Coin flipping with revision of probability

Question

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?

Answer 1

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!