# 15 Independence

Suppose that \(X=\{a, b, c\}\) and that a decision maker strictly prefers \(a\) to \(b\) (i.e., \(a\mathrel{P} b\)). Now, consider the following two lotteries: \[L_1=[a:0.6, c:0.4]\qquad\mbox{and}\qquad L_2=[b:0.6, c:0.4].\] Notice that both \(L_1\) and \(L_2\) involve the same probability of getting item \(c\). Indeed, the *only* difference between \(L_1\) and \(L_2\) is that in \(L_1\) the outcome that will obtain with probability \(0.6\) is \(a\) while in \(L_2\) the outcome that will obtain with probability \(0.6\) is \(b\). It seems irrational for a decision maker to strictly prefers \(a\) to \(b\), but not strictly prefer lottery \(L_1\) over \(L_2\). Imagine that a decision maker is about to play the lottery \(L_2\) and is offered the chance to trade \(L_2\) for \(L_1\). There seems to be something irrational about a decision maker that strictly prefers \(a\) to \(b\) yet is unwilling to trade \(L_2\) for \(L_1\). The independence axiom rules out this type of irrationality. Indeed, if the decision maker strictly prefers \(a\) to \(b\), then she should strictly prefer a 60% chance to get her preferred outcome, assuming that the other outcome is the same in both lotteries. Before stating the independence axiom formally, we note the following:

It is important that the probabilities for item \(c\) is the same in both lotteries. Indeed, there is nothing irrational about a decision maker that strictly prefers \(a\) to \(b\) yet also strictly prefers the lottery \([b:0.8, c:0.2]\) to the lottery \([a:0.05, c:0.95]\) (this may hold for a decision maker that strictly prefers \(a\) to \(b\), but prefers a good chance of getting outcome \(b\) to a small chance to getting outcome \(a\)).

It is important that the only difference in the outcomes in the lotteries is \(a\) and \(b\). Indeed, there is nothing irrational about a decision maker that strictly prefers \(a\) to \(b\) yet also strictly prefers the lottery \([b:0.6,c:0.4]\) to \([a:0.6, d:0.4]\) (this may hold for a decision maker with the strict preference \(a\mathrel{P} b \mathrel{P} c \mathrel{P} d\), and she prefers a lottery with a chance to get her second- and third- favorite outcomes to a lottery with a chance of getting her first and least- favorite outcome).

We can also conclude something about the decision maker’s preference of \(a\) and \(b\) given her preference about the lotteries \(L_1\) and \(L_2\). For instance, if the decision maker strictly prefers \(L_2\) to \(L_1\), then it would be irrational for the decision maker to not strictly prefer \(b\) to \(a\).

The independence axiom generalizes the above reasoning to any compound lotteries.

- Independence Axiom
- Suppose that \(\mathcal{L}\) is a set of lotteries and \((P, I)\) is a rational preference over \(\mathcal{L}\). For all \(L, L', L''\in \mathcal{L}\) and \(r\in (0, 1]\), \[L\mathrel{P} L'\quad\mbox{if, and only if,}\quad [L:r,\ L'':(1-r)]\mathrel{P} [L':r,\ L'':(1-r)].\] \[L\mathrel{I} L'\quad\mbox{if, and only if,}\quad [L:r,\ L'':(1-r)]\mathrel{I} [L':r,\ L'':(1-r)].\]

To illustrate the Independence Axiom, recall the following preferences over \(\mathcal{L}(\{a, b\})\) discussed in ?sec-vnm-overview for a decision maker that prefers lotteries that are closer to being a fair lottery: \[[a:\frac{1}{2}, b:\frac{1}{2}]\mathrel{P}[a:\frac{1}{4}, b:\frac{3}{4}]\mathrel{I}[a:\frac{3}{4}, b:\frac{1}{4}]\mathrel{P}[a:1, b: 0]\mathrel{I}[a:0, b:1]\] Assuming the Compound Lottery Axiom, the above preference violates the Independence Axiom: Let \(L = [a:\frac{1}{2}, b:\frac{1}{2}]\), \(L' = [a:1, b:0]\), and \(L'' = [a:0, b: 1]\). Then, \(L\mathrel{P} L'\). Now, we have the following:

- \([L: \frac{1}{2}, L'':\frac{1}{2}] \mathrel{I} s(L) = [a: (\frac{1}{4} + 0), b: (\frac{1}{4} + \frac{1}{2})] = [a: \frac{1}{4}, b: \frac{3}{4}]\); and
- \([L': \frac{1}{2}, L'':\frac{1}{2}] \mathrel{I} s(L')= [a:\frac{1}{2}, b:\frac{1}{2}]\).

Then, we have that \(L\mathrel{P} L'\), but, since \([a:\frac{1}{2}, b:\frac{1}{2}]\mathrel{P} [a: \frac{1}{4}, b: \frac{3}{4}]\), we have that \([L': \frac{1}{2}, L'':\frac{1}{2}] \mathrel{P} [L: \frac{1}{2}, L'':\frac{1}{2}]\), contrary to the Independence Axiom.

Generalizing the above example, to show that a decision maker does *not* satisfy the Independence Axiom, there must be three lotteries \(L\), \(L'\), and \(L''\) and a number a number \(r\) such that \(0<r\leq 1\) such that at least one of the following is true:

- \(L\mathrel{P} L'\), but it is not the case that \([L':r,\ L'':(1-r)]\mathrel{P} [L':r,\ L'':(1-r)]\);
- \([L:a,\ L'':(1-a)]\mathrel{P} [L':r,\ L'':(1-r)]\), but it is not the case that \(L_1\mathrel{P} L'\);
- \(L\mathrel{I} L'\), but it is not the case that \([L:r,\ L'':(1-r)]\mathrel{I} [L':r,\ L'':(1-r)]\); or
- \([L:r,\ L'':(1-r)]\mathrel{I} [L':r,\ L'':(1-r)]\), but it is not the case that \(L\mathrel{I} L'\).

## 15.1 Exercises

For all \(a, b, c, d\in X\) and all \(p,q,r\in (0, 1]\), if \(p+q+r=1\) and \(c\mathrel{P} d\), then \([a:p, b:q, c:r]\mathrel{P}[a:p, b:q, d:r]\)

Suppose that \(\mathcal{L}(X)\) is a set of lotteries on a set \(X\) and that \((P, I)\) is a rational preference on \(\mathcal{L}(X)\). Using the Independence Axiom, explain why the following is true: For all \(a, b\in X\) and all \(r\in (0, 1)\), if \(a\mathrel{P} b\), then \([a:1]\mathrel{P}[a:r, b:1-r]\).

Suppose that \(\mathcal{L}(X)\) is a set of lotteries on a set \(X\) and that \((P, I)\) is a rational preference on \(\mathcal{L}(X)\). Using the Independence Axiom, explain why the following is true: For all \(a, b\in X\) and all \(p, q\in (0, 1)\) if \(p > q\) and \(a\mathrel{P} b\), then \([a:p, b:1-p]\mathrel{P} [a:q, b:(1-q)]\).

Suppose that \(u:\{a, b, c\}\rightarrow\mathbb{R}\) is a utility function with \(u(a)=2\), \(u(b)=1\) and \(u(c)=0\). Let \(U\) be a utility function on \(\mathcal{L}(X)\) where for all \(L\in\mathcal{L}(X)\), \(U(L) = EU(L, u) + 0.5\) if \(L\) is not a sure-thing, \(U(L) = EU(L, u)\) if \(L\) is a sure-thing.

- Show that \(U\) is not a linear utility function on \(\mathcal{L}(X)\).
- Show that the preference generated from this utility function violates the Independence Axiom.

Suppose that \(u:\{a, b, c\}\rightarrow\mathbb{R}\) is a utility function with \(u(a)=2\), \(u(b)=1\) and \(u(c)=0\). Let \(U\) be a utility function on \(\mathcal{L}(X)\) where for all \(L\in\mathcal{L}(X)\), \(U(L) = EU(L, u)\) if \(L\) is not a sure-thing, \(U(L) = 2*EU(L, u)\) if \(L\) is a sure-thing.

- Show that \(U\) is not a linear utility function on \(\mathcal{L}(X)\).
- Show that the preference generated from this utility function violates the Independence Axiom.

For all \(a, b, c, d\in X\) and all \(p,q,r\in (0, 1]\), if \(p+q+r=1\) and \(c\mathrel{P} d\), then \([a:p, b:q, c:r]\mathrel{P}[a:p, b:q, d:r]\)

First, note that since \(p+q+r=1\), we have that \(p+q=1-r\). Thus, \[\frac{p}{1-r} + \frac{q}{1-r} = \frac{p+q}{1-r} = \frac{1-r}{1-r} = 1.\]

Thus, \([a:\frac{p}{1-r}, b:\frac{q}{1-r}]\) is a lottery.

Then, since \(c\mathrel{P}d\) (i.e., \([c:1]\mathrel{P}[d:1]\)), by the Independence Axiom:

\[[c:r, [a:\frac{p}{1-r}, b:\frac{q}{1-r}]:1-r]\mathrel{P}[d:r, [a:\frac{p}{1-r}, b:\frac{q}{1-r}]:1-r].\]

Note the following:

- \(s([c:r, [a:\frac{p}{1-r}, b:\frac{q}{1-r}]:1-r]) = [a:p, b:q, c:r]\), and
- \(s([d:r, [a:\frac{p}{1-r}, b:\frac{q}{1-r}]:1-r]) = [a:p, b:q, d:r].\)

So, by the Compound Lottery Axiom: \[[a:p, b:q, c:r] \mathrel{P} [a:p, b:q, d:r].\]

Suppose that \(\mathcal{L}(X)\) is a set of lotteries on a set \(X\) and that \((P, I)\) is a rational preference on \(\mathcal{L}(X)\). Using the Independence Axiom, explain why the following is true: For all \(a, b\in X\) and all \(r\in (0, 1)\), if \(a\mathrel{P} b\), then \([a:1]\mathrel{P}[a:r, b:1-r]\).

Since \(1=r+1-r\), By the Compound Lottery Axiom, we have that \[[a:1]\mathrel{I} [a:r, a:1-r].\]

Then, since \([a:1]\mathrel{P}[b:1]\), by the Independence Axiom, we have that:

\[[[a:1]:1-r, [a:1]:r]\mathrel{P}[[b:1]:1-r, [a:1]:r]\].

Since we have the following:

- \(s([[a:1]:1-r, [a:1]:r])= [a:r, a:1-r]\), and
- \(s([[b:1]:1-r, [a:1]:r]) = [a:r, b:1-r]\).

By the Compound Lottery Axiom, we have that \[[a:1]\mathrel{P}[a:r, b:1-r].\]

Suppose that \(\mathcal{L}(X)\) is a set of lotteries on a set \(X\) and that \((P, I)\) is a rational preference on \(\mathcal{L}(X)\). Using the Independence Axiom, explain why the following is true: For all \(a, b\in X\) and all \(p, q\in (0, 1)\) if \(p > q\) and \(a\mathrel{P} b\), then \([a:p, b:1-p]\mathrel{P} [a:q, b:(1-q)]\).

Suppose that \(p>q\). Then there is some \(x>0\) such that \(p=q+x\).

The first thing to note is that: \[\frac{q}{1-x} + \frac{1-p}{1-x} = \frac{q+(1-(q+x))}{1-x}=\frac{1-x}{1-x}=1.\] So, \([a:\frac{q}{1-x}, b:\frac{1-p}{1-x}]\) is a lottery.

Since \(a\mathrel{P} b\) (i.e., \([a:1]\mathrel{P}[b:1]\)), by the Independence Axiom we have that: \[[[a:1]:x, [a:\frac{q}{1-x}, b:\frac{1-p}{1-x}]:(1-x)]\mathrel{P}[[b:1]:x, [a:\frac{q}{1-x}, b:\frac{1-p}{1-x}]:(1-x)].\]

Then, we have that:

\(s([[a:1]:x, [a:\frac{q}{1-x}, b:\frac{1-p}{1-x}])= [a:(x + \frac{q}{1-x} * (1-x)), b: \frac{1-p}{1-x}*(1-x)]\) \(= [a:(x+q), b:1-p] = [a:p, b:1-p]\)

\(s([[b:1]:x, [a:\frac{q}{1-x}, b:\frac{1-p}{1-x}])= [a:\frac{q}{1-x} * (1-x), b: (x + \frac{1-p}{1-x}*(1-x))]\) \(= [a:q, b:((1-p) + x)] = [a:q, b:((1-q-x) + x)] = [a:q, b:1-q]\)

So, by the Compound Lottery axiom: \[[a:p, b:1-p] \mathrel{P} [a:q, b:1-q].\]

Suppose that \(u:\{a, b, c\}\rightarrow\mathbb{R}\) is a utility function with \(u(a)=2\), \(u(b)=1\) and \(u(c)=0\). Let \(U\) be a utility function on \(\mathcal{L}(X)\) where for all \(L\in\mathcal{L}(X)\), \(U(L) = EU(L, u) + 0.5\) if \(L\) is not a sure-thing, \(U(L) = EU(L, u)\) if \(L\) is a sure-thing.

Show that \(U\) is not a linear utility function on \(\mathcal{L}(X)\).

\[\begin{align*} U([a:0.5, b:0.5]) &= EU([a:0.5, b:0.5], u) + 0.5\\ &= 0.5 * u(a) + 0.5 * u(b) + 0.5 \\ &= 0.5*2 + 0.5 * 1 + 0.5 \\ &= 2 \end{align*}\]

\[\begin{align*} 0.5*U([a:1]) + 0.5* U([b:1]) &= 0.5* EU([a:1], u) + 0.5* EU([b:1], u)\\ &= 0.5 * u(a) + 0.5 * u(b) \\ &= 0.5*2 + 0.5 * 1 \\ &= 1.5 \end{align*}\]

Thus, \(U([a:0.5, b:0.5])\neq 0.5*U([a:1]) + 0.5* U([b:1])\), and so \(U\) is not a linear utility function.

Show that the preference generated from this utility function violates the Independence Axiom.

Since \(U([a:0.5, b:0.5]) = 2 = U([a:1])\), we have that \[[a:0.5, b:0.5]\mathrel{I} [a:1].\]

Now, consider the compound lotteries \([[a:0.5, b:0.5]:0.4, c:0.6]\) and \([[a:1]:0.4, c:0.6]\):

\[\begin{align*} U([[a:0.5, b:0.5]:0.4, c:0.6]) &= EU([[a:0.5, b:0.5]:0.4, c:0.6], u) + 0.5\\ &= 0.2 * u(a) + 0.2 * u(b) + 0.6*u(c) + 0.5 \\ &= 0.2*2 + 0.2 * 1 + 0.6 * 0 + 0.5 \\ &= 1.1 \end{align*}\]

\[\begin{align*} U([[a:1]:0.4, c:0.6]) &= EU([[a:1]:0.4, c:0.6], u) + 0.5\\ &= 0.4 * u(a) + 0.6*u(c) + 0.5 \\ &= 0.4*2 + 0.6 * 0 + 0.5 \\ &= 1.3 \end{align*}\]

Then, \[[[a:1]:0.4, c:0.6]\mathrel{P}[[a:0.5, b:0.5]:0.4, c:0.6].\]

This contradicts the Indpendence Axiom:

- \([a:0.5, b:0.5]\mathrel{I} [a:1]\), but
- it is not the case that \([[a:0.5, b:0.5]:0.4, c:0.6]\mathrel{I}[[a:0.5, b:0.5]:0.4, c:0.6]\).

Suppose that \(u:\{a, b, c\}\rightarrow\mathbb{R}\) is a utility function with \(u(a)=2\), \(u(b)=1\) and \(u(c)=0\). Let \(U\) be a utility function on \(\mathcal{L}(X)\) where for all \(L\in\mathcal{L}(X)\), \(U(L) = EU(L, u)\) if \(L\) is not a sure-thing, \(U(L) = 2*EU(L, u)\) if \(L\) is a sure-thing.

Show that \(U\) is not a linear utility function on \(\mathcal{L}(X)\).

\[\begin{align*} U([a:0.5, b:0.5]) &= EU([a:0.5, b:0.5], u)\\ &= 0.5 * u(a) + 0.5 * u(b)\\ &= 0.5*2 + 0.5 * 1 \\ &= 1.5 \end{align*}\]

\[\begin{align*} 0.5*U([a:1]) + 0.5* U([b:1]) &= 0.5* 2* EU([a:1], u) + 0.5* 2* EU([b:1], u)\\ &= 0.5 * 2 * u(a) + 0.5 * 2 * u(b) \\ &= 0.5 * 2 * 2 + 0.5 * 2 * 1 \\ &= 3 \end{align*}\]

Thus, \(U([a:0.5, b:0.5])\neq 0.5*U([a:1]) + 0.5* U([b:1])\), and so \(U\) is not a linear utility function.

Show that the preference generated from this utility function violates the Independence Axiom.

Since \(U([a:0.5, c:0.5]) = 1 < U([b:1]) = 2\), we have that \[[b:1] \mathrel{P} [a:0.5, c:0.5].\]

Now, consider the compound lotteries \([[b:1]:0.8, c:0.2]\) and \([[a:0.5, c:0.5]:0.8, c:0.2]\):

\[\begin{align*} U([[b:1]:0.8, c:0.2]) &= EU([[b:1]:0.8, c:0.2], u)\\ &= 0.8 * u(b) + 0.2 * u(c) \\ &= 0.8 * 1 + 0.2 * 0 \\ &= 0.8 \end{align*}\]

\[\begin{align*} U([[a:0.5, c:0.5]:0.8, c:0.2]) &= EU([[a:0.5, c:0.5]:0.8, c:0.2], u) + 0.5\\ &= 0.4 * u(a) + 0.6*u(c) \\ &= 0.4*2 + 0.6 * 0 \\ &= 0.8 \end{align*}\]

Then, \[[[b:1]:0.8, c:0.2]\mathrel{I}[[a:0.5, c:0.5]:0.8, c:0.2].\]

This contradicts the Indpendence Axiom:

- \([b:1] \mathrel{P} [a:0.5, c:0.5]\), but
- it is not the case that \([[b:1]:0.8, c:0.2]\mathrel{P}[[a:0.5, c:0.5]:0.8, c:0.2]\).