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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.

owen88
  • 4,670
  • Hint: the question is not asking yu to find the likelihood of the observations $x_1,\ldots, x_4$, but rather the likelihood of the maximum $\max(x_1,\ldots,x_4)$. To do this you will need to derive (or look up!) the probability distribution for the maximum of a set of independent uniform variables on an interval $[0,\theta]$. – owen88 Jan 01 '18 at 17:12
  • @owen88 if I do what is suggested on the linked thread, i.e. to find the CDF, when i differentiate back to find the pdf I am getting $f(x)=\frac{4x^{3}}{\theta^{4}}$, then the likelihood function ends up a big mess and far for the posterior needed. I have no idea what's goin' on there! – BayesApprentice Jan 01 '18 at 19:52

1 Answers1

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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.

owen88
  • 4,670
  • Hmmm, I though that the likelihood should be $\prod\limits_{i=1}^{4} (\frac{4x_i^3}{\theta^4})$ But you only do it for $i=1$. This part is unclear to me. Is it because we consider only one from the $x_i$'s , that is the max ? – BayesApprentice Jan 02 '18 at 00:48
  • The likelihood above (eg. the density of the maximum, conditioned on $\theta$) is for the maximum $x_{\text{max}} = max \left(x_1,\ldots, x_4\right)$; so we have already taken account of the different $i$ values in computing the density. – owen88 Jan 02 '18 at 12:58