Timing and Decision Making

Abstract

The paper presents a normative model which accounts for timing aspects of decision making under uncertainty. We build a model for a decision making process and then solve it, seeking the best timing to make a decision. It is based on several assumptions: information resources become more accurate over time (due to systematic improvements of a given information system), the value of utility or payoff functions deteriorates over time (moreover, loss of opportunities over time is also analyzed), and decision makers observe information only once, before making a decision. In addition, the assumptions of Bounded Rationality are integrated in the model, that is, decision makers stick to a rigid decision rule. The proposed analytical framework allows examining a decision rule over time. A cost of information gathering and a cost of decision delays are also incorporated in the model.

Appendix:

Theorem 1

• 1. Let Q₁ and Q₂ be two information structures operating on the same set of states of nature S = {S₁,…,S_n}, and producing the same set of signals Y = {Y₁,…,Y_m}.

$$ \forall p,0 \le p \le 1,\begin{array}{*{20}c} {} \\ \end{array} p*Q_{1} + (1 - p)*Q_{2} \ge Q_{2} $$

• 2. Let V ₂ the expected payoff (or expected utility) of Q ₂ under an optimal decision rule. Let V ₁ the expected payoff (or expected utility) of Q ₁ of V₂, by using the same given decision rule, $V_{1} > V_{2}$

$$ trace(\varPi *Q_{1} *D_{Q2} *U) = V_{1} $$

$$ trace(\varPi *Q_{2} *D_{Q2} *U) = V_{2} $$

$$ V_{1} > V_{2} , \begin{array}{*{20}c} {} & {} \\ \end{array} \varDelta V = V_{1} - V_{2} > 0 $$

• 3. Let Qt an information structure, which represent the transformation over time from Q₂ to Q ₁.

$$ Q_{t} = p(t)*Q_{1} + (1 - p(t))*Q_{2} $$

$$ \left\{ \begin{aligned} & p(0) = 0 \\ & p(t1) = 1 \\ & t1 > t2 > t3 \ge 0 \Rightarrow 1 > p(t2) \ge p(t3) \ge 0, \\ & t4 > t1 \Rightarrow p(t4) = 1 \\ \end{aligned} \right. $$

p(t) is a monotonic Continuous function with a second derivative in the range $0\le {\text{t}} \le {\text{t1}}$.

• 4. Let U(t) the utility matrix, which differs over time

$$ U_{\left( t \right)} = u\left( t \right)^{ * } \left( {\begin{array}{*{20}c} {U_{1,1} } & \ldots & {U_{1,n} } \\ . & {} & {} \\ . & {} & {} \\ . & {} & {} \\ {U_{k,1} } & \ldots & {U_{k,n} } \\ \end{array} } \right) = u\left( t \right) * U $$

• 5. Let f(t), a function that calculates a tradeoff between availability and outcomes over time. $f(t) = p(t)*u(t)$

• 6. Let C(t) the cost of collection of information over time. C(t) is a monotonic (non- decreased) derivative function over time.

Hence, a sufficient condition for a local maximum of expected utility over time is:

• i. $\frac{\partial f(t)}{\partial t}*\varDelta V - \frac{\partial C(t)}{\partial t} = - \frac{\partial u(t)}{\partial t}*V_{2}$

• ii. $\frac{{\partial^{2} u(t)}}{{\left( {\partial t} \right)^{2} }}*V_{2} + \frac{{\partial^{2} f(t)}}{{\left( {\partial t} \right)^{2} }}*\varDelta V - \frac{{\partial^{2} C(t)}}{{\left( {\partial t} \right)^{2} }} < 0$

Proof:

• (1) $trace(\varPi *Q_{t} *D_{Q2} *U(t)) =$

• (2) $= trace(\varPi *(p(t)*Q_{1} + (1 - p(t))*Q_{2} )*D_{Q2} *U(t)) =$

• (3) $= p(t)*trace(\varPi *Q_{1} *D_{Q2} *u(t)*U) + (1 - p(t))*trace(\varPi *Q_{2} *D_{Q2} *u(t)*U) =$

• (4) $= p(t)*u(t)*trace(\varPi *Q_{1}*D_{Q2}*U) + (1 - p(t))*u(t)*trace(\varPi *Q_{2}*D_{Q2}*U)$

• (5) It is given that

$$ trace(\varPi *Q_{1} *D_{Q2} *U) = V_{1} $$

$$ trace(\varPi *Q_{2} *D_{Q2} *U) = V_{2} $$

• (6) Hence

$$ p(t)*u(t)*trace(\varPi *Q_{1} *D_{Q2} *U) + (1 - p(t))*u(t)*trace(\varPi *Q_{2} *D_{Q2} *u(t)*U) = $$

$$ = p(t)*u(t)*V_{1} + (1 - p(t))*u(t)*V_{2} $$

• (7) $= p(t)*u(t)*\left( {V_{1} - V_{2} } \right) + u(t)*V_{2} = f(t)*\varDelta V + u(t)*V_{2}$

• (8) A necessary condition for existence of a local extremum is that the first degree derivative is 0:

$$ \frac{\partial }{\partial t}\left[ {\left( {trace(\varPi *Q_{t} *D_{Q2} *U(t))} \right) - C(t)} \right] = 0 $$

Thus: $\frac{\partial u(t)}{\partial t}*V_{2} + \frac{\partial f(t)}{\partial t}*\varDelta V - \frac{\partial C(t)}{\partial t} = 0$

• (9) $\frac{\partial f(t)}{\partial t}*\varDelta V - \frac{\partial C(t)}{\partial t} = - \frac{\partial u(t)}{\partial t}*V_{2}$

• (10) A sufficient condition for a local maximum is that the second degree derivative is negative, hence: $\frac{{\partial^{2} u(t)}}{{\left( {\partial t} \right)^{2} }}*V_{2} + \frac{{\partial^{2} f(t)}}{{\left( {\partial t} \right)^{2} }}*\varDelta V - \frac{{\partial^{2} C(t)}}{{\left( {\partial t} \right)^{2} }} < 0$

Q.E.D