Bayesian decision support for complex systems with many distributed experts

Abstract

Complex decision support systems often consist of component modules which, encoding the judgements of panels of domain experts, describe a particular sub-domain of the overall system. Ideally these modules need to be pasted together to provide a comprehensive picture of the whole process. The challenge of building such an integrated system is that, whilst the overall qualitative features are common knowledge to all, the explicit forecasts and their associated uncertainties are only expressed individually by each panel, resulting from its own analysis. The structure of the integrated system therefore needs to facilitate the coherent piecing together of these separate evaluations. If such a system is not available there is a serious danger that this might drive decision makers to incoherent and so indefensible policy choices. In this paper we develop a graphically based framework which embeds a set of conditions, consisting of the agreement usually made in practice of certain probability and utility models, that, if satisfied in a given context, are sufficient to ensure the composite system is truly coherent. Furthermore, we develop new message passing algorithms entailing the transmission of expected utility scores between the panels, that enable the uncertainties within each module to be fully accounted for in the evaluation of the available alternatives in these composite systems.

Appendices

Appendix 1: Proof of Theorem 1

We develop this proof via backward induction both through the vertices of the DAG and through time. For the purpose of this proof define for $t=T$

$$\begin{aligned} \bar{u}^{T,i}\left( \varvec{y}^T_1,\ldots , \varvec{y}^T_{i-1},\varvec{y}^{T-1}_{i},\ldots ,\varvec{y}^{T-1}_{n},d^T\right) \!=\!\!\int _{\mathscr {Y}_i}\cdots \int _{\mathscr {Y}_n}u^{\mathscr {G}}f_{T,i}\cdots f_{T,n}\text {d}\varvec{y}_i(T)\cdots \text {d}\varvec{y}_n(T), \end{aligned}$$

(21)

and note that $\bar{u}^{T,1}\equiv \bar{u}^T$.

First, without any loss of generality, fix a policy $d^T$. Then start the backward induction from $\varvec{Y}_n(T)$, which, by construction, is a leaf of the time slice DAG at time T. For a leaf, $\varvec{Y}_i(T)$ say, it follows from (4) that $\tilde{u}_{T,i}=u^{\mathscr {G}}_i(\varvec{r}_{A_i})$ and note that consequently $u^{\mathscr {G}}_i$ is a function of $\varvec{Y}_n(T)$ only through $\tilde{u}_{T,n}$. Therefore $\tilde{u}_{T,n}$ can then be simply marginalized as in Eq. (5) to obtain $\bar{u}_{T,n}$. Furthermore

$$\begin{aligned} \bar{u}^{T,n}=\sum _{i\in \{Le(\mathscr {G})\setminus \{n\}\}}u^{\mathscr {G}}_i(\varvec{r}_{A_i})+\bar{u}_{T,n}\left( \varvec{y}^T_{A'_n},\varvec{y}^{T-1}_n,d^T\right) . \end{aligned}$$

(22)

Now consider $\varvec{Y}_{n-1}(T)$. The vertex associated with this random vector in the time slice DAG is either the father of $\varvec{Y}_{n}(T)$ or a leaf of the DAG. In the latter case, since by construction $n-1\in U$, the exact same method followed for $\varvec{Y}_n(T)$ can be applied to $\varvec{Y}_{n-1}(T)$, and thus

$$\begin{aligned} \bar{u}^{T,n-1}=\sum _{i\in \{Le(\mathscr {G})\setminus \{n,n{-1}\}\}}u^{\mathscr {G}}_i(\varvec{r}_{A_i})+\sum _{j=n-1}^n\bar{u}_{T,j}\left( \varvec{y}^T_{A'_j},\varvec{y}^{T-1}_j,d^T\right) . \end{aligned}$$

(23)

If on the other hand $\varvec{Y}_{n-1}(T)$ is the father of $\varvec{Y}_n(T)$, then by construction $\varvec{Y}_{n-1}(T)$ has only one son. Thus from Eq. (4) $\bar{u}_{T,n}\equiv \tilde{u}_{T,n-1}$ and Eq. (22) is a function of $\varvec{Y}_{n-1}(T)$ only through $\bar{u}_{T,n}$. In order to deduce $\bar{u}^{T,n-1}$ only $\tilde{u}_{T,n-1}$ has to be marginalized with respect to $f_{T,n-1}$ and therefore

$$\begin{aligned} \bar{u}^{T,n-1}=\sum _{i\in \{Le(\mathscr {G})\setminus \{n,n{-1}\}\}}u^{\mathscr {G}}_i(\varvec{r}_{A_i})+\bar{u}_{T,n-1}\left( \varvec{y}^T_{A'_{n-1}},\varvec{y}^{T-1}_{n-1},\varvec{y}^{T-1}_n,d^T\right) . \end{aligned}$$

(24)

We can note from Eqs. (23) and (24) that $\bar{u}^{T,n-1}$ consists of the linear combination of two summations: the first over the leaves of the graphs with index j smaller than $n-1$ of utility terms $u^{\mathscr {G}}_j$; the second over the indices j bigger or equal than $n-1$ of the terms $\bar{u}_{T,j}$ such that the father of $\varvec{Y}_j(T)$ has an index smaller than $n-1$ in the time slice DAG. So for example in Eq. (23) the second summation is over both n and $n-1$ since the associated vertices are both leaves of the graphs. On the other hand in Eq. (24) there is no term $\bar{u}_{T,n}$ since its father has index $n-1$. More generally, for $j\in [n]$, $\bar{u}^{T,j}$ can be written as the linear combination of the following two summations:

• The first over the indices i in $Le(\mathscr {G})\cap [j{-1}]$ of $u^{\mathscr {G}}_i$;

• The second over the indices k in $B_j=\{k\ge j: F_k<j\}$ of $\bar{u}_{T,k}$, where $F_k$ is the index of the father of $\varvec{Y}_k^\text {T}$.

Therefore, for a $j\in [n]$, we have that

$$\begin{aligned} \bar{u}^{T,j}=\sum _{i\in \{Le(\mathscr {G})\cap [j-1]\}}u^{\mathscr {G}}_i(\varvec{r}_{A_i})+\sum _{k\in B_j}\bar{u}_{T,k}\left( \varvec{y}^T_{A'_k},\varvec{y}^{T-1}_k,\varvec{y}^{T-1}_{Dn_k},d^T\right) , \end{aligned}$$

(25)

where $Dn_k$ is the set of the indices of the descendants of $\varvec{Y}^\text {T}_k$. In particular for $\varvec{Y}_2(T)$ we can write Eq. (25) as

$$\begin{aligned} \bar{u}^{T,2}=\sum _{k\in S_1}\bar{u}_{T,k}\left( \varvec{y}^T_{1},\varvec{y}^{T-1}_k,\varvec{y}^{T-1}_{S_k},d^T\right) , \end{aligned}$$

(26)

since, by the connectedness of the time slice DAG, $\varvec{Y}_1(T)$ is the father of all the vertices whose father’s index is not $[n]{\setminus }\{1\}$. It then follows that Eq. (26) corresponds to $\tilde{u}_{T,1}$, as defined in Eq. (4), and therefore $\bar{u}^T$ can be written as in Eq. (3). Thus Theorem 1 holds for time T.

Now, since $\varvec{Y}_1(T)$ is the unique root of the time slice DAG, if $i,j\in S_1$, then

$$\begin{aligned} A'_i\cap A'_j =\{1\}. \end{aligned}$$

(27)

Suppose that any vertex $\varvec{Y}_j(T)$, for $j\in S_1$, is either connected by a path to one only leaf of the DAG or is a leaf of the graph itself. Because of the identity in Eq. (27) and because of the algebraic form of Eq. (26), which consists of a linear combination of the terms $\bar{u}_{T,j}$, for $j\in S_1$, we can deduce that Eq. (6) holds for the last time slice. Now, consider the case where one vertex $\varvec{Y}_j(T)$ with index in $S_1$ is connected to more than one leaf. Equation (4) guarantees the existence of a vertex $\varvec{Y}_i(T),\, i>j$, connected to both $\varvec{Y}_j(T)$ and the above mentioned leaves, such that $\tilde{u}_{T,i}$ can be written as a linear combination of terms $\bar{u}_{T,k}$, for which each of these terms is a function of one of the leaves only. It therefore follows that Eq. (6) also holds in this case.

Therefore Eq. (6) guarantees that $\bar{u}^{T,1}$ can be written as a linear combination of terms involving only variables in the same ancestral set. Since also the probability factorisation does not change as formalised in Proposition 1, the exact same recursions we explicated at time T can then be followed at time $T-1$ by substituting $u^{\mathscr {G}}_i$ with $\hat{u}_{T-1,i},\, i\in Le(\mathscr {G})$, in Eqs. (21)–(25) and by changing the time index. This then also holds for any time slice t, $1\le t\le T-1$, since $\bar{u}^{t,1}$ will be again a linear combination of terms $\hat{u}_{t-1,i},\, i\in Le(\mathscr {G})$, and the probability density function factorizes as in Proposition 1.

Appendix 2: Proof of Theorem 2

To prove Theorem 2 we proceed as follows:

• We relate the lines of the pseudo-code of Algorithm 3.1 to the Eqs. (3)–(6) of Theorem 1 and their variations which include optimization steps in Eqs. (7) and (8);

• We then show that each panel and the SB have sufficient information to perform the steps of the algorithm they are responsible for;

• We conclude by showing that the optimization steps, which in the algorithm correspond to lines (8) and (15), are able to identify optimal decisions using only combinations of quantities individual panels are able to calculate.

We start with the first two bullets. Line (1) describes the backward induction step over the time index, t, while line (2) does the same over the index of the vertices of the graph, i. Now note that in lines (5)–(7), Panel $G_i:\tilde{u}_{t,i}$ using Eq. (4). Each panel has enough information to do this, since line (10) guarantees that the scores are communicated to the panels overseeing father vertices and line (14) denotes the fact that the SB transmits $\hat{u}_{t,i}$ to the appropriate panels. The functions $\tilde{u}_{t,i}$ are then sent to the SB, who performs an optimization step in line (8) and communicates the result back to the panel. We address the validity of this step below.

Since the $SB:u^*_{t,i}\longrightarrow G_i$, each panel is able to compute $\bar{u}^*_{t,i}$ (lines 10–11) following Eq. (8). As noted before, if i is not the root of the DAG, $\bar{u}^*_{t,i}$ is sent to the appropriate panel, whilst if $i=1$, as specified by the if statement in line (9), $\longrightarrow SB$. For each time slice with time index $t\ne 1$ lines (13)–(14) compute $\hat{u}_{t,i}$, as in Eq. (6). These are sent to the appropriate panels, which can then continue the backward inductive process from the time slice with a lower time index. If on the other hand $t=1$, then the expected utility is a function of the initial decision space $\mathscr {D}(0)$ only. The SB can then perform a final optimization step over this space and thus conclude the algorithm (line 15).

We now address the optimization steps. The influence on the scores associated with time slices with index bigger than t of a decision space $\mathscr {D}_i(t)$ are included, by construction, only in the terms $\hat{u}_{t,k}$, where k is either the index of a descendant $\varvec{Y}_k(t)$ of $\varvec{Y}_i(t)$ or $k=i$. Further note that the same decision space $\mathscr {D}_i(t)$ can affect the scores of terms including descendants of $\varvec{Y}_i(t)$ at the same time point. Thus the whole contribution of $\mathscr {D}_i(t)$ is summarized within $\tilde{u}_{t,i}$, as it can be seen by recursively using Eqs. (4) and (5).

Now, as specified by Eq. (7), the optimization step over $\mathscr {D}_i(t)$ is performed by maximizing $\tilde{u}_{t,i}$, which carries all the information concerning this decision space. More specifically, no other term is an explicit function of $\mathscr {D}_i(t)$ at this stage of the algorithm, as guaranteed by Eqs. (1). Finally, Structural Assumption 2 guarantees that all the elements that appears as arguments of $\tilde{u}_{t,i}$ are observed and therefore known at the time the decision associated to this decision space needs to be made.