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The hare or the tortoise? Modeling optimal speed-accuracy tradeoff settings
van Ravenzwaaij, D.
Publication date
2012
Link to publication
Citation for published version (APA):
van Ravenzwaaij, D. (2012). The hare or the tortoise? Modeling optimal speed-accuracy
tradeoff settings.
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Appendix
B
Appendix to Chapter 4: “Do the
Dynamics of Prior Information Depend
on Task Context? An Analysis of
Optimality and an Empirical Test”
B.1
Derivation of Equation 4.6
To understand the derivation of Equation 4.6 from Equation 4.5 (see section 4.3 for details), let us focus on the first part Φh(ξ+vc)T +z
s√T
i
(the treatment of the second part is analogous) and write it as Φ [Aξ + B], with A =
√ T
s and B = vcT +z
s√T . We then have the
following integral I:
I =
Z +∞
−∞
Φ [Aξ + B] φ(ξ; v, η2)dξ,
where φ(ξ; v, η2) stands for the normal density function for ξ with mean v and variance η2. We can continue as follows:
I = Z +∞ −∞ Φ [Aξ + B] φ(ξ; v, η2)dξ = Z +∞ −∞ Z Aξ+B −∞ φ(x; 0, 1)dxφ(ξ; v, η2)dξ = Z +∞ −∞ Z B −∞ φ(x; −Aξ, 1)φ(ξ; v, η2)dxdξ = Z +∞ −∞ Z B/A −∞ φ(x; −ξ, 1/A2)φ(ξ; v, η2)dxdξ = Z B/A −∞ Z +∞ −∞ φ(x; −ξ, 1/A2)φ(ξ; v, η2)dξdx.
The inner integral is a known integral in Bayesian statistics (see Gelman, Carlin, Stern, & Rubin, 2004). The kernel of the product of the two normals contains an exponent with
B. Appendix B
quadratic terms in ξ and x (Gelman et al., 2004). Thus, ξ and x have a bivariate normal distribution and thus marginalizing over ξ results in a normal distribution for x.
Using the double expectation theorem (see Gelman et al., 2004), we can find the marginal mean of x:
E(x) = E[E(x|ξ)]
= E[−ξ]
= −v.
Applying a similar theorem for the marginal variance of x gives (Gelman et al., 2004): Var(x) = E[Var(x|ξ)] + Var[E(x|ξ)]
= E[1/A2] + Var(−ξ) = 1/A2+ η2.
Therefore, we can simplify the inner integral to φ(x; −v, 1/A2+ η2):
I = Z B/A −∞ φ(x; −v, 1/A2+ η2)dx = Z √B/A+v 1/A2 +η2 −∞ φ(x; 0, 1)dx = Φ " B/A + v p1/A2+ η2 # = Φ " v cT +z T + v ps2/T + η2 # = Φ " (vc+ v)T + z spT + T2η2/s2 # ,
which is, after multiplication with β, equal to the first term of Equation 4.6. The second part can be found in a similar way.
