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Suppose two coins $x$ and $y$ have "H" heads probability $p_x$ and $p_y$. $p_x$ and $p_y$ are independently drawn according to a uniform distribution over $[0,1]$.

Say that we know $p_x\geq p_y$. So, we update using order statistics:

$p_x$ follows the largest order statistics, whose CDF is $$p^2_x$$ while $p_y$ follows the smallest order statistics, whose CDF is $$2p_y-p_y^2.$$

The joint probability of $(p_x,p_y)$ will be given by $2$ as can be found here.

My question is what happens we flipped coin $x$ once and the outcome is $H$?

Should we or shoudn't we update the distribution of $p_x$ and $p_y$ after the observation? If we should, what would be the posterior distribution?

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  • $\begingroup$ not sure you should use order statistics...likelihood is unchanged...change is only in the prior...your marginal posterior for $p_x$ will probably be some kind of incomplete beta function $\endgroup$ Sep 25, 2019 at 8:09
  • $\begingroup$ By the prior, you mean the uniform distribution? I was thinking about a similar problem.. In the same setting, suppose somehow that $p_x$ is revealed. Then we should revise the cdf of $p_y$ in a way that it is uniformly distributed below $p_x$.. which is the distribution of the smallest order statistics when the higher one's value is given by $p_x$. $\endgroup$
    – Andeanlll
    Sep 25, 2019 at 8:21

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Taking your prior, which is the uniform distribution, $f(p_x,p_y)=1$, re-normalised over $0\leq p_y\leq p_x\leq 1$. This is still "uniform" in the region $p_y\leq p_x$. So we can say

$$f(p_x, p_y|I)\propto I_{0\leq p_y\leq p_x\leq 1}$$

The likelihood of the data observed is equal to

$$f(x=H|p_x, p_y,I)=p_x$$

Now multiply them together and we have

$$f(p_x, p_y|x=H,I)\propto (p_x, p_y|I) f(x=H|p_x, p_y,I)\propto p_x I_{0\leq p_y\leq p_x\leq 1}$$

If we want the normalizing constan, $Z$, just integrate. in this case it is easy

$$Z=\int_0^1 \int_0^1 p_x I_{p_x\geq p_y} dp_x dp_y=\int_0^1 \int_{p_y}^1 p_x dp_x dp_y=\frac{1}{2}\int_0^1 (1-p_y^2) dp_y=\frac{1}{3}$$

So the full posterior is $$f(p_x, p_y|x=H,I)=3 p_x I_{0\leq p_y\leq p_x\leq 1}$$

To show how it's different, if we calculate means, we have...$E(p_x|I)=\frac{2}{3}$ and $E(p_x|x=H,I)=\frac{3}{4}$....similarly we aslso have...$E(p_y|I)=\frac{1}{3}$ and $E(p_y|x=H,I)=\frac{3}{8}$. So the parameters both have the posterior mass shifted upwards. This is because the evidence provides "direct indication" that $p_x$ is more likely to be larger. As this is an upper bound on $p_y$ we have to allow for larger values of this parameter - the evidence provides "indirect indication" that $p_y$ is larger..

hope this helps!

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  • $\begingroup$ Thank you for your detailed answer. Could you explain the third expression from the top a bit more? I don't get how $f(p_x, p_y|x=H,I)\propto f(p_x, p_y|I) f(x=H|p_x, p_y,I)$ is derived. Denoting $p_x,p_y=A,~"x=H"=B,~I=C$, isn't it $f(A|BC)=\frac{f(ABC)}{f(BC)}=\frac{f(AC)}{f(BC)}\frac{f(ABC)}{f(AC)}$ which is different from your expression $\frac{f(AC)}{f(C)}\frac{f(ABC)}{f(AC)}$? $\endgroup$
    – Andeanlll
    Sep 25, 2019 at 15:59
  • $\begingroup$ Another thing is that why can't we start with the pdf of order statistics? When $p_y<p_x$ and when they are iid draws, I think the distribution of them can be represented using orders statistics.. I apologize if I'm slow. $\endgroup$
    – Andeanlll
    Sep 25, 2019 at 16:27
  • $\begingroup$ Oh, sorry. I got your point. It's just the way how to derive the posterior when we take $I$ as given. But I don't still get why the order statistics won't work.. $\endgroup$
    – Andeanlll
    Sep 26, 2019 at 2:32
  • $\begingroup$ the order statistics work for the prior, just not sure the posterior is an order statistic density. You could think of first generating the 2 order statistics $(p_x^{(b)},p_y^{(b)})$, and then accepting that sample with probability $p_x^{(b)}$, and rejecting it otherwise. The accepted samples have the same distribution as the posterior $\endgroup$ Sep 26, 2019 at 12:00
  • $\begingroup$ Got it. Thank you so much. Your answer really helped me a lot! $\endgroup$
    – Andeanlll
    Sep 26, 2019 at 14:19

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