モンティ・ホール問題好きのホームページ    privacy policy

Return to the list of my pages written in English about the two envelopes problem
2016/04/01 18:55:42
First edition 2016/02/20

Two methods for the proof of the equivalence of the envelopes of the two envelopes problem

Caution
I who am a Japanese wrote this page in English, but I am not so good at English.

Introduction

There are two method to prove the equivalence of the envelopes of the two envelopes problem.
One method uses the law of total expectation.
The other method uses calculating formulas of conditional expectations.

A method using the law of total expectation.

The law of total expectation is the theorem that the mean value of the conditional expectation is equal to the unconditional expectation.
For reference. Using this law we can prove the equivalence of the two envelopes.

Let x and y be the amounts in the envelope selected by you and the amount in the another envelope respectively.
Let X and Y be a random variables which take x and y as their value respectively.

step amount of money
chosen envelope
amount of money
oposit envelope
conditional expectations
under the condition a=x
x E(Y | X=x)
mean values of
conditional expectations
E(X) E[ E(Y | X=x) ]
law of total expectation   E(Y)
because
E(X) = E(Y)
  same as the chosen envelope

For reference.

A method using calculating formulas of conditional expectations.

Using the calculation formula of conditional expectation, we can prove the equivalence of the two envelopes.
For reference. I tried to make a concise proof by myself, and for this purpose I created an original concept as follows.

summantegra
Summantegra is made of summation and integral and denoted by a symbol .

Using this symbol I tried to prove the equivalence of the two envelopes.

Notations
Let x be the amount of money in the envelope selected by you.
Let X be a random variable which take x as its value.
Let a be the lesser amounts in the two envelopes.
Let r ( > 1) is the ratio of the greater amount to the lesser amount.
Let g(a) be the probability or probability density of the event of the pair of amount of money is (a, ra).
Let 1/k be the factor of integration by substitution by the change of variables from y (=rx) to x.
If the distribution of amount of money is discreet, 1/k = 1.
If the distribution of amount of money is continuous, dy/dx = r. Therefore 1/k = r.
Let P(x) be the probability or probability density of the event that X = x.

using P the equivalence of the mean values of X and Y can be proven. ( ← Added on April 1, 2016)

step amount of money
chosen envelope
amount of money
oposit envelope
conditional expectations
under the condition X=x
x k g(x/r) (x/r) + g(x) rx
-------------------------------
g(x) + k g(x/r)
mean values of
conditional expectations
xP(x)
(k g(x/r) (x/r) + g(x) rx)
-------------------------------
g(x) + k g(x/r)
P(x)
because
P(x) =
(1/2)(g(x) + k g(x/r))
1/2(k g(x/r) x + g(x) x)
1/2(k g(x/r) (x/r) + g(x) rx)
partition of summantegras
1/2(k g(x/r) x)
+
1/2g(x) x
1/2(k g(x/r) (x/r)
+
1/2g(x) rx
change of variables of the first summantegras
(x/r → x)
(k g(x) r x(1/k))
+
g(x) x
(k g(x) x(1/k))
+
g(x) rx
deformation
1/2g(x) r x
+
1/2g(x) x
1/2g(x) x
+
1/2g(x) rx
case of
r = 2
3/2g(x) x
3/2g(x) x


Remarkable thing
The steps from the partition through change of variables correspond to deformation of series.
This fact reminded me of the following articles.
If I illustrate the steps from the partition through change of variables, it will be as follows.

Before



After



↑ These drawings were added on February 28, 2016.

Addition
Indirect method using the equivalence of probability distributions

Notations
Let x be the amount of money in the envelope selected by you.
Let X be a random variable which take x as its value.
Let Y be a random variable which denotes the amount of money in your new envelope after switching under the condition you always switch.
Let a be the lesser amounts in the two envelopes.
Let r ( > 1) is the ratio of the greater amount to the lesser amount.
Let g(a) be the probability or probability density of the event of the pair of amount of money is (a, ra).
Let 1/k be the factor of integration by substitution by the change of variables from y (=rx) to x.
If the distribution of amount of money is discreet, 1/k = 1.
If the distribution of amount of money is continuous, dy/dx = r. Therefore 1/k = r.
Let P(x) be the probability or probability density of the event that X = x.
Let O(x) be the probability or probability density of the event that Y = x.
Let G(x) be the probability or probability density of the event that X = x/2 and Y = x.
Let L(x) be the probability or probability density of the event that X = 2x and Y = x.

using P the equivalence of the distribution of X and Y can be proven.

L(x) = g(x)
-------------------------------
g(x) + k g(x/r)
P(x)
=g(x)
-------------------------------
g(x) + k g(x/r)
(1/2)(g(x) + kg(x/r))
 
=
 
(1/2)g(x)
G(x) =g(x/r)
-------------------------------
g(x/r) + k g((x/r)/r)
P(x/r)
=g(x/r)
-------------------------------
g(x/r) + k g((x/r)/r)
(1/2)(k(g(x/r) + k2g((x/r)/r))
 
=
 
(1/2)kg(x/r)
∴ O(x) = L(x) + G(x) = (1/2)(g(x) + k(g(x/r)) = P(x).

In other words, the distribution of the amount of money before switch is same as it after switch.
So if mean values are finite they are same.

Reference



Return to the list of my pages written in English about the two envelopes problem