Appendix A

Key Equations

Week 1: Introduction to Modeling

Linear Cumulative Response Model

R=k⋅tR = k \cdot t

where RR is the cumulative number of responses, kk is the response rate (responses/min), and tt is elapsed time (min). In words: total responses grow at a constant rate over time.

Week 2: The Matching Law

Herrnstein's Single-Alternative Hyperbola

R=k⋅ReRe+reR = \frac{k \cdot R_e}{R_e + r_e}

where RR is response rate, kk is the asymptotic response rate, ReR_e is the programmed reinforcement rate, and rer_e is the extraneous reinforcement rate. In words: response rate is a hyperbolic function of reinforcement rate, approaching a ceiling kk.

Generalized Matching Equation

log⁡ ⁣(B1B2)=a⋅log⁡ ⁣(R1R2)+log⁡(b)\log\!\left(\frac{B_1}{B_2}\right) = a \cdot \log\!\left(\frac{R_1}{R_2}\right) + \log(b)

where aa is sensitivity to reinforcement ratios and bb is bias. In words: the log ratio of responses matches the log ratio of reinforcers, scaled by sensitivity, with a constant bias.

Week 3: Discounting

Mazur's Hyperbolic Model

V=A1+kDV = \frac{A}{1 + kD}

where VV is subjective value, AA is the undiscounted amount, kk is the discount rate, and DD is delay. In words: value decreases hyperbolically as delay increases.

Exponential Model

V=A⋅e−kDV = A \cdot e^{-kD}

In words: value decreases exponentially with delay. The exponential model predicts consistent preferences over time; the hyperbolic model predicts preference reversals.

Hyperboloid Model

V=A(1+kD)sV = \frac{A}{(1 + kD)^s}

Adds a scaling exponent ss that controls the curvature of discounting.

Week 4: Demand

Hursh-Silberberg Exponential Demand Equation

log⁡Q=log⁡Q0+k⋅(e−αC−1)\log Q = \log Q_0 + k \cdot (e^{-\alpha C} - 1)

where QQ is consumption, Q0Q_0 is demand intensity (consumption at zero price), α\alpha is the essential value parameter (rate of decline), kk is the range of consumption in log units, and CC is price. In words: consumption declines exponentially with price, at a rate determined by how essential the commodity is.

Pmax⁡P_{\max} (approximate)

Pmax⁡≈1α⋅Q0⋅k1.5P_{\max} \approx \frac{1}{\alpha \cdot Q_0 \cdot k^{1.5}}

The price at which response output is maximized.

Week 5: Respondent Conditioning

Rescorla-Wagner Model

ΔV=α⋅β⋅(λ−Vtotal)\Delta V = \alpha \cdot \beta \cdot (\lambda - V_{\text{total}})

where ΔV\Delta V is the change in associative strength, α\alpha is CS salience, β\beta is US processing rate, λ\lambda is asymptotic associative strength, and VtotalV_{\text{total}} is the sum of associative strengths of all CSs present. In words: learning is proportional to prediction error -- the discrepancy between what is expected and what occurs.

Week 6: Model Comparisons

Akaike Information Criterion

AIC=−2ln⁡(L)+2k\text{AIC} = -2 \ln(L) + 2k

where LL is the maximum likelihood and kk is the number of estimated parameters. In words: AIC balances fit against complexity by penalizing each additional parameter.

Bayesian Information Criterion

BIC=−2ln⁡(L)+k⋅ln⁡(n)\text{BIC} = -2 \ln(L) + k \cdot \ln(n)

where nn is the number of observations. BIC imposes a stronger penalty for complexity than AIC when n>7n > 7.

Corrected AIC

AICc=AIC+2k(k+1)n−k−1\text{AIC}_c = \text{AIC} + \frac{2k(k+1)}{n - k - 1}

Use when the ratio n/k<40n/k < 40.

Week 8: Probabilistic Models

Poisson Distribution

P(X=k)=(λt)k⋅e−λtk!P(X = k) = \frac{(\lambda t)^k \cdot e^{-\lambda t}}{k!}

In words: the probability of observing exactly kk events in time tt, given a constant rate λ\lambda.

Bayes' Theorem

P(H∣D)=P(D∣H)⋅P(H)P(D)P(H \mid D) = \frac{P(D \mid H) \cdot P(H)}{P(D)}

In words: the posterior probability of a hypothesis given data equals the likelihood of the data given the hypothesis, times the prior, divided by the marginal probability of the data.

Week 9: Multilevel Models

Random-Intercept, Random-Slope Model

yij=(γ00+u0j)+(γ10+u1j)xij+eijy_{ij} = (\gamma_{00} + u_{0j}) + (\gamma_{10} + u_{1j}) x_{ij} + e_{ij}

where γ\gamma terms are fixed effects, uu terms are random effects (subject-level deviations), and ee is the residual. In words: each subject gets their own intercept and slope, drawn from a group-level distribution.

Week 10: Dynamical Systems

Logistic Growth Model

dxdt=r⋅x⋅(1−xK)\frac{dx}{dt} = r \cdot x \cdot \left(1 - \frac{x}{K}\right)

In words: the rate of change in responding equals growth proportional to the current rate, braked by proximity to the ceiling KK.

Analytical Solution

x(t)=K1+(K−x0x0)e−rtx(t) = \frac{K}{1 + \left(\frac{K - x_0}{x_0}\right) e^{-rt}}

Stability of a One-Variable Equilibrium

dudt=f′(x∗)⋅u,u=x−x∗\frac{du}{dt} = f'(x^*) \cdot u, \qquad u = x - x^*

In words: near an equilibrium, a displacement grows or decays as ef′(x∗)te^{f'(x^*)t}. Negative derivative means stable, positive means unstable, and the magnitude sets how fast the system returns.

Coupled Competition Model (Two Responses)

dxdt=rx⋅x⋅(1−x+a⋅yKx),dydt=ry⋅y⋅(1−y+b⋅xKy)\frac{dx}{dt} = r_x \cdot x \cdot \left(1 - \frac{x + a \cdot y}{K_x}\right), \qquad \frac{dy}{dt} = r_y \cdot y \cdot \left(1 - \frac{y + b \cdot x}{K_y}\right)

where aa and bb are cross-suppression coefficients. The interior equilibrium solves x+ay=Kxx + ay = K_x and bx+y=Kybx + y = K_y, and it is stable when ab<1ab < 1. In words: two responses settle at a shared allocation when each suppresses the other less than it suppresses itself.

Stability from the Trace and Determinant

λ=τ±τ2−4Δ2,τ=J11+J22,Δ=J11J22−J12J21\lambda = \frac{\tau \pm \sqrt{\tau^2 - 4\Delta}}{2}, \qquad \tau = J_{11} + J_{22}, \qquad \Delta = J_{11}J_{22} - J_{12}J_{21}

In words: Δ<0\Delta < 0 is a saddle; Δ>0\Delta > 0 with τ<0\tau < 0 is stable and with τ>0\tau > 0 is unstable; τ2<4Δ\tau^2 < 4\Delta makes the approach a spiral.

Week 11: Computational Models

Q-Learning Update Rule

Q(s,a)←Q(s,a)+α[r+γmax⁡a′Q(s′,a′)−Q(s,a)]Q(s, a) \leftarrow Q(s, a) + \alpha \left[ r + \gamma \max_{a'} Q(s', a') - Q(s, a) \right]

where Q(s,a)Q(s,a) is the expected value of action aa in state ss, α\alpha is the learning rate, rr is reward, and γ\gamma is the discount factor. In words: update the value estimate by a fraction of the prediction error.

Softmax Action Selection

P(ai)=eQ(ai)/τ∑jeQ(aj)/τP(a_i) = \frac{e^{Q(a_i) / \tau}}{\sum_j e^{Q(a_j) / \tau}}

where τ\tau is the temperature. For two actions, P(a1)/P(a2)=e[Q(a1)−Q(a2)]/τP(a_1)/P(a_2) = e^{[Q(a_1) - Q(a_2)]/\tau}. In words: choice probability increases exponentially with an action's value, and τ\tau sets how large a value difference must be to change choice.

Week 12: Machine Learning

Neural Network (Single Layer)

y^=f(W⋅x+b)\hat{y} = f(W \cdot x + b)

where WW is the weight matrix, xx is the input, bb is the bias vector, and ff is an activation function. In words: the output is a nonlinear transformation of a weighted sum of inputs.

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