Richard_Kennaway comments on Have I Solved the Two Envelopes Problem Once and For All?

[Richard_Kennaway]

Here’s the general calculation.

Take any probability distribution defined on the set of all values $(x,y)$ where $x$ and $y$ are non-negative reals and $y=2x$. It can be discrete, continuous, or a mixture.

Let $p\left(x\right)$ be the marginal distribution over $x$. This method of defining $p$ avoids the distinction between choosing $x$ and then doubling it, or choosing $y$ and then halving it, or any other method of choosing $(x,y)$ such that $y=2x$.

Assume that $p$ has an expected value, denoted by $E\left(p\right)$.

The expected value of switching when the amount in the first envelope is in the range $[x-dx/2,x+dx/2]$ consists of two parts:

(i) The first envelope contains the smaller amount. This has probability $p\left(x\right)dx/2+O\left(d{x}^2\right)$. The division by 2 comes from the 50% chance of choosing the envelope with the smaller amount.

(ii) The first envelope contains the larger amount. This has probability $p\left(x/2\right)dx/4+O\left(d{x}^2\right)$. The extra factor of 2 comes from the fact that when the contents are in an interval of length $dx$, half of that (the amount chosen by the envelope-filler) is in an interval of length $dx/2$.

In the two cases the gain from switching is respectively $x+O\left(dx\right)$ or $-x/2+O\left(dx\right)$.

The expected gain given the contents is therefore $xp\left(x\right)/2-\left(x/2\right)p\left(x/2\right)/4+O\left(dx\right)$.

Multiply this by $dx$, let $dx$ tend to 0 (eliminating the term in $O\left(d{x}^2\right)$) and integrate over the real line:

${\int }_0^{\infty }\left(xp\right(x)/2-(x/2\left)p\right(x/2\left)/4\right)dx$

$={\int }_0^{\infty }xp\left(x\right)/2dx-{\int }_0^{\infty }\left(x/2\right)p\left(x/2\right)/4dx$

The first integral is $E\left(p\right)/2$. In the second, substitute $y=x/2$ (therefore $dx=2dy$), giving ${\int }_0^{\infty }yp\left(y\right)/2dy=E\left(p\right)/2$. The two integrals cancel.