13  Independence

Suppose that \(X=\{a, b, c\}\) and that a decision maker strictly prefers \(a\) to \(b\) (\(a\mathrel{P} b\)). Now, consider the following two pairs of lotteries:

How should a decision maker rank \(L_1\) and \(L_2\)? What about \(L_3\) and \(L_4\)? Take a moment to think about these questions before reading on.

\(L_1\) should be strictly preferred to \(L_2\). Both \(L_1\) and \(L_2\) involve the same probability of obtaining item \(c\). The only difference between \(L_1\) and \(L_2\) is that in \(L_1\), the outcome that will occur with probability 0.6 is \(a\), while in \(L_2\), the outcome that will occur with probability 0.6 is \(b\). It seems irrational for a decision maker to strictly prefer \(a\) to \(b\) but not to strictly prefer \(L_1\) over \(L_2\). Imagine that a decision maker is about to play lottery \(L_2\) and is offered the chance to trade \(L_2\) for \(L_1\). There would be something irrational about a decision maker who strictly prefers \(a\) to \(b\) yet is unwilling to trade \(L_2\) for \(L_1\).

Similarly, \(L_3\) should be strictly preferred to \(L_4\). Both \(L_2\) and \(L_4\) have the same outcomes (either \(a\) or \(b\)). In \(L_3\), there is a greater chance of obtaining outcome \(a\). Since the decision maker strictly prefers \(a\) to \(b\), they should strictly prefer a lottery that offers a higher probability of getting their preferred outcome. Therefore, the decision maker should strictly prefer \(L_3\) to \(L_4\), and it would be irrational for them to prefer \(L_4\) over \(L_3\).

The basic assumptions of transitivity and completeness alone do not provide sufficient information to determine how a rational decision maker would rank \(L_1\) and \(L_2\), or how they would rank \(L_3\) and \(L_4\). To ensure that it is irrational not to strictly prefer \(L_1\) to \(L_2\), or not to strictly prefer \(L_3\) to \(L_4\), we need an additional constraint on the decision maker’s preferences over lotteries.

The independence axiom rules out the type of irrationality discussed above. Before formally introducing the independence axiom, it is useful to consider the following points:

With these points in mind, we can now formally state the Independence Axiom.

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 \(r\in (0, 1]\), \[L\mathrel{P} L'\quad\mbox{if, and only if,}\quad (r\cdot L + (1-r)\cdot L'')\mathrel{P} (r\cdot L' + (1-r)\cdot L'').\] \[L\mathrel{I} L'\quad\mbox{if, and only if,}\quad (r\cdot L + (1-r)\cdot L'')\mathrel{I} (r\cdot L' + (1-r)\cdot L'').\]

It is useful to see how the Independence Axiom can be violated. Recall the example of Carol from Section 12.1, who preferred lotteries that are closer to being more fair. We have seen that her preferences are not expected utility representable. We can also demonstrate that her preferences violate the Independence Axiom. To do this, we must identify three lotteries \(L\), \(L'\), and \(L''\) and a number \(r\) such that \(0 < r < 1\) where:

  1. \(L\mathrel{P} L'\) but
  2. it is not the case that \((r\cdot L + (1-r)\cdot L'')\mathrel{P} (r\cdot L' + (1-r)\cdot L'')\).

The following lotteries illustrate how Carol’s preferences violate the Independence Axiom:

  1. Let \(L = 0.5 \cdot a + 0.5 \cdot b\) and \(L' = 1 \cdot a\). According to Carol, \(L \mathrel{P} L'\). In other words: \[(0.5\cdot a + 0.5\cdot b) \mathrel{P} (1\cdot a)\]

  2. Now, let \(L'' = 1 \cdot b\) and \(r = 0.5\). We will show that it is not the case that \((0.5 \cdot L + 0.5 \cdot L'') \mathrel{P} (0.5 \cdot L' + 0.5 \cdot L'')\). We have that \(0.5 \cdot L + 0.5 \cdot L'' = 0.25 \cdot a + 0.75 \cdot b\): \[\begin{align*} 0.5\cdot L + 0.5\cdot L' &= 0.5\cdot (0.5\cdot a + 0.5 \cdot b) + 0.5\cdot (1\cdot b) \\ &= 0.25\cdot a + (0.25+0.5) \cdot b \\ &= 0.25\cdot a + 0.75 \cdot b \end{align*}\] Furthermore, \(0.5 \cdot L' + 0.5 \cdot L'' = 0.5 \cdot a + 0.5 \cdot b\). Therefore, according to Carol, it is not the case that \((0.5 \cdot L + 0.5 \cdot L'') \mathrel{P} (0.5 \cdot L' + 0.5 \cdot L'')\). In other words:

    it is not the case that \((0.25\cdot a + 0.75\cdot b)\mathrel{P}(0.5\cdot a + 0.5\cdot b)\).

More generally, to demonstrate that a decision maker does not satisfy the Independence Axiom, we must find three lotteries \(L\), \(L'\), and \(L''\) and a number \(r\) such that \(0 < r < 1\) where at least one of the following holds:

  1. \(L\mathrel{P} L'\), but it is not the case that \((r\cdot L + (1-r)\cdot L'')\mathrel{P} (r\cdot L' + (1-r)\cdot L'')\);
  2. \((r\cdot L + (1-r)\cdot L'')\mathrel{P} (r\cdot L' + (1-r)\cdot L'')\), but it is not the case that \(L\mathrel{P} L'\);
  3. \(L\mathrel{I} L'\), but it is not the case that \((r\cdot L + (1-r)\cdot L'')\mathrel{I} (r\cdot L' + (1-r)\cdot L'')\); or
  4. \((r\cdot L + (1-r)\cdot L'')\mathrel{I} (r\cdot L' + (1-r)\cdot L'')\), but it is not the case that \(L\mathrel{I} L'\).

Returning to the example lotteries that introduced this section, a direct application of the Independence Axiom ensures that \(L_1 \mathrel{P} L_2\). In other words, starting with lottery \(L_2\) and replacing an outcome in \(L_2\) (specifically, \(b\)) with a strictly preferred prize (\(a\)) results in a lottery \(L_1\) that is strictly preferred to \(L_2\). More generally, we have the following principles that follow directly from the Independence Axiom:

Better Prizes: For any lottery \(L\), replacing any outcome \(b\) in \(L\) with a better prize (i.e., an outcome \(a\) such that \(a \mathrel{P} b\)) results in a lottery that is strictly preferred to \(L\).

There are two things to note about applying the Better Prizes property:

  1. If \(a\) is strictly preferred to \(b\), then receiving \(a\) for sure is always strictly preferred to any mixture of \(a\) and \(b\): Suppose that \(a\) and \(b\) are two outcomes and \(0<r<1\). Then, \(a\mathrel{P} b\) if, and only if, \((1\cdot a)\mathrel{P} (r\cdot a + (1-r) \cdot b)\).

  2. Better Prizes can be applied to lotteries with any number of outcomes. For instance, suppose that \(p+q+r=1\). Then, \(b\mathrel{P} b'\) iff \((p\cdot a + q\cdot b + r\cdot c)\mathrel{P}(p\cdot a + q\cdot b' + r\cdot c)\).

Turning to a related point, when we compared the lotteries \(L_3\) and \(L_4\), we observed that \(L_3\) is preferred to \(L_4\) because it offers better chances of receiving a preferred outcome. This observation leads to the following principle, which is a direct consequence of the Independence Axiom:

Better Chances: For all outcomes \(a\) and \(b\) and \(p, q\in (0, 1)\), if \(p > q\) and \(a\mathrel{P} b\) then \[(p\cdot a + (1-p)\cdot b)\ \mathrel{P}\ (q\cdot a + (1-q)\cdot b).\]

To understand why this holds, suppose \(p > q\) and that \(a\mathrel{P}b\). Then, \(p-q > 0\) and so \[p\cdot a + (1-p)\cdot b = (p-q)\cdot a + q\cdot a + (1-p) \cdot b.\] Since \(a\mathrel{P}b\), applying the Better Prizes principle we have that: \[((p-q)\cdot a + q\cdot a + (1-p) \cdot b)\mathrel{P}((p-q)\cdot b + q\cdot a + (1-p) \cdot b)\] Finally, note that: \[(p-q)\cdot b + q\cdot a + (1-p) \cdot b = q\cdot a + (1-q)\cdot b.\]

13.1 Exercises

  1. 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 the preference generated from this utility function violates the Independence Axiom.

  2. 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 the preference generated from this utility function violates the Independence Axiom.

  1. 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 the preference generated from this utility function violates the Independence Axiom.

    First, note that we have

    \[\begin{align*} U(0.5\cdot a + 0.5\cdot b) &= EU(0.5\cdot a + 0.5\cdot b, u) + 0.5 \\ &= 0.5\times u(a) + 0.5\times u(b) + 0.5 \\ &= 0.5\times 2 + 0.5\times 1 + 0.5 \\ &= 2 \\ &= u(a) \\ &= U(1\cdot a) \\ \end{align*}\] Thus, we have that: \[(0.5\cdot a + 0.5\cdot b)\mathrel{I} (1\cdot a).\]

    Now, let \(L=0.5\cdot a + 0.5\cdot b\), \(L'= 1\cdot a\) and \(L'' = 1\cdot c\) and consider the lotteries \(0.4\cdot L + 0.6\cdot L''\) and \(0.4\cdot L' + 0.6\cdot L''\):

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

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

    Then, \[(0.4\cdot (1\cdot a) + 0.6 \cdot c)\mathrel{P}(0.4\cdot (0.5\cdot a + 0.5\cdot b) + 0.6\cdot c).\]

    This contradicts the Indpendence Axiom:

    1. \(0.5\cdot a + 0.5\cdot b\mathrel{I} (1\cdot a:1)\), but
    2. it is not the case that \((0.4\cdot L + 0.6\cdot L'')\mathrel{I}(0.4\cdot L' + 0.6\cdot L'')\).

  2. 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, \(U(L) = 2\times EU(L, u)\) if \(L\) is a sure-thing. Show that the preference generated from this utility function violates the Independence Axiom.

    First, note that we have

    \[\begin{align*} U(0.5\cdot a + 0.5\cdot c) &= EU(0.5\cdot a + 0.5\cdot c, u) \\ &= 0.5\times u(a) + 0.5\times u(c) \\ &= 0.5\times 2 + 0.5\times 0 \\ &= 1 \\ &< 2\\ &= 2\times u(b) \\ &= 2\times EU(1\cdot b, u) \\ &= U(1\cdot b) \\ \end{align*}\] Thus, we have that: \[(1\cdot b)\mathrel{P}(0.5\cdot a + 0.5\cdot b).\]

    Now, let \(L= 1\cdot b\) , \(L'=0.5\cdot a + 0.5\cdot c\), and \(L'' = 1\cdot c\) and consider the lotteries \(0.8\cdot L + 0.2\cdot L''\) and \(0.8\cdot L' + 0.2\cdot L''\):

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

    \[\begin{align*} U(0.8\cdot L' + 0.2\cdot L'') &= U(0.8\cdot (0.5\cdot a + 0.5\cdot c) + 0.2\cdot (1\cdot c))\\ &= U(0.4\cdot a + 0.4\cdot c + 0.2\cdot c)\\ &= EU(0.4\cdot a + 0.4\cdot c + 0.2\cdot c)\\ &= 0.4 \times u(a) + 0.4\times u(c) + 0.2\times u(c) \\ &= 0.4\times 2 + 0.4 \times 0 + 0.2\times 0 \\ &= 0.8 \end{align*}\]

    Then, we have that \[(0.8\cdot L + 0.2\cdot L'')\mathrel{I}(0.8\cdot L' + 0.2\cdot L'')\]

    This contradicts the Indpendence Axiom:

    1. \((1\cdot b) \mathrel{P} (0.5\cdot a + 0.5\cdot c)\), but
    2. it is not the case that \((0.8\cdot (1\cdot b) + 0.2\cdot (1\cdot c))\mathrel{P}(0.8\cdot (1\cdot (0.5\cdot a + 0.5\cdot c)) + 0.2\cdot (1\cdot c))\).