I need to find the Bayes estimator for a $U[0, \theta]$ distribution, with $\theta$ distribution being $q(t) = \frac{1}{t^2}$ for $t \geqslant 1$.
This is my first time attempting to do something of this sort, so I still don't understand some things. I know the formulas:
$\theta_n^* = \int\limits_{\Theta}t \cdot q(t|x_1, \ldots, x_n) \ dt$
$q(t|x_1, \ldots, x_n) = \frac{f_t(x_1, \ldots, x_n) \ q(t)}{\int\limits_{\Theta}f_s(x_1, \ldots, x_n) \ q(s) \ ds}$
So, in this case, $f_t(x_1, \ldots, x_n) = \frac{1}{\theta^n} I_{[0, \theta]}(X_{n)})$, where $I$ is the indicator function.
First of all, I don't really get how to integrate expressions containing indicator functions? Also, what are the integration limits, is $\Theta$ the $[1; \infty)$ half-interval?
I know the answer for this problem, and it contains a $\frac{n+1}{n}$ fraction, but how do I even obtain this here, if the only thing containing a degree of $n$ is $\frac{1}{\theta^n}$, but do they not cancel out in the numerator and the denumenator?
Basically, I'm kind of lost trying to apply these formulas, so all my questions are technical and overly specific, but maybe this could serve as an example of finding the Bayes estimator for other people.
This problem is rather tricky, and requires some sleight of hand to deal with the presence of indicator functions. To work with indicator functions [and note that $q$ also involves an indicator: $q(t)=\frac1{t^2}I_{[1;\infty)}(t)$ ], always recall the interpretation: $I_A(u)$ equals $1$ when $u\in A$, and equals $0$ otherwise.
You are right, the parameter space $\Theta$ here is the half-interval $[1;\infty)$. However, note that the dummy variable for $\theta$ is written $t$, so your joint density $f_t(x_1,\ldots,x_n)$ for the observed $X_1,\ldots,X_n$ is: $$ f_t(x_1,\ldots,x_n)=\frac1{t^n}I_{[0,t]}(x_{(n)}). $$ To obtain $\theta_n^*$ you'll need to evaluate the expression $$ \theta_n^*:=\frac{\int_\Theta t\cdot f_t(x_1,\ldots,x_n)q(t)\,dt}{\int_\Theta f_t(x_1,\ldots,x_n)q(t)\,dt},\tag{*} $$ which is the ratio of two integrals. Each integral is handled the same way, by plugging in $f_t$ and $q$. Here's how to handle the denominator:
To plug in $f_t$ and $q$, you will need to include the indicators: $$ \int_\Theta f_t(x_1,\ldots,x_n)q(t)\,dt=\int_\Theta \frac1{t^n}I_{[0,t]}(x_{(n)})\frac1{t^2}I_{[1;\infty)}(t)\,dt.\tag1 $$ The confusing part here is that we are integrating over $t$, yet the first indicator looks like a function of $x_{(n)}$ which is a constant as far as $t$ is concerned. The trick is to rewrite $I_{[0,t]}(x_{(n)})$ as $I_{[x_{(n)};\infty)}(t)$ (convince yourself that this is true!). Now the integral (1) has the form $$ \int_\Theta\frac1{t^{n+2}}I_{[x_{(n)};\infty)}(t)I_{[1;\infty)}(t)\,dt.\tag2 $$ You can combine the two indicators into one, since $I_A(u)I_B(u) = I_{A\cap B}(u)$. Now, since $$[x_{(n)};\infty)\cap[1;\infty)=[\max(x_{(n)},1);\infty),\tag3$$ you can see the integral (2) equals $$ \int_\Theta\frac1{t^{n+2}}I_{[\max(x_{(n)},1);\infty)}(t)\,dt=\int_{ \max(x_{(n)},1)}^\infty\frac1{t^{n+2}}\,dt,\tag4 $$ which I'll leave to you to evaluate.