If someone choose a digit $\alpha$ and a digit $\beta$ independently. Each one can be in $0,1, ...,9$. So $\mu = \alpha \beta$ (e.g. if $\alpha = 5$ and $\beta = 3$ then $\mu =53$). And I observe a sample: $\{10,12,45,50\}$. How can I attribute a likelihood distribution and a prior distribution to $\alpha$ and $\beta$ to test the hypothesis that $\alpha < \beta$? The sample for μ is {10,12,45,50},for α is {1,1,4,5} and for β is {0,2,5,0}. μ can any number between (0;99), α and β can be any number between (0;9). The person choosed only one time α and β, after this μ is fixed for a number of balls in an urn that can have (0;99) balls depending on the one's choice. Only after this, I observe the sample, with μ fixed. My estimative will depends on the sample that I observed.
Here is my answer:
$p(\alpha) = \frac{1}{10}$
For estimate $\alpha$, the most important digit of $\mu$ is the left one. So I call $a_{i}$ the left digit of the $n_{i}$ observation. And I call $a = max(a_{i}), i=1,...,n$. I don't know what is the pdf of $a = max(a_{i})$, but I know its cdf:
$p(a \leq a_{i} |\alpha) = p(a_{1} \leq a |\alpha)p(a_{2} \leq |\alpha)...p(a_{n} \leq a |\alpha) = (\frac{max(a_{i})}{\alpha})^n = F(a|\alpha) $
To find the pdf: $\frac{dF(a|\alpha)}{da} = \frac{n*max(a_{i})^{n-1}}{\alpha^{n}}$
So the posterior of $\alpha$: $p(\alpha | a) \propto 1 \times \frac{n*max(a_{i})^{n-1}}{\alpha^{n}}$
The same for $\beta$: $p(\beta| b) \propto 1 \times \frac{n*max(b_{i})^{n-1}}{\beta^{n}}$
My only problem is how to test the hypothesis that $\alpha < \beta$?
The Bayes Fator will be:
$BF = \frac{\frac{n*max(a_{i})^{n-1}}{\alpha^{n}}}{\frac{n*max(b_{i})^{n-1}}{\beta^{n}}} = \frac{\alpha^{n}}{\beta^{n}} \times \frac{n*max(a_{i})^{n-1}}{n*max(b_{i})^{n-1}} $
That still depends on $\alpha$ and $\beta$, unknown