*credences*or

*subjective probabilities*of a group of individuals to give their collective credences (here, here, here). In some of those posts, this one in particular, I asked how we should combine credences if we wish to use them to make a group decision. And I gave an argument for aggregating by linear pooling that is based on a mathematical theorem.

So what's linear pooling? Suppose there are $n$ individuals with probabilistic credence functions $P_1, \ldots, P_n$; and suppose $P_G$ is their aggregate. Then $P_G$ is a result of linear pooling if there are weights $0 \leq \alpha_1, \ldots, \alpha_n \leq 1$ with $\sum^n_{i=1} \alpha_i =1$ such that $$P_G(-) = \sum^n_{i=1} \alpha_iP_i(-)$$

Next, a little notation from decision theory. Suppose $a$ is an act, $P$ is a credence function over a set of possible states of the world $S$, and $U$ is a utility function. Then, according to

*expected utility theory*(

*EU theory*), an agent with $P$ and $U$ evaluates $a$ as its expected utility relative to those functions. The expected utility of $a$ relative to $P$ and $U$ is $$\mathrm{EU}_{P, U}(a) = \sum_{s \in S} P(s||a)U(a\ \&\ s)$$where: (i) $P(s||a)$ is the agent's credence in state $s$ given that act $a$ is performed, and (ii) $U(a\ \&\ s)$ is the agent's utility for the outcome of act $a$ at state $s$. We will write $P^a(s)$ for $P(s||a)$ and $U^a(s)$ for $U(a\ \&\ s)$, so that $$\mathrm{EU}_{P, U}(a) = \sum_{s \in S} P^a(s)U^a(s)$$

And now the mathematical result:

**Theorem 1**Suppose $P_1, \ldots, P_n$ and $P_G$ are probabilistic credence functions over the same set of possibilities.

(I) Suppose $P_G$

*is not*a result of linear pooling -- that is, $P_G$ is not a weighted average of $P_1, \ldots, P_n$. Then there is a utility function $U$ and a pair of acts $a$ and $b$ such that:

- $\mathrm{EU}_{P_i, U}(a) < \mathrm{EU}_{P_i, U}(b)$, for $1 \leq i \leq n$.
- $\mathrm{EU}_{P_G, U}(b) < \mathrm{EU}_{P_G, U}(a)$

(II) Suppose $P_G$

*is*a result of linear pooling -- that is, $P_G$ is not a weighted average of $P_1, \ldots, P_n$. Then for any utility function $U$ and pair of acts $a$ and $b$: if

- $\mathrm{EU}_{P_i, U}(a) < \mathrm{EU}_{P_i, U}(b)$, for $1 \leq i \leq n$.

- $\mathrm{EU}_{P_G, U}(a) < \mathrm{EU}_{P_G, U}(b)$

Now, notice: all of this what I've just described takes place in the context of expected utility theory. But many decision theorists believe that expected utility theory is too demanding as a set of norms of rational choice. Most often, they cite the so-called

*Allais paradox*to support their claims. In that, the French economist, Maurice Allais describes a pair of decisions, $A$ vs $B$, and $C$ vs $D$. When presented with these decisions, people often prefer $A$ to $B$, and $D$ to $C$. However, there is no way to assign utilities to the outcomes in these cases so that those preferences may be recommended by expected utility theory. There are a number of responses to this, and the literature is now somewhat crowded with

*non-expected utility theories*. In this note, I'll be interested only in Lara Buchak's

*risk-weighted expected utility theory*(

*REU theory*), which she lays out in her 2013 book,

*Risk and Rationality*.

I have described Buchak's REU theory in a lot of detail here, so I won't try to motivate it in this post. In short, while EU theory demands that an agent evaluate an act as its expected utility, REU says that they should evaluate it as its risk-weighted expected utility. While an agent's expected utility for an act is determined only by her credence and utility functions, her risk-weighted expected utility is determined by her credence, utility,

*and risk*functions. For Buchak, an agent's risk function is a strictly increasing, continuous function $r : [0, 1] \rightarrow [0, 1]$ with $r(0) = 0$ and $r(1) = 1$. It represents that agent's attitudes to risk.

Suppose $a$ is an act, $U$ is a utility function, and $S = \{s_1, \ldots, s_n\}$ is the set of states of the world. And now let $S^* = \{S_1, \ldots, S_k\}$ be the coarse-graining of the states of the world that (i) collects together states on which the outcome of $a$ has the same utility and (ii) orders those states from worst to best relative to $U$. Thus: (i) $s_i, s_j$ are in the same state in $S^*$ iff $U^a(s_i) = U^a(s_j)$; (ii) $U^a(S_1) < U^a(S_2) < \ldots < U^a(S_k)$. Then the risk-weighted expected utility of $a$ relative to $P$, $U$, and $r$ is:

$$\mathrm{REU}_{P, r, u}(a) = U^a(S_1) + \sum^k_{i=2} r(P^a(S_i \vee \ldots \vee S_k))[U^a(S_i) - U^a(S_{i-1})]$$

With that in hand, we can now state the result:

**Theorem 2**Suppose $r_1, \ldots, r_n$ and $r_G$ are risk functions.

(I) Suppose $r_G$

*is not*a result of linear pooling -- that is, $r_G$ is not a weighted average of $r_1, \ldots, r_n$. Then there is a set of states $S$, a credence function $P$ over $S$, a utility function $U$, and a pair of acts $a$ and $b$ such that

- $\mathrm{REU}_{P, r_i, U}(a) < \mathrm{REU}_{P, r_i, U}(b)$, for $1 \leq i \leq n$
- $\mathrm{REU}_{P, r_G, U}(b) < \mathrm{REU}_{P, r_G, U}(a)$

(II) Suppose $r_G$

*is*a result of linear pooling -- that is, $r_G$ is not a weighted average of $r_1, \ldots, r_n$. Then for any set of states $S$, any credence function $P$, any utility function $U$, and pair of acts $a$ and $b$: if

- $\mathrm{EU}_{P, r_i, U}(a) < \mathrm{EU}_{P, r_i, U}(b)$, for $1 \leq i \leq n$.

- $\mathrm{EU}_{P, r_G, U}(a) < \mathrm{EU}_{P, r_G, U}(b)$

*Proof of Theorem 2.*We present a series of lemmas, which we tie together at the end.

First, we present a lemma that allows us to redescribe risk-weighted expected utilities as standard expected utilities calculated with respect to different credence functions.

**Lemma 1**Suppose $r$ is a risk function, $P$ is a credence function over $S_m = \{s_1, \ldots, s_m\}$, and $U$ is a utility function. Now suppose $a$ is an act and $U^a(s_1) < \ldots < U^a(s_m)$. Then define the credence function $P_{r, U, a}$ as follows:

\begin{eqnarray*}

P_{r, U, a}(s_1) & = & r(P(s_1 \vee \ldots \vee s_m)) - r(P(s_2 \vee \ldots \vee s_m)) \\

P_{r, U, a}(s_2) & = & r(P(s_2 \vee \ldots \vee s_m)) - r(P(s_3 \vee \ldots \vee s_m)) \\

\vdots & \vdots & \vdots \\

P_{r, U, a}(s_{m-1}) & = & r(P(s_{m-1} \vee s_m)) - r(P(s_m)) \\

P_{r, U, a}(s_m) & = & r(P(s_m))

\end{eqnarray*}

Then$$\mathrm{REU}_{P, r, U}(a) = \mathrm{EU}_{P_{r, U, a}, U}(a)$$

*Proof of Lemma 1.*Suppose $U^a(s_1) < \ldots < U^a(s_m)$. Then

\begin{eqnarray*}

& & \mathrm{REU}_{P, r, U}(a) \\

& = & U^a(s_1) + \sum^m_{k=2} r(P(s_k \vee \ldots \vee s_m))[U^a(s_k) - U^a(s_{k-1})] \\

& = & r(P(s_m))U^a(s_m) + \sum^{m-1}_{k=1} \left [ r(P(s_k \vee \ldots \vee s_m)) - r(P(s_{k+1} \vee \ldots \vee s_m)) \right ]U^a(s_k) \\

& = & P_{r, u, a}(s_m)U^a(s_m) + \sum^{m-1}_{k=1} P_{r, u, a}(s_k)U^a(s_k) \\

& = & \mathrm{EU}_{P_{r, U, a}, U}(a)

\end{eqnarray*}

as required.

**Definition 1**Let $P^m$ be the uniform distribution over $S_m = \{s_1, \ldots, s_m\}$. That is, for all $1 \leq k \leq m$,$$P^m(s_k) = \frac{1}{m}$$

We now state two lemmas about $P^m$:

**Lemma 2**Suppose $r$ is a risk function, $S_m = \{s_1, \ldots, s_m\}$ is a set of states, $U$ is a utility function, and $a$, $b$ are acts (which possibly order the states differently, so that $U^a(s_i) < U^a(s_j)$ but $U^b(s_i) > U^b(s_j)$ for some $i, j$). Then:$$P^m_{r, U, a}(-) = P^m_{r, U, b}(-)$$

**Lemma 3**If $r_G$ is not a linear pool of $r_1, \ldots, r_n$, then there is $m$ such that $P^m_{r_G, U, a}$ is not a linear pool of $P^m_{r_1, U, a}, \ldots, P^m_{r_n, U, a}$ for any utility function $U$ and act $a$.

*Proof of Lemma 3*. First, we show that, if $U$ is a utility function and $a$ is an act and $P^m_{r_G, u, a}$ is a linear pool of $P^m_{r_1, U, a}, \ldots, P^m_{r_n, U, a}$, then, for all $1 \leq k \leq m$, $r_G(\frac{k}{m}) = \sum_i \alpha_i r_i(\frac{k}{m})$.

To that end, suppose $P^m_{r_G, u, a}$ is a linear pool of $P^m_{r_1, u, a}, \ldots, P^m_{r_n, u, a}$. That is, there are $\alpha_1, \ldots, \alpha_n$ such that

$$P^m_{r_G, u, a}(-) = \sum_i \alpha_i P^m_{r_i, u, a}(-)$$

Then

\begin{eqnarray*}

r_G(\frac{1}{m}) & = & r_G(P^m(s_m)) = P^m_{r_G, u, a}(s_m) = \\

& & \sum_i P^m_{r_i, u, a}(s_m) = \sum_i \alpha_i r_i(P^m(s_m)) = \sum_i \alpha_i r_i(\frac{1}{m})

\end{eqnarray*}

So $r_G(\frac{1}{m}) = \sum_i \alpha_i r_i(\frac{1}{m})$. And

\begin{eqnarray*}

r_G(\frac{2}{m}) - r_G(\frac{1}{m}) & = & r_G(P^m(s_{m-1} \vee s_m)) - r_G(P^m(s_m)) \\

& = & P^m_{r_G, u, a}(s_{m-1}) \\

& = & \sum_i \alpha_i P^m_{r_i, u, a}(s_{m-1}) \\

& = & \sum_i \alpha_i \left [ r_i(P^m(s_{m-1} \vee s_m)) - r_i(P^m(s_m)) \right ] \\

& = & \sum_i \alpha_i r_i(\frac{2}{m})- \sum_i \alpha_i r_i(\frac{1}{m}) \\

& = & \sum_i \alpha_i r_i(\frac{2}{m}) - r_G(\frac{1}{m})

\end{eqnarray*}

Thus, $r_G(\frac{2}{m}) = \sum_i \alpha_i r_i(\frac{2}{m})$. And similarly for $r_G(\frac{3}{m}), \ldots, r_G(\frac{m-1}{m}), r_G(\frac{m}{m})$. So, for all $1 \leq k \leq m$,

$$r_G(\frac{k}{m}) = \sum_i \alpha_i r_i(\frac{k}{m})$$

Next, we note that, since each risk function $r_1, \ldots, r_n$, $r_G$ is continuous, then if $r_G$ is not a linear pool of $r_1, \ldots, r_n$, there must be $m$ such that there are no $\alpha_1, \ldots, \alpha_n$ such that, for $1 \leq k \leq m$, $$r_G(\frac{k}{m}) = \sum_i \alpha_i r_i(\frac{k}{m})

$$And that completes our proof of Lemma 3.

We are now ready to piece our proof together. We'll start by proving Theorem 2(I). So suppose $r_G$ is not a linear pool of $r_1, \ldots, r_n$. Then, by Lemma 3, for any utility function $u$ and act $d$, there is $m$ such that $P^m_{r_G, u, d}$ is not a linear pool of $P^m_{r_1, u, d}, \ldots, P^m_{r_n, u, d}$. Then, by Theorem 1, there are acts $a$ and $b$ such that

- $\mathrm{EU}_{P^m_{r_i, u, d}, u}(a) < \mathrm{EU}_{P^m_{r_i, u, d}}(b)$, for $1 \leq i \leq n$; but
- $\mathrm{EU}_{P^m_{r_G, u, d}, u}(b) < \mathrm{EU}_{P^m_{r_G, u, d}}(a)$

- $\mathrm{EU}_{P^m_{r_i, u, a}, u}(a) < \mathrm{EU}_{P^m_{r_i, u, b}}(b)$, for $1 \leq i \leq n$; but
- $\mathrm{EU}_{P^m_{r_G, u, b}, u}(b) < \mathrm{EU}_{P^m_{r_G, u, a}}(a)$

- $\mathrm{REU}_{P^m, r_i, u}(a) < \mathrm{REU}_{P^m, r_i, u}(b)$, for $1 \leq i \leq n$; but
- $\mathrm{REU}_{P^m, r_G, u}(b) < \mathrm{REU}_{P^m, r_G, u}(a)$

Next, we prove Theorem 2(II). So suppose $r_G$ is not a linear pool of $r_1, \ldots, r_n$. That is, there are $\alpha_1, \ldots, \alpha_n$ such that $r_G(-) = \sum_i \alpha_i r_i(-)$. Then, if $u^a(s_1) < \ldots < u^a(s_n)$, then

\begin{eqnarray*}

& & \mathrm{REU}_{P, r_G, U}(a) \\

& = & U^a(s_1) + \sum^m_{k=2} r_G(P(s_k \vee \ldots \vee s_m))[U^a(s_k) - U^a(s_{k-1})] \\

& = & U^a(s_1) + \sum^m_{k=2} \left (\sum_i \alpha_i r_i(P(s_k \vee \ldots \vee s_m)) \right ) [U^a(s_k) - U^a(s_{k-1})] \\

& = & \sum_i \alpha_i \left ( U^a(s_1) + \sum^m_{k=2} r_i(P(s_k \vee \ldots \vee s_m)) [U^a(s_k) - U^a(s_{k-1})] \right ) \\

& = & \sum_i \alpha_i \mathrm{REU}_{P, r_i, U}(a)

\end{eqnarray*}

And similarly for $b$. So, if

- $\mathrm{REU}_{P, r_i, U}(a) < \mathrm{REU}_{P, r_i, U}(b)$, for $1 \leq i \leq n$

- $\mathrm{REU}_{P, r_G, U}(a) < \mathrm{REU}_{P, r_G, U}(b)$