Question

Two envelopes, each containing a check, are placed in front of you. You are to choose one of the envelopes, open it, and see the amount of the check. At this point, either you can accept that amount or you can exchange it for the check in the unopened envelope. What should you do? Is it possible to devise a strategy that does better than just accepting the first envelope? Let $A$and $B$, $A<B$, denote the (unknown) amounts of the checks and note that the strategy that randomly selects an envelope and always accepts its check has an expected return of $(A+B)/2$. Consider the following strategy: Let $F(·)$be any strictly increasing (that is, continuous) distribution function. Choose an envelope randomly and open it. If the discovered check has the value $x$, then accept it with probability $F\left(x\right)$and exchange it with probability $1-F\left(x\right)$.

(a) Show that if you employ the latter strategy, then your expected return is greater than $(A+B)/2$. Hint: Condition on whether the first envelope has the value $A$or $B$. Now consider the strategy that fixes a value x and then accepts the first check if its value is greater than x and exchanges it otherwise. (b) Show that for any $x$, the expected return under the$x$-strategy is always at least $(A+B)/2$and that it is strictly larger than $(A+B)/2$if $x$lies between $A$and $B$.

(c) Let $X$ be a continuous random variable on the whole line, and consider the following strategy: Generate the value of$X$, and if $X=x$, then employ the $x$-strategy of part (b). Show that the expected return under this strategy is greater than $(A+B)/2$.

Short answer

a). $F$is a non-decreasing function.

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

c).$P\left(\Omega \right)=1$

Step-by-step solution

Given Information(Part a)

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

Explanation (Part a)

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

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

Explanation (Part a)

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

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

Explanation (Part a)

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

$\frac{1}{2}\left(F\right(A)·A+(1-F\left(A\right))·B+F(B)·B+(1-F\left(B\right))·A)$$=\frac{A+B}{2}+\frac{1}{2}(A-B)\left(F\right(A)-F(B\left)\right)\ge \frac{A+B}{2}$

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

Final Answer (Part a)

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

Given Information (Part b)

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

Explanation (Part b)

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 $\frac{A+B}{2}$.

Explanation (Part b)

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

$\frac{A+B}{2}<\frac{B+B}{2}=B$

Final Answer (Part b)

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

Given Information (Part c)

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

Explanation (Part c)

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

$E\left(R\right)=E(R\mid X<A)P(X<A)+E(R\mid A<X<B)P(A<X<B)$

$+E(R\mid B<X)P(B<X)$

$=(*)$

Explantion (Part c)

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

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, $P(X<A)+P(A<X<B)+P(B<X)=P\left(\Omega \right)=1$.

Final Answer (Part c)

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