,t(\mathbf{x}))$$ depends on the data only through the statistic "t=product".

The function $$\frac{1}{\prod_{i}X_{i}}$$ you wrongly identfied as the sufficient statistic is the function of "x alone".

Then the posterior is the following (hint: when calculating the posterior discard any quantity that does not depend on $$\theta$$)

$$\pi(\theta|y) \propto e^{-\beta \theta}\theta^n y^{\theta-1}$$

$$\propto e^{-\beta \theta}\theta^n e^{(\theta-1) log y}$$

$$\propto \theta^ne^{-(\beta-logy)\theta}$$

...we immediately recognize in this posterior a Gamma distribution...

now you can kill the problem by yourself without solving the integral analitically

• The expectation will then give $\displaystyle \frac{n+1}{\beta-\log(Y)}$ correct? Based off of being a Gamma distribution. – oliverjones Aug 19 at 23:20
• @oliverjones : correct! – tommik Aug 20 at 6:27