Operant Conditioning

Quantum Mechanics Postulates

• Quantum mechanics is built upon postulates that are accepted without proof.
• Postulates are validated by the theory's ability to accurately depict and predict physical system behaviors.
• There are five main postulates in quantum mechanics.

State of a System (Postulate 1)

• Postulate 1: A system's state is comprehensively defined by a wave function $\Psi(\mathbf{r}_1, \mathbf{r}_2,..., \mathbf{r}_N, t)$.
  • The function depends on the coordinates of particles and time.
  • $\Psi^*(\mathbf{r}_1, \mathbf{r}_2,..., \mathbf{r}_N, t)\Psi(\mathbf{r}_1, \mathbf{r}_2,..., \mathbf{r}_N, t)d\tau$ represents the probability at time t of finding particle 1 in the volume element $d\mathbf{r}_1$ at $\mathbf{r}_1$, particle 2 in $d\mathbf{r}_2$ at $\mathbf{r}_2$, and so forth for N particles.
• Wave functions must be single-valued, continuous, and quadratically integrable.
• For a single particle, the wave function $\Psi(\mathbf{r}, t)$ relies on the particle's position $\mathbf{r}$ and time t.
• $\Psi^*(\mathbf{r}, t)\Psi(\mathbf{r}, t)d\mathbf{r} = |\Psi(\mathbf{r}, t)|^2d\mathbf{r}$ indicates the probability of finding the particle in volume $d\mathbf{r}$ at time t.
• $|\Psi(\mathbf{r}, t)|^2$ stands for the probability density.
• The probability of finding a particle in volume V at time t is described by: $P = \int_V |\Psi(\mathbf{r}, t)|^2 d\mathbf{r}$.
• The normalization condition, where the particle is somewhere in space, is: $\int_{-\infty}^{\infty} |\Psi(\mathbf{r}, t)|^2 d\mathbf{r} = 1$.
• A wave function that meets the normalization condition is considered normalized.
• If $\Psi$ is a solution to the Schrödinger equation, then $c\Psi$ is also a solution, where c is an arbitrary constant.
• The constant c is selected to fulfill the normalization condition.

Observables and Linear Hermitian Operators (Postulate 2)

• Postulate 2: Each observable in classical mechanics corresponds to a linear Hermitian operator in quantum mechanics.
• An observable is any measurable physical quantity such as position, momentum, energy, or angular momentum.
• A linear operator $\hat{A}$ satisfies these conditions:
  • $\hat{A}(f(x) + g(x)) = \hat{A}f(x) + \hat{A}g(x)$
  • $\hat{A}(cf(x)) = c\hat{A}f(x)$
  • $f(x)$ and $g(x)$ are arbitrary functions and c is an arbitrary constant.
• An operator $\hat{A}$ is Hermitian if it meets the following condition: $\int_{-\infty}^{\infty} f^(x) \hat{A}g(x) dx = \int_{-\infty}^{\infty} g(x)[\hat{A}f(x)]^ dx$, where $f(x)$ and $g(x)$ are arbitrary functions.
• The eigenvalues of a Hermitian operator are real.
• The eigenfunctions of a Hermitian operator relating to different eigenvalues are orthogonal.

Measurement Outcome (Postulate 3)

• Postulate 3: Measuring an observable with operator $\hat{A}$ will only yield values that are eigenvalues a, satisfying $\hat{A}\Psi = a\Psi$.

Average Value of an Observable (Postulate 4)

• Postulate 4: For a system in state $\Psi$, the average value of the observable corresponding to $\hat{A}$ is: $\langle A \rangle = \int_{-\infty}^{\infty} \Psi^* \hat{A} \Psi d\tau$, where $\Psi$ is normalized.
• $\langle A \rangle$ is the expectation value of A.

Time Evolution of the System (Postulate 5)

• Postulate 5: The time evolution of a system's state is ruled by the time-dependent Schrödinger equation: $i\hbar \frac{\partial \Psi}{\partial t} = \hat{H} \Psi$.
• $\hat{H}$ is the Hamiltonian operator of the system.
• $\hat{H} = -\frac{\hbar^2}{2m} \nabla^2 + V(\mathbf{r}, t)$.
• $\nabla^2$ is the Laplacian operator.
• $\hbar = h/2\pi$, where h is Planck's constant.
• $V(\mathbf{r}, t)$ represents the potential energy.
• For a time-independent potential, the wave function can be written as: $\Psi(\mathbf{r}, t) = \Psi(\mathbf{r})e^{-iEt/\hbar}$.
• $\Psi(\mathbf{r})$ satisfies the time-independent Schrödinger equation: $\hat{H}\Psi(\mathbf{r}) = E\Psi(\mathbf{r})$.

Hidden Markov Models (HMM) vs. Bayesian Networks

• HMMs are a chain structure where each state depends only on the previous state.
• Bayesian Networks use a directed acyclic graph (DAG) which allows for more complex dependencies.
• HMMs are generative models.
• Bayesian Networks can be generative or discriminative.

Inference

• HMMs use the Viterbi algorithm and Kalman filtering.
• Bayesian Networks use exact inference for small networks and approximate inference for larger networks.

Learning

• HMMs use Baum-Welch algorithm (expectation-maximization).
• Bayesian Networks use maximum likelihood estimation, Bayesian inference.

Representation

• HMMs represent hidden states, observations, transition probabilities, emission probabilities.
• Bayesian Networks represent nodes (variables), edges (dependencies), conditional probabilities.

Typical Applications

• HMMs are used in speech recognition, bioinformatics, and finance.
• Bayesian Networks are used in medical diagnosis, recommendation engines, and natural language processing.

Strengths

• HMMs are simple, efficient for sequential data and modeling temporal processes.
• Bayesian Networks represent complex dependencies, integrate prior knowledge, reason in uncertain situations.

Weaknesses

• HMMs have a strict Markov assumption, are hard to model long-range dependencies, and are computationally costly for long sequences.
• Bayesian Networks inference is computationally costly for large networks, learning is hard with incomplete data, and are subject to overfitting.

Examples

• HMMs model disease progression using hidden states for different stages.
• Bayesian Networks model the relationships between symptoms, diseases, and treatments.

Complexity

• HMM complexity depends on number of states and sequence length, increasing either will significantly increase complexity.
• Bayesian Networks complexity increases significantly with the number of nodes and connections.

Summary

• HMMs are a special case of Bayesian networks with chain structure.
• HMMs are simpler to implement and more efficient for sequential data.
• Bayesian networks can represent more complex dependencies and are better adapted for reasoning in uncertain situations.

Black-Scholes Model Assumptions

• The stock price follows a geometric Brownian motion, $dS = \mu S dt + \sigma S dW_t$.
• The risk-free interest rate, r, is constant.
• No dividends are paid during the option's life.
• The market is efficient, eliminating arbitrage.
• Trading in the underlying asset is continuous.
• Short selling is allowed.
• The option is European and can only be exercised at its expiration date.

Black-Scholes Equation Derivation

• Define a portfolio, $\Pi = V - \Delta S$, consisting of one option and $-\Delta$ shares of the underlying asset.
• Change in portfolio value: $d\Pi = dV - \Delta dS$.
• Using Itô's lemma: $dV = \frac{\partial V}{\partial t} dt + \frac{\partial V}{\partial S} dS + \frac{1}{2} \frac{\partial^2 V}{\partial S^2} (dS)^2$.
• Substitute $dS = \mu S dt + \sigma S dW_t$ into Itô's lemma to find $dV$.
• Substitute $dV$ and $dS$ into the equation for $d\Pi$.
• Eliminate the stochastic term by choosing $\Delta = \frac{\partial V}{\partial S}$, creating a riskless portfolio.
• Riskless portfolio must earn the risk-free rate, $d\Pi = r \Pi dt = r \left(V - S \frac{\partial V}{\partial S} \right) dt$.
• Combining results, the Black-Scholes equation is: $\frac{\partial V}{\partial t} + \frac{1}{2} \sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} + r S \frac{\partial V}{\partial S} - r V = 0$.

Black-Scholes Formula for European Call Option

• Formula: $C(S, t) = S N(d_1) - K e^{-r(T-t)} N(d_2)$
  • $d_1 = \frac{\ln(S/K) + (r + \frac{1}{2} \sigma^2)(T-t)}{\sigma \sqrt{T-t}}$
  • $d_2 = \frac{\ln(S/K) + (r - \frac{1}{2} \sigma^2)(T-t)}{\sigma \sqrt{T-t}} = d_1 - \sigma \sqrt{T-t}$
  • $N(x)$ is the cumulative distribution function of the standard normal distribution.

Black-Scholes Formula for European Put Option

• Formula: $P(S, t) = K e^{-r(T-t)} N(-d_2) - S N(-d_1)$
  • $d_1$ and $d_2$ are defined as in the call option formula.

Matrix Transformations Definition

• A matrix transformation is a function $T: \mathbb{R}^n \rightarrow \mathbb{R}^m$ such that $T(\overrightarrow{\mathbf{x}}) = A\overrightarrow{\mathbf{x}}$ for some $m \times n$ matrix $A$.

Remarks

• Matrix transformations are linear transformations.
• Most linear transformations from $\mathbb{R}^n$ to $\mathbb{R}^m$ are matrix transformations.

Examples

• Reflection through the $x$-axis with $A = \begin{bmatrix} 1 & 0 \ 0 & -1 \end{bmatrix}$. Then $T\begin{pmatrix} 2 \ 1 \end{pmatrix} = \begin{pmatrix} 2 \ -1 \end{pmatrix}$.
• Rotation about the origin through an angle $\theta$ with $A = \begin{bmatrix} \cos{\theta} & -\sin{\theta} \ \sin{\theta} & \cos{\theta} \end{bmatrix}$. When $\theta = \frac{\pi}{2}$, $T\begin{pmatrix} 2 \ 2 \end{pmatrix} = \begin{pmatrix} -2 \ 2 \end{pmatrix}$.
• Projection onto the $x$-axis with $A = \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix}$. Then $T\begin{pmatrix} 2 \ 1 \end{pmatrix} = \begin{pmatrix} 2 \ 0 \end{pmatrix}$.
• Mapping $\mathbb{R}^2$ into $\mathbb{R}^3$ with $A = \begin{bmatrix} 1 & 0 \ 0 & 1 \ 0 & 0 \end{bmatrix}$. Then $T\begin{pmatrix} 2 \ 1 \end{pmatrix} = \begin{pmatrix} 2 \ 1 \ 0 \end{pmatrix}$.
• Mapping $\mathbb{R}^3$ into $\mathbb{R}^2$ with $A = \begin{bmatrix} 1 & 0 & 3 \ 0 & 1 & 2 \end{bmatrix}$. Then $T\begin{pmatrix} 1 \ 2 \ 3 \end{pmatrix} = \begin{pmatrix} 10 \ 8 \end{pmatrix}$.

Theorem

• If $T: \mathbb{R}^n \rightarrow \mathbb{R}^m$ is a linear transformation, then $T$ is a matrix transformation where $T(\overrightarrow{\mathbf{x}}) = A\overrightarrow{\mathbf{x}}$
• The standard matrix for the linear transformation $T$ is: $A = \begin{bmatrix} T(\overrightarrow{\mathbf{e}}_1) & T(\overrightarrow{\mathbf{e}}_2) & \cdots & T(\overrightarrow{\mathbf{e}}_n) \end{bmatrix}$ and $\overrightarrow{\mathbf{e}}_i = \begin{pmatrix} 0 \ \vdots \ 1 \ \vdots \ 0 \end{pmatrix} \leftarrow i^{th} \text{ position }$.
• For $x = \begin{pmatrix} x_1 \ x_2 \ \vdots \ x_n \end{pmatrix} \in \mathbb{R}^n$, $T(\overrightarrow{\mathbf{x}}) = A\overrightarrow{\mathbf{x}}$.

Linear Models Examples

• Maple syrup box sells for $20, production cost is $12.
  • The cost function: $C(x) = 12x$.
  • The revenue function: $R(x) = 20x$.
  • The profit function: $P(x) = R(x) - C(x) = 8x$.

First-Order Ordinary Differential Equations (ODEs)

• A differential equation relates a function to its derivatives.
• An ODE involves a function of a single variable.
• The order of the ODE is the order of the highest derivative.
• The general solution of a first-order ODE contains an arbitrary constant.
• An initial condition determines the value of the constant, giving a particular solution.

Linear ODEs

• A first-order linear ODE has the form: $\frac{dy}{dx} + p(x)y = q(x)$
• Solve by multiplying by the integrating factor: $\mu(x) = e^{\int p(x) dx}$
• The general solution is: $y(x) = \frac{1}{\mu(x)} \int \mu(x) q(x) dx$

Example

Solve $\frac{dy}{dx} + 2y = e^{-x}$ with $y(0) = 3$

• $p(x) = 2$, so $\mu(x) = e^{\int 2 dx} = e^{2x}$
• $y(x) = \frac{1}{e^{2x}} \int e^{2x} e^{-x} dx = e^{-2x} \int e^x dx = e^{-2x} (e^x + C) = e^{-x} + Ce^{-2x}$
• $y(0) = 3 \Rightarrow 3 = e^0 + Ce^0 \Rightarrow C = 2$
• $y(x) = e^{-x} + 2e^{-2x}$

Separable ODEs

• A separable ODE has the form: $\frac{dy}{dx} = f(x)g(y)$
• Rewrite as: $\frac{dy}{g(y)} = f(x) dx$
• Integrate both sides: $\int \frac{dy}{g(y)} = \int f(x) dx$

Example

Solve $\frac{dy}{dx} = xy^2$ with $y(0) = 1$

• $\frac{dy}{y^2} = x dx$
• $\int \frac{dy}{y^2} = \int x dx \Rightarrow -\frac{1}{y} = \frac{x^2}{2} + C$
• $y(0) = 1 \Rightarrow -1 = 0 + C \Rightarrow C = -1$
• $-\frac{1}{y} = \frac{x^2}{2} - 1 \Rightarrow y = \frac{1}{1 - \frac{x^2}{2}} = \frac{2}{2 - x^2}$

Exact ODEs

• An ODE is exact if it has the form: $M(x, y) dx + N(x, y) dy = 0$
  • where $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$
• The solution is $F(x, y) = C$, where $\frac{\partial F}{\partial x} = M$ and $\frac{\partial F}{\partial y} = N$

Example

Solve $(2x + y) dx + (x + 3y^2) dy = 0$

• $M(x, y) = 2x + y$, $N(x, y) = x + 3y^2$
• $\frac{\partial M}{\partial y} = 1$, $\frac{\partial N}{\partial x} = 1$, so it is exact.
• $F(x, y) = \int (2x + y) dx = x^2 + xy + g(y)$
• $\frac{\partial F}{\partial y} = x + g'(y) = N = x + 3y^2 \Rightarrow g'(y) = 3y^2 \Rightarrow g(y) = y^3$
• $F(x, y) = x^2 + xy + y^3 = C$

Applications of ODEs

• Population growth and decay
• Electrical circuits
• Mixtures
• Newton's law of cooling

Summary Table of First-Order ODE Types

• Linear: $\frac{dy}{dx} + p(x)y = q(x)$ - Find the integrating factor and use the formula.
• Separable: $\frac{dy}{dx} = f(x)g(y)$ - Separate the variables and integrate both sides.
• Exact: $M(x, y) dx + N(x, y) dy = 0$ - Find $F(x, y)$ such that partial derivaties equal to M and N.