Two Envelopes Paradox and Improper Prior Resolution

Scenario

You are given two indistinguishable envelopes. One contains $X$ dollars and the other contains $2X$ dollars. You choose one envelope at random and open it to find \A$. The naive expected value calculation suggests:

• With probability $0.5$, the other envelope contains $2A$.
• With probability $0.5$, the other envelope contains $A/2$.
• Expected value of switching: $\mathbb{E}[\text{Other}] = 0.5(2A) + 0.5(A/2) = 1.25A > A$. This suggests one should always switch, yet before opening the envelope, the envelopes are symmetric.

Your Task

1. Identify the flawed implicit assumption of an improper prior over $\mathbb{R}^+$ in the naive calculation.
2. Let $p(x)$ be any proper continuous probability density for the smaller amount $X$. Formulate the exact Bayesian posterior probability $\mathbb{P}(X = A/2 \mid \text{observed } A)$.
3. Prove that for any valid proper probability distribution, the expected gain from switching integrated over all possible observed values $A$ is identically zero: $\int_0^\infty \mathbb{E}[\text{Gain} \mid A] f(A) dA = 0$

Constraints

• Kolmogorov probability axioms: Proper probability distributions must integrate to 1.

Examples A uniform prior over $[0, \infty)$ is improper (cannot integrate to 1). Under any proper distribution (e.g., $X \sim \text{Exp}(\lambda)$), observing large $A$ increases the posterior probability that $A$ is the larger envelope.