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The first envelope has the value $A$ or$B$.

Observe that $x$can assume only two values. If $x=A$(that means we have chosen the envelope with the amount $A$in it), we accept it with the probability $F\left(A\right)$and reject it with probability $1-F\left(A\right)$and take the second envelope, in which is amount $B$. The expected return in this case is

If $x=B$(that means we have chosen the envelope with the amount $B$in it), we accept it with the probability $F\left(B\right)$and reject it with probability $1-F\left(B\right)$and take the second envelope, in which is amount $A$. The expected return in this case is

$F\left(B\right)路B+(1-F(B\left)\right)路A$

Both envelopes are equivalent priorly, we have that the expected return is

Term $(A-B)\left(F\right(A)-F(B\left)\right)$ is always greater or equal to zero. Why? If $A鈮�B$, we have that$F\left(A\right)鈮�F\left(B\right)$ since $F$ is a non-decreasing function. Similarly if $A<B$.

$F$ is a non-decreasing function, if$A<B$.

The expected return under the $x$-strategy is always at least $(A+B)/2$and that it is strictly larger than$(A+B)/2$

We have three cases.

If $x<A<B$, we will accept the first envelope no matter what the value inside is. Since both envelopes are equivalent priorly, we have that the expected return is $\frac{A+B}{2}$

If $A<B<x$, no matter which envelope we take first, we will change it for the other one. Similarly, as in the first case, the expected return is also $\frac{A+B}{2}$.

If $A<x<B$ we have a bit different situation. If we chose envelope with amount $A$ in it, we will reject it and take the second one which has $B$ in it since $A<x$. If we chose the $B$ one, we will accept it since we have that $x<B$. In both cases, we have that we get amount $B$, so the expected return is equal to $B$.

Now we see that no matter which is the value of $x$, we have that the expected return is greater or equal to $\frac{A+B}{2}$.

Furthermore, if $A<x<B$ we have that

The expected return in that case is strictly greater than$\raisebox{1ex}{$A+B$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$.

Let$X$be a continuous random variable on the whole line.

Define$R$as the return. Using the law of the total expectation, we have that

$+E(R鈭�B<X)P(B<X)$

$=(*)$

Observe that $E(R|X<A)$, $E(R鈭�A<X<B)$, $E(R|B<X)$are always greater than equal to $\frac{A+B}{2}$and we know that $E(R鈭�A<X<B)>\frac{A+B}{2}$.

So, we have that,

$\left({}^{*}\right)>\frac{A+B}{2}\left(P\right(X<A)+P(A<X<B)+P(B<X\left)\right)=\frac{A+B}{2}$

since we have that, $P(X<A)+P(A<X<B)+P(B<X)=P\left(惟\right)=1$.

The law of total expectation is$P\left(惟\right)=1$.