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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

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Reminded me of Benford's Law (en.wikipedia.org/wiki/Benford's_law) –  Sim May 20 '13 at 3:18
    
Interesting. But I was looking for a categoric probability distribution, like Dirchlet. –  Filipe Ferminiano May 20 '13 at 18:48
    
I don't think this is well posed. What is the sample you observe? Has the person chosen $\alpha$ and $\beta$ anew each time you take a sample? Otherwise you should always get the same $\mu$ –  Ross Millikan May 20 '13 at 20:49
    
The sample for $\mu$ is {10,12,45,50},for $\alpha$ is {1,1,4,5} and for $\beta$ is {0,2,5,0}. $\mu$ can any number between (0;99), $\alpha$ and $\beta$ can be any number between (0;9). The person choosed only one time $\alpha$ and $\beta$, after this $\mu$ 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 $\mu$ fixed. My estimative will depends on the sample that I observed. –  Filipe Ferminiano May 20 '13 at 20:57
2  
Original poster asks: what is the value of μ?. Do you mean ... what is the expected value of $\mu$? –  wolfies May 26 '13 at 8:36

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