Two envelopes are given. Envelope 1 contains $x$ dollars and envelope 2 contains $2x$ dollars. We opened one of them and found in it $100$$. Now we have the option to change envelopes or not.
Formulate the problem as a Baisian estimation problem.
Is it worthwhile to exchange the envelopes?
This is the original question. In order to simplify it, I added the condition that the envelopes can contain only $5{$},10{$},20{$}$ and $100{$}$ bills.
As I understood, a Baisian estimation problem formulation contains the following terms:
$\Omega$-the set of possible states, $X$-the set of observations,$P$-a probablistic model, $A$-possible actions, $\Gamma$-cost for any action.
I also understood that there is a risk function: $R(\alpha_k|x_j)=\Sigma_{\omega_i\in \Omega}\lambda(\alpha_k|\omega_i)P(\omega_i|x_j)$, a probability to be in a state $\omega_i$: $P(\omega_i|x_j)=\frac{P(x_j|\omega_i)}{P(x_j)}P_0(\omega_i)$, and a total risk function: $R[\alpha(x)]=\Sigma_jR(\alpha(x_j)|x_j)P(x_j)$.
But, I am not sure how to build this risk function for the above exercise. It seems like the set of optional actions is $a_0$-not exchanging envelopes,$a_1$-exchanging envelopes. What are the possible states set $\Omega$ for example? What is a possible state in this example? If I choose to exchange envelopes and there are $5+5+10$ bills in envelope 1 and $5+5+10+20$ in envelope 2, is this considred one possible state? Also, what is the cost of being wrong?
Any ideas? Thanks!