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%TCIDATA{Created=Thu Oct 07 14:29:47 2004}
%TCIDATA{LastRevised=Fri Oct 15 22:19:39 2004}
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\begin_body
\begin_layout Title
Uncertainty
\end_layout
\begin_layout Author
Michael Peters
\end_layout
\begin_layout Date
\begin_inset ERT
status collapsed
\begin_layout Plain Layout
\backslash
today
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\end_layout
\begin_layout Section
Lotteries
\end_layout
\begin_layout Standard
In many problems in economics, people are forced to make decisions without
knowing exactly what the consequences will be.
For example, when you buy a lottery ticket, you don't know whether or not
you will win when you buy the ticket.
There are many important problems like this.
If you buy car insurance, you hope you won't need it, but you aren't sure.
A politician who spends time and money running for public office is not
sure whether or not they will be elected.
A drug company that invests in developing a new drug is not sure whether
or not it will actually work.
A corporate insider who sells shares using inside information is never
sure whether or not they will be caught.
\end_layout
\begin_layout Standard
One way to think about such problems is to use the concept of a
\emph on
lottery
\emph default
.
A lottery is a pair of objects.
The first,
\begin_inset Formula $\mathcal{X}$
\end_inset
, is a list of possible consequences of a decision.
The second is a list
\begin_inset Formula $p=\left\{ p_{1},p_{2},\dots p_{n}\right\} $
\end_inset
of probabilities with which you think that each of the consequences will
occur.
The number of probabilities you list,
\begin_inset Formula $n$
\end_inset
, is exactly equal to the number of consequences in
\begin_inset Formula $\mathcal{X}$
\end_inset
.
For example, if you buy a lottery ticket, the set of consequences consists
of two things, you either win or lose.
Each of these consequences occurs with probability
\begin_inset Formula $\frac{1}{2}$
\end_inset
.
If you sell stock based on an insider's tip, you either get away with it,
or you don't.
More generally, there could be many consequences.
If you open a new restaurant you might sell 1 or 2 or 3, or any number
of meals per week.
The set of consequences could be very large.
\end_layout
\begin_layout Standard
In this definition, the set of consequences could be very general.
In particular, each of the consequences could be a lottery.
A lottery over lotteries is called a
\emph on
compound lottery
\emph default
.
Lotteries over anything else are called
\emph on
simple
\emph default
lotteries.
As an abstract example, consider the following game.
I will flip a coin and if the coin comes up heads right away, I will give
you $2.
If it comes up tails, I will flip the coin again.
If it comes up heads on the second flip, I will pay you $4.
If it comes up tails, we stop and you get nothing.
This is a simple example of a compound lottery.
The first of the two lotteries has a single consequence, you receive $2
(of course you get this consequence with probability 1).
The second lottery has two possibly consequences - either you get $4 or
you get nothing.
In this second lottery, each consequence occurs with equal probability.
The compound lottery involves
\begin_inset Formula $\frac{1}{2}$
\end_inset
chance that you will play the first lottery and get $2 for sure.
Then there is
\begin_inset Formula $\frac{1}{2}$
\end_inset
chance that you will play the second lottery and get either $4 or $0.
\end_layout
\begin_layout Standard
It should be clear to you that this compound lottery is pretty much the
same thing as a simple lottery where the consequences are that you receive
either $2, $4 or $0 with probabilities
\begin_inset Formula $\frac{1}{2}$
\end_inset
,
\begin_inset Formula $\frac{1}{4}$
\end_inset
, and
\begin_inset Formula $\frac{1}{4}$
\end_inset
.
This type of lottery is sometimes referred to as the
\emph on
reduced
\emph default
lottery associated with the original compound lottery.
\end_layout
\begin_layout Subsection
Monty Hall
\end_layout
\begin_layout Standard
We normally assume that compound lotteries and the reduced lotteries associated
with them can be used interchangeably.
This isn't always as straightforward as in the example above.
In the following problem, it is very easy to make a mistake calculating
the reduced lottery.
You are a contestant in a game show, and are given the choice of three
doors.
Behind one door is $1 million which you will win if you happen to open
this door.
There is nothing behind the other doors, and if you pick one of them you
will win nothing.
Once you choose one of the doors, the host will open one of the remaining
doors and show you that it contains no prize.
You are then given the option to change from the door you picked in the
first place, to the remaining door.
The problem is to decide whether or not to switch doors.
\end_layout
\begin_layout Standard
This is a compound lottery in which the prize is first placed randomly behind
a door, then after observing where the prize is, the host randomly opens
one of the remaining doors.
It appears that the prize is equally likely to be behind either of the
three doors, so it can't matter whether you switch or not after the host
opens the door.
However, you will do better on average in this game if you always switch
doors.
You might be able to see this from the following casual reasoning - the
prize could be behind any of the three doors.
If it is behind the door that you chose, then it would, of course, be a
mistake to switch.
In either of the other alternatives, switching doors will win you the prize.
To put it another way, the host will actually tell you which door contains
the prize in two of the three situations you might face.
Following his advice won't always work, but it will most of the time.
\end_layout
\begin_layout Standard
Another way to think through the problem is to try to compute the reduced
lottery that you actually face when you hold on to your original choice,
and the one you face when you switch.
The first part of the lottery involves the placement of the prize.
With probability
\begin_inset Formula $\frac{1}{3}$
\end_inset
it is placed behind either of the three doors.
\begin_inset space \space{}
\end_inset
The second part of the lottery is the host's announcement about the door
that doesn't contain the prize.
We might as well assume that you pick door
\begin_inset Formula $A$
\end_inset
, since the thing works the same way no matter which door you pick.
The lottery you get when you stick with your original choice is depicted
in Figure 1.
\end_layout
\begin_layout Standard
\begin_inset Float figure
placement tbh
wide false
sideways false
status open
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\begin_inset space \hspace*{\fill}
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begin{egame}(600,380)
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%
\end_layout
\begin_layout Plain Layout
% put the initial branch at (300,240), with (x,y) direction
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% (2,1), and horizontal length 200
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\backslash
putbranch(300,340)(1,1){200}
\end_layout
\begin_layout Plain Layout
%
\end_layout
\begin_layout Plain Layout
% give the branch two actions, label it for player 1,
\end_layout
\begin_layout Plain Layout
% and label the actions $L$ and $R$
\end_layout
\begin_layout Plain Layout
\backslash
iiib{Door A,B or C}{$
\backslash
frac{1}{3}$}{$
\backslash
frac{1}{3}$}{$
\backslash
frac{1}{3}$}
\end_layout
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%
\end_layout
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\backslash
putbranch(300,140)(1,1){75}
\end_layout
\begin_layout Plain Layout
\backslash
ib{}{$1$}[$B,Lose$]
\end_layout
\begin_layout Plain Layout
\backslash
putbranch(500,140)(1,1){75}
\end_layout
\begin_layout Plain Layout
\backslash
ib{}{$1$}[$C,Lose$]
\end_layout
\begin_layout Plain Layout
\backslash
putbranch(100,140)(1,1){75}
\end_layout
\begin_layout Plain Layout
\backslash
iib{}{$
\backslash
frac{1}{2}$}{$
\backslash
frac{1}{2}$}[$B,Win$][$C,Win$]
\end_layout
\begin_layout Plain Layout
\backslash
end{egame}
\end_layout
\begin_layout Plain Layout
\end_layout
\end_inset
\begin_inset space \hspace*{\fill}
\end_inset
\begin_inset Caption
\begin_layout Plain Layout
You stick with your initial choice
\end_layout
\end_inset
\end_layout
\begin_layout Plain Layout
\begin_inset CommandInset label
LatexCommand label
name "f:one"
\end_inset
\end_layout
\end_inset
\end_layout
\begin_layout Standard
The important part to understanding the correct strategy in this game is
to observe that the outcome of the second lottery depends both on the outcome
of the first lottery (the door where the prize is placed) and on your choice.
In Figure 1, the first set of branches shows the various locations where
the prize can be placed.
The second set of branches shows the doors that the host can open.
Notice that if your choice is
\begin_inset Formula $A$
\end_inset
and the prize is actually there, then the host can choose completely randomly
to open either door
\begin_inset Formula $B$
\end_inset
or
\begin_inset Formula $C$
\end_inset
.
On the other hand, if the prize is behind doors
\begin_inset Formula $B$
\end_inset
or
\begin_inset Formula $C$
\end_inset
, then the host doesn't have any choice and is forced to open a door that
effectively reveals the location of the prize.
\end_layout
\begin_layout Standard
Figure 2 shows what the lottery looks like in the case where you switch
doors.
\end_layout
\begin_layout Standard
\begin_inset Float figure
placement tbh
wide false
sideways false
status open
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\begin_inset space \hspace*{\fill}
\end_inset
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status open
\begin_layout Plain Layout
\backslash
begin{egame}(600,380)
\end_layout
\begin_layout Plain Layout
%
\end_layout
\begin_layout Plain Layout
% put the initial branch at (300,240), with (x,y) direction
\end_layout
\begin_layout Plain Layout
% (2,1), and horizontal length 200
\end_layout
\begin_layout Plain Layout
\backslash
putbranch(300,340)(1,1){200}
\end_layout
\begin_layout Plain Layout
%
\end_layout
\begin_layout Plain Layout
% give the branch two actions, label it for player 1,
\end_layout
\begin_layout Plain Layout
% and label the actions $L$ and $R$
\end_layout
\begin_layout Plain Layout
\backslash
iiib{Door A,B or C}{$
\backslash
frac{1}{3}$}{$
\backslash
frac{1}{3}$}{$
\backslash
frac{1}{3}$}
\end_layout
\begin_layout Plain Layout
%
\end_layout
\begin_layout Plain Layout
\backslash
putbranch(300,140)(1,1){75}
\end_layout
\begin_layout Plain Layout
\backslash
ib{}{$1$}[$B,Win$]
\end_layout
\begin_layout Plain Layout
\backslash
putbranch(500,140)(1,1){75}
\end_layout
\begin_layout Plain Layout
\backslash
ib{}{$1$}[$C,Win$]
\end_layout
\begin_layout Plain Layout
\backslash
putbranch(100,140)(1,1){75}
\end_layout
\begin_layout Plain Layout
\backslash
iib{}{$
\backslash
frac{1}{2}$}{$
\backslash
frac{1}{2}$}[$B,Lose$][$C,Lose$]
\end_layout
\begin_layout Plain Layout
\backslash
end{egame}
\end_layout
\begin_layout Plain Layout
\end_layout
\end_inset
\begin_inset space \hspace*{\fill}
\end_inset
\begin_inset Caption
\begin_layout Plain Layout
You switch choices after the host opens a door
\end_layout
\end_inset
\end_layout
\begin_layout Plain Layout
\begin_inset CommandInset label
LatexCommand label
name "f:otwo"
\end_inset
\end_layout
\end_inset
\end_layout
\begin_layout Standard
Then just glancing at the outcomes in these two figures, it should be clear
that you will win two thirds of the time if you switch, but only one third
of the time if you stick with your initial choice.
\end_layout
\begin_layout Subsection
\begin_inset VSpace bigskip
\end_inset
St.
Petersburg
\end_layout
\begin_layout Standard
Here is a famous reduced lottery involving coins that actually provides
most of the motivation for the approach that we currently use in economics.
This resembles the previous coin flipping problem.
As before, if the coin comes up heads on the first flip, I give you $2,
and if it comes up tails on the first flip, I flip again.
If it comes up heads on the second flip I give you $4, otherwise I flip
again.
If it comes up heads on the third flip, I pay you $8, otherwise I flip
again.
\begin_inset space \space{}
\end_inset
We keep going until I flip a head, then if it takes me
\begin_inset Formula $k$
\end_inset
flips to get the head, I pay you
\begin_inset Formula $\$2^{k}$
\end_inset
.
The set of consequences in this reduced lottery is the set
\begin_inset Formula
\[
\left\{ 1,2,3,\dots,k,\dots\right\} .
\]
\end_inset
It won't take you too much thinking to see that the probabilities are
\begin_inset Formula
\[
\left\{ \frac{1}{2},\frac{1}{4},\dots,\frac{1}{2^{k}},\dots\right\}
\]
\end_inset
\end_layout
\begin_layout Standard
I am going to ask you how much you would be willing to pay me to play this
game.
You could refuse to play - then you would get nothing for sure.
Or you could offer to pay me $2 or more (you can't lose from this choice
unless you pay more than $2).
Both choices would involve different lotteries, though one of them (not
playing) is sort of degenerate.
\end_layout
\begin_layout Standard
You might try to decide whether or not to play this game by figuring out
how much you would win on average from the game.
This calculation is straightforward.
\begin_inset space \space{}
\end_inset
With probability
\begin_inset Formula $\frac{1}{2}$
\end_inset
you win $2 right away, with probability
\begin_inset Formula $\frac{1}{4}$
\end_inset
you win $4, with probability
\begin_inset Formula $\dots$
\end_inset
\begin_inset Formula $\frac{1}{2^{k}}$
\end_inset
you win $2
\begin_inset Formula $^{k}$
\end_inset
.
Averaging all these gives
\begin_inset Formula
\[
\sum_{i=1}^{\infty}\frac{1}{2^{i}}2^{i}=\sum_{i=1}^{\infty}1=\infty
\]
\end_inset
You will never find anyone who is willing to pay an infinite amount of money
to play this game.
\end_layout
\begin_layout Section
\begin_inset VSpace bigskip
\end_inset
Choosing among lotteries
\end_layout
\begin_layout Standard
In each of the problems above, you need to choose among lotteries.
There is no obvious way to do this.
However, as you actually make the choice, I am probably safe in thinking
that you will be able to express a preference over any pair of lotteries,
and that the preferences you express will be transitive (in other words,
if you say that you prefer lottery
\begin_inset Formula $\left(\mathcal{X},p\right)$
\end_inset
to lottery
\begin_inset Formula $\left(\mathcal{X}^{\prime},p^{\prime}\right)$
\end_inset
, and you say you like lottery
\begin_inset Formula $\left(\mathcal{X}^{\prime},p^{\prime}\right)$
\end_inset
more than lottery
\begin_inset Formula $\left(\mathcal{X}^{\prime\prime},p^{\prime\prime}\right)$
\end_inset
, then I should be sure that you will prefer lottery
\begin_inset Formula $\left(\mathcal{X},p\right)$
\end_inset
to lottery
\begin_inset Formula $\left(\mathcal{X}^{\prime\prime},p^{\prime\prime}\right)$
\end_inset
.
\end_layout
\begin_layout Standard
Now, if there is some set of lotteries
\begin_inset Formula $\mathcal{L}$
\end_inset
from which a choice is to be made, I
\begin_inset space \space{}
\end_inset
can ask for pairwise comparisons between all the lotteries and eventually
learn all of your preferences.
To make the notation a little simpler, let me suppose that every lottery
in my set of alternatives
\begin_inset Formula $\mathcal{L}$
\end_inset
has the same set
\begin_inset Formula $\mathcal{X}$
\end_inset
of consequences.
Then I can think of a lottery as a simple list of probabilities with which
these various outcomes occur.
The outcomes don't have to be amounts of money, they can be anything imaginable
, including lotteries as you have seen.
Yet, as with all choice problems I am probably not too far off the mark
by assuming you can express a preference between
\emph on
every
\emph default
pair of lotteries in
\begin_inset Formula $\mathcal{L}$
\end_inset
, and that the preferences you express will be transitive (which means that
if
\begin_inset Formula $p\succeq p^{\prime}$
\end_inset
, and
\begin_inset Formula $p^{\prime}\succeq p^{\prime\prime}$
\end_inset
then it must be that
\begin_inset Formula $p\succeq p^{\prime\prime}$
\end_inset
).
If your preferences are also continuous in an appropriate way, I will be
able to represent with a utility function
\begin_inset Formula $u$
\end_inset
in the sense that
\begin_inset Formula $p\succeq p^{\prime}$
\end_inset
if and only if
\begin_inset Formula $u\left(p\right)\geq u\left(p^{\prime}\right)$
\end_inset
.
\end_layout
\begin_layout Standard
One of the more important discoveries in economics is that if your preferences
satisfy a third condition, referred to as the
\emph on
independence axiom
\emph default
\begin_inset Foot
status collapsed
\begin_layout Plain Layout
This is a bit of a misnomer.
It isn't really an axiom, since it is far from self evident.
\begin_inset space \space{}
\end_inset
It is more like an assumption.
\end_layout
\end_inset
, then this utility function will, in fact, be linear in probabilities.
The independence axiom say this: suppose that for any three lotteries
\begin_inset Formula $p$
\end_inset
,
\begin_inset Formula $p^{\prime}$
\end_inset
, and
\begin_inset Formula $p^{\prime\prime}$
\end_inset
,
\begin_inset Formula $p\succeq p^{\prime}$
\end_inset
implies that for any
\begin_inset Formula $\lambda\in\left[0,1\right]$
\end_inset
\begin_inset Formula
\[
\lambda p+\left(1-\lambda\right)p^{\prime\prime}\succeq
\]
\end_inset
\begin_inset Formula
\[
\lambda p^{\prime}+\left(1-\lambda\right)p^{\prime\prime}.
\]
\end_inset
These last two objects are compound lotteries in which you are given lottery
\begin_inset Formula $\left(\mathcal{X},p\right)$
\end_inset
(or
\begin_inset Formula $\left(\mathcal{X},p^{\prime}\right)$
\end_inset
) with probability
\begin_inset Formula $\lambda$
\end_inset
and lottery
\begin_inset Formula $\left(\mathcal{X},p^{\prime\prime}\right)$
\end_inset
with probability
\begin_inset Formula $\left(1-\lambda\right)$
\end_inset
.
If you can rank two lotteries, you will rank them the same way if they
are mixed with a common third lottery.
\end_layout
\begin_layout Standard
For the utility function to be linear in probabilities, it means that we
will be able to find numbers
\begin_inset Formula $u_{i}$
\end_inset
, one for each of the
\begin_inset Formula $n$
\end_inset
outcomes in the set
\begin_inset Formula $\mathcal{X}$
\end_inset
such that
\begin_inset Formula
\[
u\left(\mathcal{X},p\right)=\sum_{i=1}^{n}u_{i}p_{i}
\]
\end_inset
Since there is one number
\begin_inset Formula $u_{i}$
\end_inset
for each of the
\begin_inset Formula $n$
\end_inset
outcomes in
\begin_inset Formula $\mathcal{X}$
\end_inset
, it is convenient to think of
\begin_inset Formula $u_{i}$
\end_inset
as the utility value associated with outcome
\begin_inset Formula $x_{i}$
\end_inset
.
If the outcomes happened to be expressed in dollars, then we could think
of the utility numbers as representing some underlying utility for wealth.
The point to be emphasized is that the existence of the utility for wealth
function is not an assumption, but an implication of a certain method of
ranking lotteries.
\end_layout
\begin_layout Standard
This is such an important idea that it is worthwhile to see how it works.
To see it, suppose that there is a
\begin_inset Formula $b\in\mathcal{L}$
\end_inset
(the best lottery) such that
\begin_inset Formula $b\succeq p$
\end_inset
for all
\begin_inset Formula $p\in\mathcal{L}$
\end_inset
; a
\begin_inset Formula $w\in\mathcal{L}$
\end_inset
(the worst lottery) such that
\begin_inset Formula $p\succeq w$
\end_inset
for all
\begin_inset Formula $p\in\mathcal{L}$
\end_inset
.
We can now try to mimic the proof of the existence of a utility function
that we did previously.
Recall that we used monotonicity of preferences and continuity.
So we will say preferences are monotonic if
\begin_inset Formula $\lambda b+\left(1-\lambda\right)w\succ\lambda^{\prime}b+\left(1-\lambda^{\prime}\right)w$
\end_inset
if and only if
\begin_inset Formula $\lambda>\lambda^{\prime}$
\end_inset
.
\begin_inset Foot
status collapsed
\begin_layout Plain Layout
The statement
\begin_inset Formula $p\succ p^{\prime}$
\end_inset
means that
\begin_inset Formula $p\succeq p^{\prime}$
\end_inset
but not
\begin_inset Formula $p^{\prime}\succeq p$
\end_inset
.
The notation
\begin_inset Formula $p\sim p^{\prime}$
\end_inset
means that
\begin_inset Formula $p\succeq p^{\prime}$
\end_inset
and
\begin_inset Formula $p^{\prime}\succeq p$
\end_inset
.
\end_layout
\end_inset
We will say that preferences are continuous if the sets
\begin_inset Formula $\left\{ \lambda\in\left[0,1\right]:\lambda b+\left(1-\lambda w\right)\succeq p\right\} $
\end_inset
and
\begin_inset Formula $\left\{ \lambda\in\left[0,1\right]:p\succeq\lambda b+\left(1-\lambda\right)w\right\} $
\end_inset
are both closed intervals.
\end_layout
\begin_layout Standard
Now we can state the result.
\end_layout
\begin_layout Standard
\begin_inset ERT
status collapsed
\begin_layout Plain Layout
\backslash
begin{theorem}
\end_layout
\end_inset
If preferences are complete, transitive, continuous, monotonic and satisfy
the independence axiom, then there is a utility function
\begin_inset Formula $u$
\end_inset
representing preferences over lotteries in
\begin_inset Formula $\mathcal{L}$
\end_inset
that is linear in probabilities.
\begin_inset ERT
status collapsed
\begin_layout Plain Layout
\backslash
end{theorem}
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset ERT
status collapsed
\begin_layout Plain Layout
\backslash
begin{proof}
\end_layout
\end_inset
The proof is constructive.
Let's start by creating the utility function.
This is a function that assigns a real number to each lottery
\begin_inset Formula $p\in\mathcal{L}$
\end_inset
.
To do so set
\begin_inset Formula
\begin{equation}
u\left(p\right)=\left\{ \lambda\in\left[0,1\right]:\lambda b+\left(1-\lambda\right)w\sim p\right\} \label{define}
\end{equation}
\end_inset
Notice that the
\begin_inset Formula $\lambda$
\end_inset
that satisfies this relation (warning - it is not an equation) always exists
and is unique.
To see why, observe that by completeness
\begin_inset Formula $\lambda^{\prime}b+\left(1-\lambda^{\prime}\right)w\succeq p$
\end_inset
or the reverse for every
\begin_inset Formula $\lambda^{\prime}\in\left[0,1\right]$
\end_inset
.
Then
\begin_inset Formula
\[
\left\{ \lambda\in\left[0,1\right]:\lambda b+\left(1-\lambda\right)w\succeq p\right\} \cup
\]
\end_inset
\begin_inset Formula
\[
\left\{ \lambda\in\left[0,1\right]:p\succeq\lambda b+\left(1-\lambda\right)w\right\}
\]
\end_inset
is all of the interval
\begin_inset Formula $\left[0,1\right]$
\end_inset
.
Since both these sets are closed by the continuity of preferences, they
must have at least one point in common.
Since preferences are monotonic they can't have more than one point in
common (Prove this by contradiction.)
\end_layout
\begin_layout Standard
Next, we should show that the function
\begin_inset Formula $u\left(\cdot\right)$
\end_inset
as constructed above, actually represents preferences
\begin_inset Formula $\succeq$
\end_inset
.
\begin_inset space \space{}
\end_inset
This relies on the monotonicity of preferences and is left as an exercise.
\end_layout
\begin_layout Standard
Finally, we come to the important step - showing that the utility function
defined in (
\begin_inset CommandInset ref
LatexCommand ref
reference "define"
\end_inset
) above is linear in probabilities, i.e., that
\begin_inset Formula
\[
u\left(\lambda p+\left(1-\lambda\right)p^{\prime}\right)=\lambda u\left(p\right)+\left(1-\lambda\right)u\left(p^{\prime}\right)
\]
\end_inset
for all
\begin_inset Formula $\lambda$
\end_inset
,
\begin_inset Formula $p$
\end_inset
, and
\begin_inset Formula $p^{\prime}$
\end_inset
.
I will not write down a series of relations - make sure that you don't
mistake them for equations.
First observe that
\begin_inset Formula
\[
\lambda p+\left(1-\lambda\right)p^{\prime}\sim
\]
\end_inset
\begin_inset Formula
\[
\lambda\left[u\left(p\right)b+\left(1-u\left(p\right)\right)w\right]+\left(1-\lambda\right)p^{\prime}
\]
\end_inset
This follows from the definition of the utility function
\begin_inset Formula $u\left(p\right)$
\end_inset
and the independence axiom (in the sense that we are mixing the lotteries
\begin_inset Formula $p$
\end_inset
and
\begin_inset Formula $u\left(p\right)b+\left(1-u\left(p\right)\right)w$
\end_inset
together with the common third lottery
\begin_inset Formula $p^{\prime}$
\end_inset
).
Do the same thing again to get
\begin_inset Formula
\[
\lambda\left[u\left(p\right)b+\left(1-u\left(p\right)\right)w\right]+\left(1-\lambda\right)p^{\prime}\sim
\]
\end_inset
\begin_inset Formula
\[
\lambda\left[u\left(p\right)b+\left(1-u\left(p\right)\right)w\right]+\left(1-\lambda\right)\left[u\left(p^{\prime}\right)b+\left(1-u\left(p^{\prime}\right)\right)w\right]
\]
\end_inset
This is a fairly complicated compound lottery (first you mix lotteries
\begin_inset Formula $b$
\end_inset
and
\begin_inset Formula $w$
\end_inset
together using
\begin_inset Formula $u\left(p\right)$
\end_inset
and
\begin_inset Formula $u\left(p^{\prime}\right)$
\end_inset
.
Then you mix the result using
\begin_inset Formula $\lambda$
\end_inset
.) The reduced lottery associated with this is
\begin_inset Formula
\[
\left[\lambda u\left(p\right)+\left(1-\lambda\right)u\left(p^{\prime}\right)\right]b+\left[1-\lambda u\left(p\right)-\left(1-\lambda\right)u\left(p^{\prime}\right)\right]w
\]
\end_inset
which, if you recall where we started, is then indifferent to
\begin_inset Formula $\lambda p+\left(1-\lambda\right)p^{\prime}$
\end_inset
.
If you glance back at (
\begin_inset CommandInset ref
LatexCommand ref
reference "define"
\end_inset
), you will see that we have just discovered
\begin_inset Formula
\[
u\left(\lambda p+\left(1-\lambda\right)p^{\prime}\right)=\lambda u\left(p\right)+\left(1-\lambda\right)u\left(p^{\prime}\right)
\]
\end_inset
which is the linearity property we wanted.
\begin_inset ERT
status collapsed
\begin_layout Plain Layout
\backslash
end{proof}
\end_layout
\end_inset
\end_layout
\begin_layout Standard
Modern economic theory is concerned largely with problems where there is
some kind of risk or uncertainty about outcomes.
In finance, this uncertainty arises from the inherent unpredictability
of asset returns.
In mechanism design (the theory of auctions and institutions), uncertainty
arises because of the inability to know the tastes of others.
In game theory, uncertainty arises because of the inability to predict
exactly how another player will behave.
Expected utility is the cornerstone of the modern approach to uncertainty,
so it is probably one of the most useful ideas that you will encounter.
\end_layout
\begin_layout Standard
It has been challenged in a number of ways.
The challenges reflect both the strength of the theory and its weakness.
The strength of the theory lies in the fact that is lays out so precisely
what can and cannot happen if the theory is true.
The 'can happen' part is good, because theories are supposed to explain
things we see.
The 'can't happen' part is also important since it shows what kinds of
behavior would allow us to reject the theory.
\end_layout
\begin_layout Standard
You might wonder why we need a theorem like the one above connecting expected
utility which is a model of the utility function, to the independence axiom,
which you might think of as a restriction on the way people behave.
Why couldn't we just write down a specific utility function then somehow
test that, instead of worrying about behavioral properties like the independenc
e axiom.
\end_layout
\begin_layout Standard
There are two reasons.
The first is, that provided you buy completeness, transitivity and continuity,
the independence axiom and expected utility are equivalent.
So the independence axiom provides all the behavioral restrictions that
come from assuming the utility function is linear in probabilities.
Assumptions about utility functions will typically make it easy to derive
some restrictions on behavior, but not all of them.
Knowing all of the implications makes it far easier to construct an effective
empirical test.
\end_layout
\begin_layout Standard
The second reason is that it is quite possible to impose assumptions on
utility that have no implications at all for behavior.
A trivial example might involve assumptions about the utility value associated
with the best or worst lotteries.
In any case, theorems like the one above lay out very clearly what the
additional implications of linearity are, and how they differ from the
implications of other assumptions.
\end_layout
\begin_layout Section
Empirical tests
\end_layout
\begin_layout Standard
I'll provide an example of a challenge to expected utility that involves
experimental tests (we already described how econometric tests could be
used to test the implications of completeness and transitivity).
To describe the test, let me simplify things a bit and suppose that the
set of consequences consists of exactly three things - i.e.,
\begin_inset Formula $\mathcal{X=}\left\{ x_{1},x_{2},x_{3}\right\} $
\end_inset
.
The set of lotteries,
\begin_inset Formula $\mathcal{L}$
\end_inset
, is then just the set of triples of probabilities
\begin_inset Formula $q=\left\{ q_{1},q_{2},q_{3}\right\} $
\end_inset
.
Since the probabilities have to sum to one, it is possible to depict
\begin_inset Formula $\mathcal{L}$
\end_inset
in a simple two dimensional diagram as in Figure
\begin_inset CommandInset ref
LatexCommand ref
reference "lotteries"
\end_inset
\begin_inset ERT
status collapsed
\begin_layout Plain Layout
%TCIMACRO{
\backslash
FRAME{ftbpFU}{3.243in}{2.5953in}{0pt}{
\backslash
Qcb{The set of lotteries}%
\end_layout
\begin_layout Plain Layout
\end_layout
\end_inset
\begin_inset ERT
status collapsed
\begin_layout Plain Layout
%}{
\backslash
Qlb{lotteries}}{undergrad_uncertainty_fig_lottery.eps}%
\end_layout
\begin_layout Plain Layout
\end_layout
\end_inset
\begin_inset ERT
status collapsed
\begin_layout Plain Layout
%{
\backslash
special{ language "Scientific Word"; type "GRAPHIC";
\end_layout
\begin_layout Plain Layout
\end_layout
\end_inset
\begin_inset ERT
status collapsed
\begin_layout Plain Layout
%maintain-aspect-ratio TRUE; display "USEDEF"; valid_file "F";
\end_layout
\begin_layout Plain Layout
\end_layout
\end_inset
\begin_inset ERT
status collapsed
\begin_layout Plain Layout
%width 3.243in; height 2.5953in; depth 0pt; original-width 3.1981in;
\end_layout
\begin_layout Plain Layout
\end_layout
\end_inset
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%original-height 2.5538in; cropleft "0"; croptop "1"; cropright "1";
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\begin_layout Plain Layout
\end_layout
\end_inset
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%cropbottom "0";
\end_layout
\begin_layout Plain Layout
\end_layout
\end_inset
\begin_inset ERT
status collapsed
\begin_layout Plain Layout
%filename 'undergrad_uncertainty_fig_lottery.eps';file-properties "XNPEU";}}}%
\end_layout
\begin_layout Plain Layout
\end_layout
\end_inset
\begin_inset ERT
status collapsed
\begin_layout Plain Layout
%BeginExpansion
\end_layout
\begin_layout Plain Layout
\end_layout
\end_inset
\begin_inset Float figure
placement ptb
wide false
sideways false
status open
\begin_layout Plain Layout
\align center
\begin_inset Graphics
filename undergrad_uncertainty_fig_lottery.eps
width 3.243in
height 2.5953in
\end_inset
\begin_inset Caption
\begin_layout Plain Layout
The set of lotteries
\end_layout
\end_inset
\end_layout
\begin_layout Plain Layout
\align center
\begin_inset CommandInset label
LatexCommand label
name "lotteries"
\end_inset
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset ERT
status collapsed
\begin_layout Plain Layout
%EndExpansion
\end_layout
\begin_layout Plain Layout
\end_layout
\end_inset
\end_layout
\begin_layout Standard
Every point in the triangle with sides of length
\begin_inset Formula $1$
\end_inset
in the diagram above is a lottery in
\begin_inset Formula $\mathcal{L}$
\end_inset
.
To see this, take a point like
\begin_inset Formula $q$
\end_inset
.
The coordinate on the horizontal axis,
\begin_inset Formula $q_{1}$
\end_inset
represents the probability with which consequence
\begin_inset Formula $x_{1}$
\end_inset
occurs.
The coordinate on the vertical axis
\begin_inset Formula $q_{2}$
\end_inset
represents the probability with which consequence
\begin_inset Formula $x_{2}$
\end_inset
occurs under lottery
\begin_inset Formula $q$
\end_inset
.
The probability of consequence
\begin_inset Formula $x_{3}$
\end_inset
is just the remainder
\begin_inset Formula $1-q_{1}-q_{2}$
\end_inset
.
This is given by the length of the horizontal line segment from
\begin_inset Formula $q$
\end_inset
over to the hypotenuse of the triangle (the dashed line with the arrow
at the end in the picture).
Now the lottery
\begin_inset Formula $q^{\prime}$
\end_inset
can be easily compared to
\begin_inset Formula $q$
\end_inset
.
\begin_inset Formula $q^{\prime}$
\end_inset
lies down, and to the right of
\begin_inset Formula $q$
\end_inset
, and is closer to the hypotenuse of the triangle.
So, it assigns lower probability to
\begin_inset Formula $x_{1}$
\end_inset
, higher probability to
\begin_inset Formula $x_{2}$
\end_inset
, and lower probability to
\begin_inset Formula $x_{3}$
\end_inset
.
\end_layout
\begin_layout Standard
Each point in the triangle represents a different simple lottery in
\begin_inset Formula $\mathcal{L}$
\end_inset
.
Compound lotteries are lotteries over lotteries.
For example, one might form a lottery with two consequences - consequence
1 is the lottery
\begin_inset Formula $q$
\end_inset
while consequence
\begin_inset Formula $2$
\end_inset
is the lottery
\begin_inset Formula $q^{\prime}$
\end_inset
.
Let
\begin_inset Formula $\lambda$
\end_inset
be the probability with which the first lottery
\begin_inset Formula $q$
\end_inset
is played.
Then the reduced lottery associated with this compound lottery is
\begin_inset Formula $\lambda q+\left(1-\lambda\right)q^{\prime}$
\end_inset
.
This lottery is just the simple lottery that lies
\begin_inset Formula $\lambda/\left(1-\lambda\right)$
\end_inset
of the way along the line segment between
\begin_inset Formula $q$
\end_inset
and
\begin_inset Formula $q^{\prime}$
\end_inset
.
This point is illustrated in Figure
\begin_inset CommandInset ref
LatexCommand ref
reference "lotteries"
\end_inset
.
\end_layout
\begin_layout Standard
Preferences over lotteries can then be represented by a family of indifference
curves that look exactly like the ones you are used to using to think about
preferences over commodity bundles.
An indifference curve through the lottery
\begin_inset Formula $q$
\end_inset
is set of lotteries
\begin_inset Formula $\tilde{q}$
\end_inset
that have the same utility value as
\begin_inset Formula $q$
\end_inset
.
Formally, an indifference curve is the set
\begin_inset Formula
\[
\left\{ \tilde{q}\in\mathcal{L}:u\left(\tilde{q}\right)=u\left(q\right)\right\}
\]
\end_inset
If we use the linearity property of preferences given in our theorem above,
then the equation that defines this indifference curve is
\begin_inset Formula
\[
q_{1}u_{1}+q_{2}u_{2}+q_{3}u_{3}=u\left(q\right)
\]
\end_inset
Since
\begin_inset Formula $q_{3}=1-q_{1}-q_{2}$
\end_inset
, this becomes
\begin_inset Formula
\[
q_{1}u_{1}+q_{2}u_{2}+\left(1-q_{1}-q_{2}\right)u_{3}=u\left(q\right)
\]
\end_inset
or
\begin_inset Formula
\[
q_{2}=\frac{u\left(q\right)-qu_{1}-\left(1-q\right)u_{3}}{u_{2}-u_{3}}
\]
\end_inset
This is a linear function of
\begin_inset Formula $q_{1}$
\end_inset
, which means that indifference curves are straight lines.
\end_layout
\begin_layout Standard
To see the argument another way, go back to the independence axiom, which
states that for
\emph on
any
\emph default
three lotteries
\begin_inset Formula $q$
\end_inset
,
\begin_inset Formula $q^{\prime}$
\end_inset
and
\begin_inset Formula $q^{\prime\prime}$
\end_inset
and any
\begin_inset Formula $\lambda$
\end_inset
, the lotteries
\begin_inset Formula $\lambda q+\left(1-\lambda\right)q^{\prime\prime}$
\end_inset
and
\begin_inset Formula $\lambda q^{\prime}+\left(1-\lambda\right)q^{\prime\prime}$
\end_inset
must be ranked the same way as
\begin_inset Formula $q$
\end_inset
and
\begin_inset Formula $q^{\prime}$
\end_inset
.
Since this must be true for
\emph on
any
\emph default
three lotteries, then it must be that we can use
\begin_inset Formula $q^{\prime}$
\end_inset
in place of
\begin_inset Formula $q^{\prime\prime}$
\end_inset
in the argument above to conclude that
\begin_inset Formula $\lambda q+\left(1-\lambda\right)q^{\prime}$
\end_inset
and
\begin_inset Formula $\lambda q^{\prime}+\left(1-\lambda\right)q^{\prime}$
\end_inset
must be ranked the same way as
\begin_inset Formula $q$
\end_inset
and
\begin_inset Formula $q^{\prime}$
\end_inset
.
Then reducing the compound lottery, if
\begin_inset Formula $q\symbol{126}q^{\prime}$
\end_inset
then
\begin_inset Formula $\lambda q+\left(1-\lambda\right)q^{\prime}\symbol{126}q^{\prime}\symbol{126}q$
\end_inset
.
Which means that every lottery on the line segment between
\begin_inset Formula $q$
\end_inset
and
\begin_inset Formula $q^{\prime}$
\end_inset
must be indifferent to
\begin_inset Formula $q$
\end_inset
.
That is just another way of saying that the indifference curves are straight
lines.
\end_layout
\begin_layout Standard
It is a bit hard to deal with indifference.
If you offer me
\begin_inset Formula $q$
\end_inset
and
\begin_inset Formula $q^{\prime}$
\end_inset
in an experiment I will choose one of the them.
I might strictly prefer the one I
\begin_inset space \space{}
\end_inset
pick, or I might be indifferent between them.
It would be hard to figure this out in practice.
Fortunately, the independence axiom provides a much stronger condition
that gets around this.
You can see this condition in Figure
\begin_inset CommandInset ref
LatexCommand ref
reference "parallel"
\end_inset
.
\begin_inset ERT
status collapsed
\begin_layout Plain Layout
%TCIMACRO{
\backslash
FRAME{ftbpFU}{3.243in}{2.5953in}{0pt}{
\backslash
Qcb{Parallel Indifference
\end_layout
\begin_layout Plain Layout
\end_layout
\end_inset
\begin_inset ERT
status collapsed
\begin_layout Plain Layout
%Curves}}{
\backslash
Qlb{parallel}}{undergrad_uncertainty_fig_parallel.eps}%
\end_layout
\begin_layout Plain Layout
\end_layout
\end_inset
\begin_inset ERT
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\end_layout
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\end_layout
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\end_layout
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\end_layout
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\end_layout
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\end_layout
\begin_layout Plain Layout
\end_layout
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%filename 'undergrad_uncertainty_fig_parallel.eps';file-properties "XNPEU";}}}%
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\begin_layout Plain Layout
\end_layout
\end_inset
\begin_inset ERT
status collapsed
\begin_layout Plain Layout
%BeginExpansion
\end_layout
\begin_layout Plain Layout
\end_layout
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\begin_inset Float figure
placement ptb
wide false
sideways false
status open
\begin_layout Plain Layout
\align center
\begin_inset Graphics
filename undergrad_uncertainty_fig_parallel.eps
width 3.243in
height 2.5953in
\end_inset
\begin_inset Caption
\begin_layout Plain Layout
Parallel Indifference Curves
\end_layout
\end_inset
\end_layout
\begin_layout Plain Layout
\align center
\begin_inset CommandInset label
LatexCommand label
name "parallel"
\end_inset
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset ERT
status collapsed
\begin_layout Plain Layout
%EndExpansion
\end_layout
\begin_layout Plain Layout
\end_layout
\end_inset
In the Figure,
\begin_inset Formula $q$
\end_inset
and
\begin_inset Formula $q^{\prime}$
\end_inset
are two lotteries on the same indifference curve.
By the previous argument the indifference curve connecting them is a straight
line.
Choose some other lottery like
\begin_inset Formula $p$
\end_inset
which strictly preferred to
\begin_inset Formula $q$
\end_inset
.
We shall try to determine what the indifference curve looks like through
the lottery
\begin_inset Formula $p$
\end_inset
.
\end_layout
\begin_layout Standard
Draw a line segment from
\begin_inset Formula $q$
\end_inset
through
\begin_inset Formula $p$
\end_inset
to some third lottery
\begin_inset Formula $q^{\prime\prime}$
\end_inset
as in the figure.
Suppose that
\begin_inset Formula $\lambda$
\end_inset
is such that
\begin_inset Formula $p=\lambda q+\left(1-\lambda\right)q^{\prime}$
\end_inset
.
Now since
\begin_inset Formula $q$
\end_inset
\begin_inset Formula $\sim q^{\prime}$
\end_inset
, we have by the independence axiom that
\begin_inset Formula
\[
p=\lambda q+\left(1-\lambda\right)q^{\prime\prime}\sim\lambda q^{\prime}+\left(1-\lambda q^{\prime\prime}\right)
\]
\end_inset
Now the line segment from
\begin_inset Formula $p$
\end_inset
to
\begin_inset Formula $\lambda q^{\prime}+\left(1-\lambda q^{\prime\prime}\right)$
\end_inset
is evidently parallel to the line segment from
\begin_inset Formula $q$
\end_inset
to
\begin_inset Formula $q^{\prime}$
\end_inset
, so the indifference curve through
\begin_inset Formula $p$
\end_inset
must be parallel to the indifference curve through
\begin_inset Formula $q$
\end_inset
.
In other words, all the indifference curves are parallel to one another
when the independence axiom holds.
\end_layout
\begin_layout Standard
Now this is something that can be tested with an experiment.
Simply present an experimental subject a choice between two lotteries and
observe their choice.
Once you see what their choice is, present them another two lotteries whose
probabilities are scaled in such a way that knowing that the indifference
curves are all straight and parallel to one another will allow you to predict
their choice in the second lottery.
\end_layout
\begin_layout Standard
This is the experiment that was suggested by Allais.
The consequences are monetary with
\begin_inset Formula $x_{1}=\$100$
\end_inset
,
\begin_inset Formula $x_{2}=\$50$
\end_inset
and
\begin_inset Formula $x_{3}=0$
\end_inset
.
The first pair of lotteries offered to the experimental subject are
\begin_inset Formula
\[
q=\left\{ 0,1,0\right\}
\]
\end_inset
and
\begin_inset Formula
\[
q^{\prime}=\left\{ .1,.89,.01\right\}
\]
\end_inset
In other words, you can have either a sure $50, or take a chance on getting
$100 with a small chance that you will lose everything.
Most people are inclined to take the sure $50 in this case.
\end_layout
\begin_layout Standard
The second pair of lotteries is
\begin_inset Formula
\[
p=\left\{ 0,.11,.89\right\}
\]
\end_inset
and
\begin_inset Formula
\[
p^{\prime}=\left\{ .1,0,.9\right\}
\]
\end_inset
In this case, you probably won't win anything with either lottery.
Lottery
\begin_inset Formula $p$
\end_inset
offers you a small chance to earn $50.
Lottery
\begin_inset Formula $p^{\prime}$
\end_inset
gives you a slightly smaller chance of earning
\begin_inset Formula $\$100$
\end_inset
, but also increases the probability with which you won't win anything.
Most people are inclined to take the chance in this case and opt for lottery
\begin_inset Formula $p^{\prime}$
\end_inset
- perhaps because they are so unlikely to win anything they feel there
is nothing to lose in going for the $100.
\end_layout
\begin_layout Standard
You should plot each of these four lotteries in a Figure like the one above.
You will see that the line segment joining
\begin_inset Formula $q$
\end_inset
and
\begin_inset Formula $q^{\prime}$
\end_inset
is parallel to the line segment joining
\begin_inset Formula $p$
\end_inset
and
\begin_inset Formula $p^{\prime}$
\end_inset
.
If indifference curves are all straight lines, parallel to one another,
anyone who chooses
\begin_inset Formula $q$
\end_inset
over
\begin_inset Formula $q^{\prime}$
\end_inset
must also choose
\begin_inset Formula $p$
\end_inset
over
\begin_inset Formula $p^{\prime}$
\end_inset
(just draw in the indifference curve that would induce them to choose
\begin_inset Formula $q$
\end_inset
over
\begin_inset Formula $q^{\prime}$
\end_inset
then shift it down to see what they will do with
\begin_inset Formula $p$
\end_inset
and
\begin_inset Formula $p^{\prime})$
\end_inset
.
\end_layout
\end_body
\end_document