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## Introduction – Toothpaste Brand Example

Consider the following scenerio:

• Brand T is a toothpaste company that controls 20% of the toothpaste market
• A market researcher predicts the following effects of an ad campaign:
• A consumer using brand T will continue using brand T with a 90% probability
• A consumer not using brand T will switch to brand T with a 70% probability
• For any given customer in the toothpaste market, let…
• $T$ denote a state of using brand T
• $T’$ denote the state of using a toothpaste brand other than T

Definitions:

• transition diagram is wieghted directed graph whose nodes denote various states a system can exist in and edges denote the probability of the system transitioning from one state to another after some time step $\Delta t$
• Transition diagram describing our toothpaste brand example:

• transition probability matrix $\mathrm{P}$ is a $n \times n$ matrix whose $(i,j)$ entries give the probabiliy a “sample” in the $i$th state of a $n$-state system will transition to its $j$th state
• Importantly, the elemental sum of all rows in any $\mathrm{P}$ must equal 1
• Here is transition probability matrix describing our toothpaste brand example:

• An initial state distribution matrix $\mathrm{S_0}$ for a $n$-state system is a $n$-dimensional column vector whose $i$th entries denote the percentage of “samples” that are in state $i$ at time $t=0$
• Here is the initial state distribution matrix for our toothpaste example:

• The $i=1$st entry indicates that 20% of customers in the toothpaste market use brand T (state $T$) and the $i=2$nd entry indicates that 80% of customers in the toothpaste market use brand T’ (state $T’$)
• probability tree gives the probability a “sample” will transition to some state after some time
• For our toothpaste example, let’s construct  a probability tree that tells us the probability a person will use brand T vs. T’ after one month

• In order to determine the probability a customer is using brand T after one week, we need to take the sum of all the possible products of the probability transitions that end in state T (Fig 5)
• Here, there are two possible state transition sequences that end in state T (highlighted in orange and green):
• The orange path gives the probability of a customer will use toothpaste brand T for the entire month ⇒ $P(T \rightarrow T) = (0.2)(0.9) = 0.18$
• The green path gives the probability of a customer using brand T’ at the beginning of the month, then switching to band T at the end of the month ⇒ $P(T’ \rightarrow T) = (0.8)(0.7) = 0.56$
• Now, we can sum the probabilities of each possible transition sequence ending in state $T$ in order to determine the total probability a custormer will be using brand T by the end of the month: $P(T) = P(T \rightarrow T) + P(T’ \rightarrow T) = (0.18)(0.56) = 0.74$
• There is a 74% chance a random customer in the toothpaste market will be using brand T by the end of the month
• Since we only have two possible states in our system, subtracting $P(T)$ from 1 will give us the probability a random customer will be using brand T’ by the end of the month: $P(T’) = 1 – P(T) = 1 – 0.74 = 0.26$
• Importantly, we can construct a state distribution matrix giving the the probabilities for the toothpaste brand a randomly sampled customer will be using after one month:

Theorem: $\mathrm{S_i} \cdot \mathrm{P} = \mathrm{S_{i+1}}$

• Specifically, this is saying that the dot product between a state distribution matrix for time $i$th time step and a corresponding transition probability matrix will return the state distribution matrix for the $i+1$th time step
• Let’s check and make sure this hold true for our example:

$\mathrm{S_0} \cdot \mathrm{P} = \begin{bmatrix} 0.2 & 0.8 \end{bmatrix} \cdot \begin{bmatrix} 0.9 & 0.1 \\ 0.7 & 0.3 \end{bmatrix} = \begin{bmatrix} (0.2)(0.9) + (0.8)(0.7) & (0.2)(0.1) + (0.8)(0.3) \end{bmatrix} = \begin{bmatrix} 0.74 & 0.26 \end{bmatrix} = \mathrm{S_{1}}$

• Results agree with our probability tree!

Assuming $\mathrm{P}$ remains valid, we can determine the expected state distribution matrix for any time step (i.e., month)

• State distribution matrix for second time step:

$\mathrm{S_1} \cdot \mathrm{P} = \begin{bmatrix} 0.74 & 0.26 \end{bmatrix} \cdot \begin{bmatrix} 0.9 & 0.1 \\ 0.7 & 0.3 \end{bmatrix}$

$= \begin{bmatrix} (0.74)(0.9) + (0.26)(0.7) & (0.74)(0.1) + (0.26)(0.3) \end{bmatrix}$

$= \begin{bmatrix} 0.848 & 0.152 \end{bmatrix} = \mathrm{S_{2}}$

• State distribution matrix for third time step

$\mathrm{S_2} \cdot \mathrm{P} = \begin{bmatrix} 0.848 & 0.152 \end{bmatrix} \cdot \begin{bmatrix} 0.9 & 0.1 \\ 0.7 & 0.3 \end{bmatrix}$

$= \begin{bmatrix} (0.848)(0.9) + (0.152)(0.7) & (0.848)(0.1) + (0.152)(0.3) \end{bmatrix}$

$= \begin{bmatrix} 0.8698 & 0.1304 \end{bmatrix} = \mathrm{S_{3}}$

## Regular Markov Chains: Stationary Matrices and Steady State Markov Chains

Example: assume a company initially has 10% of the market share

• Using an advertising campaign, the transition probability matrix is given by
• Notations:
• $A =$ state where a customer is using brand A
• $A’ =$ state where a customer is using brand A’
• Question: what happens to the company’s market shar over a long period of time (assuming $\mathrm{P}$ continues to be valid)
• Solution:

$\mathrm{S_0} = \begin{bmatrix} 0.1 & 0.9 \end{bmatrix}$

$\mathrm{S_1} = \mathrm{S_0} \cdot \mathrm{P} = \begin{bmatrix} 0.1 & 0.9 \end{bmatrix} \cdot \begin{bmatrix} 0.8 & 0.2 \\ 0.6 & 0.4 \end{bmatrix} = \begin{bmatrix} 0.62 & 0.38 \end{bmatrix}$

$\mathrm{S_2} = \mathrm{S_1} \cdot \mathrm{P} = \begin{bmatrix} 0.62 & 0.38 \end{bmatrix} \cdot \begin{bmatrix} 0.8 & 0.2 \\ 0.6 & 0.4 \end{bmatrix} = \begin{bmatrix} 0.724 & 0.276 \end{bmatrix}$

$\mathrm{S_3} = \mathrm{S_2} \cdot \mathrm{P} = \begin{bmatrix} 0.7448 & 0.2552 \end{bmatrix}$

$\mathrm{S_4} = \mathrm{S_3} \cdot \mathrm{P} = \begin{bmatrix} 0.74896 & 0.25104 \end{bmatrix}$

$\mathrm{S_5} = \mathrm{S_4} \cdot \mathrm{P} = \begin{bmatrix} 0.749792 & 0.250208\end{bmatrix}$

$\mathrm{S_6} = \mathrm{S_5} \cdot \mathrm{P} = \begin{bmatrix} 0.7499584 & 0.2500416\end{bmatrix}$

• Here, the state distribution matrices $\mathrm{S_i}$ get closer and closer to $\begin{bmatrix} 0.75 & 0.25 \end{bmatrix}$ as $i \rightarrow \infty$
• Moreover, if we take the dot product between $\mathrm{S} = \begin{bmatrix} 0.75 & 0.25 \end{bmatrix}$ and $\mathrm{P}$, we get $\mathrm{S}$:
• $\mathrm{S} = \mathrm{S} \cdot \mathrm{P} = \begin{bmatrix} 0.75 & 0.25 \end{bmatrix} \cdot \begin{bmatrix} 0.8 & 0.2 \\ 0.6 & 0.4 \end{bmatrix} = \begin{bmatrix} 0.75 & 0.25 \end{bmatrix}$
• No change occurs!
• The matrix $\mathrm{S}$ is called a stationary matrix and the system is said to be at steady state

Questions:

1. Does every Markov chain have a unique stationary matrix?
2. If a Markov chain does have a unique stationary matrix, will the successive state matrices always approach this stationary matrix?

• Generally, the answer to both questions is no
• However, if a Markov chain is a regular Markov chain, then the answer to both questions is yes

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## Overview

### Broca’s Area

Generally located in the inferior frontal gyrus (located in the frontal lobe) of the dominant hemisphere

• Defined in terms of the pars opercularis (Brodmann’s area 44) and the pars triangularis (Brodmann’s area 45)

Both case fMRI and case studies of patients with chronic aphasia (speech impairment) suggest the Broca’s area primarily controls…

• Speech production
• Facial neuron control
• Language processing

Note: slow destruction of the Broca’s area (e.g., via tumor growth) allows for these neural connections to shift to nearby brain areas and leaving speech intact for the most part

### Wernicke’s Area

Generally located in the dominant posterior temporal lobe (below the primary auditoy complex) of the brain

• Traditionally defined as Brodmann’s area 22

Damage to the Wernicke’s area usually results with fluent aphasia (patient can connect words, but phrases lack meaning)

Functions of the Wernicke’s area include,

• Language comprehension
• Semantic processing
• Language recognition
• Language interpretation

### Arcuate Fascilicus

The acuate fascilicus is the group of nerve fiber bundles connecting the Broca’s and Wernicke’s areas

### Angular Gyrus

Located in the anterolateral region of the parietal lobe

Primary functions are associated with