Posterior distribution given maximum of uniform variables This is the question: (I am doing currently the a) )
https://i.stack.imgur.com/82F7X.png
So far, I say that:
$\text{Posterior probability} \propto \text{Likelihood}\times \text{Prior probability}.$
Then, we have Prior already, so we just need to compute Likelihood. I assume that samples are independent, so
$f_{X1,X2,X3,X4}(x_1,x_2,x_3,x_3\mid \theta) = \frac{1}{\theta} \times\frac{1}{\theta} \times\frac{1}{\theta} \times\frac{1}{\theta} = \frac{1}{\theta^{4}}$.
Therefore, $\text{Posterior probability} \propto \frac{1}{\theta^{4}}\times \frac{1}{10}.$
I am not sure how to derive the expression from the question:
$\pi(\theta | D) = c \cdot \frac{1}{\theta^{4}} ,   8<=\theta<10.$ 
I know that $\pi$ should integrate to 1 to be a valid density.
So, I say that $\int_{8}^{10} c \frac{0.1}{\theta^{4}} d\theta= 1. $
And I find that c=31475.44.  
 A: As per my comment, and your response: the likelihood of $\theta$ given that the maximum of four i.i.d. uniform $[0,\theta]$ variables is $x$ ($=8$ in your question), is given by:
$$L(\theta \, | \, x) = \frac{4 x^3}{\theta^4},$$
which clearly only holds if $\theta > x$ (else the maximum could not equal $x$).
Then, the product of the likelihood and the prior distribution $p(\theta)$ (which is non-zero only on $[0,10]$) is
$$ L(\theta \, | \, x) \times p(\theta) = \frac{4x^3}{10 \,\theta^4}, \qquad \theta \in [x,10], $$
note that the domain $[x,10]$ is obtained from both the condition that the likelihood is only non-zero on $\theta > x$, and that the prior distribution is supported on $[0,10]$; combined these make the interval $[x,10]$.
Hence we find that the posterior distribution is proportional to:
$$P(\theta\,|\,x) \propto \frac{4x^3}{10 \,\theta^4} \propto \frac{1}{\theta^4}, \qquad \theta \in [x,10]$$
where we have discarded the terms in the constants, and $x$, as these are absorbed into the normalizing constant of the distribution.
