How to achieve the conditional probability distribution function in this case?

Question

I am looking this sample Bayesian problem. I do not get how they get this formula in the second case.

$f_{X\mid\Theta}(x\mid\theta ') = 1/ \theta^n$

Shouldn't it be just $1/\theta_{max}$? Please help me understand. TIA!

Answer 1

$f_{X|\Theta}(x|\theta)$ is the likelihood function of observing $X$ conditioned on the fact that distribution is uniform with parameter $\theta$. Now consider the likelihood of observing $n$ variables $X_1,X_2,\cdots,X_n$, when $\forall i: X_i$'s are i.i.d. (identically independently distributed). because of independence we have

$$f_{X_1,\cdots,X_n|\Theta}(x_1,\cdots,x_n|\theta) = f_{X_1|\Theta}(x_1|\theta)\times \cdots \times f_{X_n|\Theta}(x_n|\theta)$$

Here a subtle fact is that all $f_{X_i|\Theta}(x_i|\theta)$ must be non-zero and the parameter $\theta$ is shared. For $X_i$ we have $$f_{X_i|\Theta}(x_i|\theta) \begin{cases} \frac{1}{\theta}, \qquad \text{if} \quad 0\le x_i \le\theta \\ 0, \qquad \text{otherwise}\end{cases}$$

When we consider all of $x_i$'s this way, the likelihood equation given emerges. In other words $\forall i: 0\le x_i \le \theta \le 1$ which is equivalent to $0 \le \max\{ x_1,\cdots,x_n \} \le \theta \le 1$.

Now down to integration, $c$ is some normalizing constant, therefore the integration in denominator is unnecessary to calculate, the only thing we need to calculate is $f_{\Theta}f_{X|\Theta}$, i.e. the nominator of fraction which with the given explanations is obvious.

Answer 2

From the uniformity of the conditional distribution of $\ X_i\ $ over $\ [0,\theta]\ $ given $\ \Theta=\theta\ $, we have $$ P\left(X_i\le x_i\right)=\cases{\frac{x_i}{\theta}&if $ 0\le x_i\le\theta\ $\\ 0 &otherwise}\ , $$ so $$ f_{X_i|\Theta}\left(x_i |\theta\right)=\cases{\frac{1}{\theta}&if $ 0\le x_i\le\theta\le1\ $\\ 0 &otherwise}\ , $$ and from the conditional independence of $\ X_1,X_2,\dots,X_n\ $ given $\ \Theta=\theta\ $, we get \begin{align} f_{X|\Theta}\left(x|\theta\right)&= f_{X_1|\Theta}\left(x_1|\theta\right) f_{X_2|\Theta}\left(x_2|\theta\right)\dots f_{X_n|\Theta}\left(x_n |\theta\right)\\ &=\cases{\frac{1}{\theta^n}&if $ 0\le x_i\le\theta\le1\ $ for all $i=1,2\dots n\ $\\ 0 &otherwise}\ . \end{align}