Two Envelopes Paradox
You are handed two sealed envelopes. You are told that one contains exactly twice as much money as the other, but you do not know which is which or what the amounts are.
You pick one envelope, open it, and see $X$ dollars inside. You are then offered the chance to switch to the other envelope.
Here is the "switching argument": the other envelope contains either
X$ or $X/2$, each with probability /2$. So the expected value of switching is
$\frac{1}{2}(2X) + \frac{1}{2}\left(\frac{X}{2}\right) = \frac{5}{4}X > X.$
This suggests you should always switch -- but by the same logic you would switch back, and so on forever.
1. Explain precisely where the switching argument goes wrong.
2. Suppose the smaller amount $A$ is drawn from a known prior distribution. How does the correct conditional analysis depend on that prior?
3. Give an example of a proper prior where, for a specific observed $X$, switching is favorable, and another $X$ where it is not.
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