What to do with a forecast?

Abstract

In the literature one finds two non-equivalent responses to forecasts; deference and updating. Herein it is demonstrated that, under certain conditions, both responses are entirely determined by one’s beliefs as regards the calibration of the forecaster. Further it is argued that the choice as to whether to defer to, or update on, a forecast is determined by the aim of the recipient of that forecast. If the aim of the recipient is to match their credence with the prevailing objective chances, they should defer to the forecast; if it is to maximize the veritistic value of their beliefs, they should update on the forecast.

Appendix 1

1.1 (a) Derivation of the deference principles

• Sharp Empirical Deference Principle: $C_{\alpha }^{t+1} (A)_{\langle F_{\sigma }^{A,x}\rangle }=\left\langle {\mathcal {F}}_{\sigma }^{{\mathbb {T}}}\right\rangle _{\alpha }^t (x)$.

The principles required to derive this deference principle are:

Principle of conditionalization: $C^{t+1}(A)_{E}=C^{t}(A|E)$

Miller’s Principle: $C(A|ch(A)=x)=x$, where $ch$ is the objective chance function and $C$ is initial or $ch(A)=x$ screens off all other evidence for $A$ in $C$,

Conditional continuous expansion theorem:

$$\begin{aligned} P_{1}(A)=\int \limits _{0}^1 P_{1}(A|P_{2}(A|B)=x, B)P_{1}(P_{2}(A|B)=x|B)dx, \end{aligned}$$

Conditionalizing out: $P_{1}(P_{2}(A|B)=x|B)=P_{1}(P_{2}(A|B)=x)$.

Direct Inference $^{**}$: If $A$ is a value of a variable in ${\mathbb {T}}$, then

$$\begin{aligned} \left\langle ch\left( A\Big |F_{\sigma }^{A,x}\right) \right\rangle _{\alpha }^t=\left\langle {\mathcal {F}}_{\sigma }^{{\mathbb {T}}}\right\rangle _{\alpha }^t (x). \end{aligned}$$

One also requires two theorems of probability theory:

• Conditional continuous expansion theorem:

  $$\begin{aligned} P_{1}(A)=\int \limits _{0}^1 P_{1}(A|P_{2}(A|B)=x, B)P_{1}(P_{2}(A|B)=x|B)dx. \end{aligned}$$

• Skyrms (1988) theorem : $P_{1}(A|B,P_{2}(A|B)=x)=x$, if $P_{1}(A|P_{2}(A)=x)=x$.

By the principle of conditionalization:

$$\begin{aligned} C_{\alpha }^{t+1} (A)_{F_{\sigma }^{A,x}}=C_{\alpha }^t \left( A\Big |F_{\sigma }^{A,x}\right) . \end{aligned}$$

(24)

By the conditional continuous expansion theorem:

$$\begin{aligned} C_{\alpha }^{t+1} (A)_{F_{\sigma }^{A,x}}\!=\!\int \limits _{0}^1C_{\alpha }^t\left( A\Big |ch\left( A\Big |F_{\sigma }^{A,x}\right) \!=\!y,F_{\sigma }^{A,x}\right) C_{\alpha }^t\left( ch\left( A\Big |F_{\sigma }^{A,x}\right) \!=\!y\Big |F_{\sigma }^{A,x}\right) dy.\qquad \end{aligned}$$

By conditionalizing out:

$$\begin{aligned} C_{\alpha }^{t+1} (A)_{F_{\sigma }^{A,x}}=\int \limits _{0}^1C_{\alpha }^t\left( A\Big |ch\left( A\Big |F_{\sigma }^{A,x}\right) =y,F_{\sigma }^{A,x}\right) C_{\alpha }^t\left( ch\left( A\Big |F_{\sigma }^{A,x}\right) =y\right) dy. \end{aligned}$$

By the Miller’s Principle with Skyrm’s theorem applied:

$$\begin{aligned} C_{\alpha }^{t+1} (A)_{\left\langle F_{\sigma }^{A,x}\right\rangle }=\int \limits _{0}^1 y C_{\alpha }^t\left( ch\left( A\Big |F_{\sigma }^{A,x}\right) =y\right) dy=\left\langle ch\left( A\Big |F_{\sigma }^{A,x}\right) \right\rangle _{\alpha }^t. \end{aligned}$$

As $A$ is a value of a variable in $T$, Direct Inference$^{**}$ applies, giving the Sharp Empirical Deference Principle:

$$\begin{aligned} C_{\alpha }^{t+1} (A)_{\left\langle F_{\sigma }^{A,x}\right\rangle }=\left\langle {\mathcal {F}}_{\sigma }^{{\mathbb {T}}}\right\rangle _{\alpha }^t (x). \end{aligned}$$

The derivation of

• Interval Empirical Deference Principle: $C_{\alpha }^{t+1} (A)_{\langle F_{\sigma }^{A,\varDelta }\rangle }=\widehat{\left\langle {\mathcal {F}}_{\sigma }^{{\mathbb {T}}}\right\rangle _{\alpha }^t}^{\varDelta }$

follows exactly the same route except $\varDelta$ is substituted for $x$ and Direct Inference$^{\ddagger \ddagger }$ is used in the final step.

1.2 (b) Derivation of the empirical update function

The principles and theorems required for this derivation are the same as those for the empirical deference principle save that one also requires Bayes’ theorem and one requires the following pair of principles rather than the Direct Inference$^{**}$ principle:

• Direct Inference $^{\ddagger \ddagger }$: If $A$ is a value of a variable in ${\mathbb {T}}$, then

  $$\begin{aligned} \left\langle ch\left( A\Big |F_{\sigma }^{A,\varDelta }\right) \right\rangle _{\alpha }^t= \widehat{\left\langle {\mathcal {F}}_{\sigma }^{{\mathbb {T}}}\right\rangle _{\alpha }^t}^{\varDelta }=\frac{1}{b-a}\int \limits _{a}^b\left\langle {\mathcal {F}}_{\sigma }^{{\mathbb {T}}}\right\rangle _{\alpha }^t (x)dx. \end{aligned}$$

• Direct Inference $^{\dagger \dagger }$: If $A$ is a value of a variable in ${\mathbb {T}}$, then

  $$\begin{aligned} \left\langle ch\left( F_{\sigma }^{A,\varDelta }\Big |F_{\sigma }^{A}\right) \right\rangle _{\alpha }^t=\left\langle {\mathcal {F}}\left( F_{\sigma }^{{\mathbb {T}},\varDelta }\Big |F_{\sigma }^{{\mathbb {T}}}\right) \right\rangle _{\alpha }^t=\int \limits _{a}^{b}\left\langle \lambda _{\sigma }^{{\mathbb {T}}}\right\rangle _{\alpha }^t (x)dx. \end{aligned}$$

This derivation begins with the principle of conditionalization and Bayes theorem.

$$\begin{aligned} C_{\alpha }^{t+1} (A)_{F_{\sigma }^{A,\varDelta }}\!=\!C_{\alpha }^t \left( A\Big |F_{\sigma }^{A,\varDelta }\right) =\frac{C_{\alpha }^t \left( F_{\sigma }^{A,\varDelta }\Big |A\right) C_{\alpha }^t (A)}{C_{\alpha }^t \left( F_{\sigma }^{A,\varDelta }\Big |A\right) C_{\alpha }^t (A)+C_{\alpha }^t \left( F_{\sigma }^{A,\varDelta }\Big |\lnot A\right) C_{\alpha }^t (\lnot A)}. \end{aligned}$$

(25)

Our task is to find useful expressions for the likelihoods: $C_{\alpha }^t (F_{\sigma }^{A,\varDelta }|A)$ and $C_{\alpha }^t (F_{\sigma }^{A,\varDelta }|\lnot A)$. To this end we again invoke the conditional expansion theorem, though this time in its discrete form.

$$\begin{aligned} C_{\alpha }^t \left( F_{\sigma }^{A,\varDelta }\Big |A\right)&= C_{\alpha }^t \left( F_{\sigma }^{A,\varDelta }\Big |F_{\sigma }^{A},A\right) C_{\alpha }^t \left( F_{\sigma }^{A}\Big |A\right) \\&+C_{\alpha }^t \left( F_{\sigma }^{A,\varDelta }\Big |\lnot F_{\sigma }^{A},A\right) C_{\alpha }^t \left( \lnot F_{\sigma }^{A}\Big |A\right) \\ C_{\alpha }^t \left( F_{\sigma }^{A,\varDelta }\Big |\lnot A\right)&= C_{\alpha }^t \left( F_{\sigma }^{A,\varDelta }\Big |F_{\sigma }^{A},\lnot A\right) C_{\alpha }^t \left( F_{\sigma }^{A}\Big |\lnot A\right) \\&+C_{\alpha }^t \left( F_{\sigma }^{A,\varDelta }\Big |\lnot F_{\sigma }^{A},\lnot A\right) C_{\alpha }^t \left( \lnot F_{\sigma }^{A}\Big |\lnot A\right) \end{aligned}$$

As $\lnot F_{\sigma }^{A}\wedge F_{\sigma }^{A,\varDelta }$ is a logical contradiction the above reduce to:

$$\begin{aligned} C_{\alpha }^t \left( F_{\sigma }^{A,\varDelta }\Big |A\right)&= C_{\alpha }^t \left( F_{\sigma }^{A,\varDelta }\Big |F_{\sigma }^{A},A\right) C_{\alpha }^t \left( F_{\sigma }^{A}\Big |A\right) ,\end{aligned}$$

(26)

$$\begin{aligned} C_{\alpha }^t \left( F_{\sigma }^{A,\varDelta }\Big |\lnot A\right)&= C_{\alpha }^t \left( F_{\sigma }^{A,\varDelta }\Big |F_{\sigma }^{A},\lnot A\right) C_{\alpha }^t \left( F_{\sigma }^{A}\Big |\lnot A\right) . \end{aligned}$$

(27)

We now impose the condition that $C_{\alpha }^t (F_{\sigma }^{A}|A)=C_{\alpha }^t (F_{\sigma }^{A}|\lnot A)=C_{\alpha }^t (F_{\sigma }^{A})$. This is eminently plausible considering that the forecaster can forecast with any probability between 0 and 1. Using this identity to substitute back into (26) and (27), and then substituting these back into (25), delivers (after cancelling terms):

$$\begin{aligned} C_{\alpha }^{t+1} (A)_{F_{\sigma }^{A,\varDelta }}\!=\!\frac{C_{\alpha }^t \left( F_{\sigma }^{A,\varDelta }\Big |F_{\sigma }^{A},A\right) C_{\alpha }^t (A)}{C_{\alpha }^t \left( F_{\sigma }^{A,\varDelta }\Big |F_{\sigma }^{A},A\right) C_{\alpha }^t (A)+C_{\alpha }^t \left( F_{\sigma }^{A,\varDelta }\Big |F_{\sigma }^{A},\lnot A\right) C_{\alpha }^t (\lnot A)}.\qquad \end{aligned}$$

(28)

Now we use the conditional continuous expansion theorem to expand these new likelihoods:

$$\begin{aligned}&\displaystyle C_{\alpha }^t \left( F_{\sigma }^{A,\varDelta }\Big |F_{\sigma }^{A},A\right) =\int \limits _{0}^1 C_{\alpha }^t \left( F_{\sigma }^{A,\varDelta }\Big |ch\left( F_{\sigma }^{A,\varDelta }\Big | F_{\sigma }^{A},A\right) =y, F_{\sigma }^{A}, A\right) \\&\displaystyle C_{\alpha }^t \left( ch\left( F_{\sigma }^{A,\varDelta }\Big |F_{\sigma }^{A},A\right) =y\Big |F_{\sigma }^{A}, A\right) dy,\\&\displaystyle C_{\alpha }^t \left( F_{\sigma }^{A,\varDelta }\Big |F_{\sigma }^{A},\lnot A\right) =\int \limits _{0}^1 C_{\alpha }^t \left( F_{\sigma }^{A,\varDelta }\Big |ch\left( F_{\sigma }^{A,\varDelta }\Big | F_{\sigma }^{A},\lnot A\right) =y, F_{\sigma }^{A}, \lnot A\right) \\&\displaystyle C_{\alpha }^t \left( ch\left( F_{\sigma }^{A,\varDelta }\Big |F_{\sigma }^{A},\lnot A\right) =y\Big |F_{\sigma }^{A}, \lnot A\right) dy. \end{aligned}$$

By the Miller’s Principle, Skyrm’s theorem and conditionalizing out, these expressions simplify to:

$$\begin{aligned} C_{\alpha }^t \left( F_{\sigma }^{A,\varDelta }\Big |F_{\sigma }^{A},A\right)&= \int \limits _{0}^1 y C_{\alpha }^t \left( ch\left( F_{\sigma }^{A,\varDelta }\Big |F_{\sigma }^{A},A\right) =y\right) dy\\&= \left\langle ch\left( F_{\sigma }^{A,\varDelta }\Big |F_{\sigma }^{A},A\right) \right\rangle _{\alpha }^t,\\ C_{\alpha }^t \left( F_{\sigma }^{A,\varDelta }\Big |F_{\sigma }^{A},\lnot A\right)&= \int \limits _{0}^1 y C_{\alpha }^t \left( ch\left( F_{\sigma }^{A,\varDelta }\Big |F_{\sigma }^{A},\lnot A\right) =y\right) dy\\&= \left\langle ch\left( F_{\sigma }^{A,\varDelta }\Big |F_{\sigma }^{A},\lnot A\right) \right\rangle _{\alpha }^t. \end{aligned}$$

By Bayes’ Theorem, the continuous expansion theorem, Direct Inference$^{\dagger }$, Direct Inference$^{*}$, Direct Inference$^{\ddagger \ddagger }$ and Direct Inference$^{\dagger \dagger }$ (20) and (21) were derived in the text ($\varDelta =[a,b]\subseteq [0,1]$)

$$\begin{aligned} \left\langle ch\left( F_{\sigma }^{A,\varDelta }\Big |F_{\sigma }^{A}, A\right) \right\rangle _{\alpha }^{t}&= 2\widehat{\left\langle {\mathcal {F}}_{\sigma }^{{\mathbb {T}}}\right\rangle _{\alpha }^{t}}^{\varDelta }\int \limits _{a}^b\left\langle \lambda _{\sigma }^{{\mathbb {T}}}\right\rangle _{\alpha }^{t}(x)dx,\\ \left\langle ch\left( F_{\sigma }^{A,\varDelta }\Big |F_{\sigma }^{A},\lnot A\right) \right\rangle _{\alpha }^{t}&= 2\left( 1-\widehat{\left\langle {\mathcal {F}}_{\sigma }^{{\mathbb {T}}}\right\rangle _{\alpha }^{t}}^{\varDelta }\right) \int \limits _{a}^b\left\langle \lambda _{\sigma }^{{\mathbb {T}}}\right\rangle _{\alpha }^{t}(x)dx. \end{aligned}$$

Making substitutions using (20) and (21) gives:

$$\begin{aligned} C_{\alpha }^t \left( F_{\sigma }^{A,\varDelta }\Big |F_{\sigma }^{A},A\right)&= 2\widehat{\left\langle {\mathcal {F}}_{\sigma }^{{\mathbb {T}}}\right\rangle _{\alpha }^{t}}^{\varDelta }\int \limits _{a}^b\left\langle \lambda _{\sigma }^{{\mathbb {T}}}\right\rangle _{\alpha }^{t}(x)dx,\end{aligned}$$

(29)

$$\begin{aligned} C_{\alpha }^t \left( F_{\sigma }^{A,\varDelta }\Big |F_{\sigma }^{A},\lnot A\right)&= 2\left( 1-\widehat{\left\langle {\mathcal {F}}_{\sigma }^{{\mathbb {T}}}\right\rangle _{\alpha }^{t}}^{\varDelta }\right) \int \limits _{a}^b\left\langle \lambda _{\sigma }^{{\mathbb {T}}}\right\rangle _{\alpha }^{t}(x)dx. \end{aligned}$$

(30)

Substituting these back into (28) and cancelling terms gives the Interval Empirical Update Function:

$$\begin{aligned} C_{\alpha }^{t+1}(A)_{F_{\sigma }^{A,\varDelta }}=\frac{ \widehat{\left\langle {\mathcal {F}}_{\sigma }^{{\mathbb {T}}}\right\rangle _{\alpha }^t}^{\varDelta }C_{\alpha }^{t}(A)}{\widehat{\left\langle {\mathcal {F}}_{\sigma }^{{\mathbb {T}}}\right\rangle _{\alpha }^t}^{\varDelta }C_{\alpha }^{t}(A)+\left( 1-\widehat{\left\langle {\mathcal {F}}_{\sigma }^{{\mathbb {T}}}\right\rangle _{\alpha }^t}^{\varDelta }\right) C_{\alpha }^{t}(\lnot A)}. \end{aligned}$$

Noting that the average value of a function over an interval at the limit where that interval contains a single real is just the value of that function at that real we have the Sharp Empirical Update Function:

$$\begin{aligned} C_{\alpha }^{t+1}(A)_{F_{\sigma }^{A,x}}=\frac{ \left\langle {\mathcal {F}}_{\sigma }^{{\mathbb {T}}}\right\rangle _{\alpha }^t(x)C_{\alpha }^{t}(A)}{\left\langle {\mathcal {F}}_{\sigma }^{{\mathbb {T}}}\right\rangle _{\alpha }^t(x)C_{\alpha }^{t}(A)+\left( 1-\left\langle {\mathcal {F}}_{\sigma }^{{\mathbb {T}}}\right\rangle _{\alpha }^t(x)\right) C_{\alpha }^{t}(\lnot A)}. \end{aligned}$$