Vtyjkyuk

I am trying to solve this problem from a past exam paper:

Let $X_1, dots, X_n$ be a random sample of size $n$ from the density:

$p(x; theta) = exp -(x-theta) , qquad x > theta$

In a Bayesian framework, assume that the prior distribution for
$theta$, $pi(theta)$, is a Uniform distribution on the interval
$(0, B)$, where $B$ is a known positive constant. Determine the
posterior distribution of $theta$, clearly stating the normalising
constant.

I know that posterior $propto$ prior $times$ likelihood.

Prior: $$pi(theta) = dfrac1B$$

Likelihood: $$p(underlinex ; theta) = prod_i=i^n exp - (x_i - theta) = exp left - left(sum_i=1^n x_i - nthetaright) right $$

EDIT: Changed equality to proportionality in response to the comment below.

Posterior: $$ pi(theta mid underlinex) propto dfrac1B times exp left - left(sum_i=1^n x_i - nthetaright) right $$

Now I am stuck. I know the posterior distribution needs to integrate to 1, and this is the way I can find the normalising constant. However, I am unsure about how to find the limits of the integration and how to actually compute this integration (I assume the integration is with respect to $theta$?)

Minor comment: posterior is proportional to the product of prior and maximum likelihood, not equal.

The integration is indeed with respect to $theta$. To find the limits refer to the definition of p.d.f., namely it's non-zero when $theta < x$. Hence,
$$
pi(theta mid underlinex) = frac 1 mathcalZ cdot dfrac1B times exp left - left(sum_i=1^n x_i - nthetaright) right
$$
and we need to find the constant $mathcalZ$. Integrating one gets that
$$
int pi(theta mid underlinex) ~dtheta = frac 1 mathcalZ cdot frac 1 nB e^-sum_i=1^n x_i int_-infty^max(x_i) e^ntheta ~ d(ntheta) = frac 1nB e^nmax(x_i) - sum_i=1^n x_i cdot frac 1 mathcalZ = 1 implies \ mathcalZ = Bn cdot e^- nmax(x_i) + sum_i=1^n x_i.
$$
Finally, the posterior distribution can be written as follows
$$
pi(theta mid underlinex) = frac 1 B^2n cdot expleftn[max(x_i) + theta] -2 sum_i=1^n x_iright.
$$