Overview of Probability Theory
and Probability Distributions

Introduction

• Today we will review probability basics.
  • Basic terminology
  • Basic properties
  • Types of probabilities
  • Types of events
• We will also review probability distributions.
  • Binomial
  • Poisson
  • Uniform
  • Normal
  • Gamma
  • Beta

Terminology for Probability

• Experiment: A process that results in one and only one of many possible observations.

• Simple outcomes: The possible results of our experiment.

• Sample space: Collection of possible outcomes of the experiment.

• Event: A collection of one or more of the outcomes of the experiment.

• Example: rolling a die once.

  • Outcome: the result of the die.
  • Sample space: {1, 2, 3, 4, 5, 6}
  • Events: rolling an odd; rolling a multiple of 3; rolling a 3 or better.

Properties of Probability

• There are two properties of probability that we must keep at the forefront of our mind.

• First, probability always falls between 0 and 1. Mathematically,

0 \le P[E_i] \le 1

• What does p=0 imply?
  
  
• What does p=0.5 imply?
  
  
• What does p=1 imply?

Properties of Probability

• There are two properties of probability that we must keep at the forefront of our mind.

• Second, the sum of all simple events for an experiment is always 1. Mathematically,

\sum_{i=1}^n P[E_i] = P[E_1] + ... + P[E_n] = 1

• If there are 2 events and we know P[E_1]=0.7, what is P[E_2]?
  
  
  

• If there are 4 events and we know P[E_1]=P[E_2]=0.1, P[E_3]=0.6, what is P[E_4]?

Assigning Probabilities

• How do we assign probabilities?
  • Subjective probability
  • Classicial probability rules
  • Relative frequency

Subjective Probability

• Subjective probability is the probability assigned to an event based on subjective judgement, experience, information, and belief.

• Examples:

  • P[UWF wins national championship]
  • P[tomato plant eaten by hornworms]
  • P[A in this course]

Classical Probability

• Let A be an event for an experiment with equally likely outcomes,

P[A] = \frac{\text{Number of outcomes favorable to $A$}}{\text{Total number of outcomes for the experiment}}

• Examples:
  • P[2 heads on 3 coin tosses]
  • P[at least 2 heads on 4 coin tosses]
  • P[even when rolling die]

Relative Frequency

• If an experiment is repeated n times and an event A is observed f times, then

P[A] = \frac{f}{n} = \frac{\text{Frequency of $A$}}{\text{Sample size}}

• Example:
  • P[car is a lemon] given 10/500 sampled cars from a factory are lemons.
  • P[person is a homeowner] given 730/1000 sampled individuals own a home.

Contingency Tables

• Suppose 100 employees at Target were asked whether they are in favor of or against extending store hours during the holiday season.

Marginal Probability

• A marginal probability is the probability of a single event occurring without considering any other variables.

  • In a contingency table, marginal probabilities are found outside the body of the table.

• It tells us the likelihood of one category happening overall, regardless of how it combines (or interacts) with other categories.

• In our Target example:

  • What is the probability that a randomly selected employee is in favor?
  • What is the probability that randomly selected employee works in the Grocery department?

Marginal Probability

• What is the probability that a randomly selected employee is in favor?

Marginal Probability

• What is the probability that randomly selected employee works in the Grocery department?

Joint Probability

• Joint probability: the probability that two events happen at the same time.

  • In a contingency table, joint probabilities are found inside the body of the table.

• It tells us the likelihood that a randomly selected observation falls into both categories simultaneously.

• In our Target example:

  • What is the probability that an employee is in the Grocery department and in favor of extended hours?
  • What is the probability that an employee is in Electronics and against extended hours?

Joint Probability

• What is the probability that an employee is in the Grocery department and in favor of extended hours?

Joint Probability

• What is the probability that an employee is in Electronics and against extended hours?

Conditional Probability

• Conditional probability: one event occurs given that we already know another event has occurred.
  • “What is the probability of Event A if we know Event B is true?”
  • In a contingency table, conditional probabilities are found by limiting yourself to a specific row or column of the table, then finding the corresponding probability.
• In our Target example,
  • What is the probability that an employee is in favor of extended hours given that they work in the Grocery department?
  • What is the probability that an employee works in the Electronics department given that they are against extended hours?

Conditional Probability

• What is the probability that an employee is in favor of extended hours given that they work in the Grocery department?

Conditional Probability

• What is the probability that an employee works in the Electronics department given that they are against extended hours?

Mutually Exclusive Events

• Events that cannot occur together are mutually exclusive or disjoint.

• Consider rolling a single die:

  • A = an even number = {2, 4, 6}
  • B = an odd number = {1, 3, 5}
  • C = a number less than 5 = {1, 2, 3, 4}

• Are events A and B mutually exclusive?

• Are events A and C mutually exclusive?

Mutually Exclusive Events

• Consider rolling a single die:
  • A = an even number = {2, 4, 6}
  • B = an odd number = {1, 3, 5}
  • C = a number less than 5 = {1, 2, 3, 4}
• Are events A and B mutually exclusive?

Mutually Exclusive Events

• Consider rolling a single die:
  • A = an even number = {2, 4, 6}
  • B = an odd number = {1, 3, 5}
  • C = a number less than 5 = {1, 2, 3, 4}
• Are events A and C mutually exclusive?

General Addition Rule

• General Addition Rule: For any two events, A and B,

P[A \cup B] = P[A] + P[B] - P[A \cap B]

• When events are mutually exclusive, P[A \cap B] = 0. In this case, the rule simplifies to

P[A] + P[B]

• In our Target example:
  • What is the probability that a randomly selected employee works in Grocery or is in favor of extended hours?
  • What is the probability that a randomly selected employee works in Electronics or Clothing?

General Addition Rule

• What is the probability that a randomly selected employee works in Grocery or is in favor of extended hours?

General Addition Rule

• What is the probability that a randomly selected employee works in Electronics or Clothing?

Independent Events

• Independent events: the occurrence of one event does not affect the probability of the occurrence of the other event.

• Mathematically,

P[A|B] = P[A] \text{ or } P[B|A] = P[B]

• In our Target example,
  • Is department independent of being in favor of extended hours?

Independent Events

• We need to examine P[department|favor] or P[favor|department]. We say that they are independent if
  • P[department|favor] = P[department]
  • P[favor|department] = P[favor]

Complementary Events

• Complement of A: the event that includes all the outcomes that are not in A.

• Mathematically,

\begin{align*} P[A] &+ P[A^c] = 1 \\ P[A] &= 1 - P[A^c] \\ P[A^c] &= 1 - P[A] \end{align*}

Complementary Events

• Find P[Electronics^c].

Complementary Events

• Find P[Electronics^c | Against].

Bayes’ Theorem

• Bayes’ Theorem: allows us to reverse a conditional probability.
  • If we know P[B|A], we can find P[A|B].

P[A|B] = \frac{P[B|A] \times P[A]}{P[B]}

• Recall from the Conditional Probability example, we found the probability that an employee is in favor of extended hours given that they work in the Grocery department.

• To move from P[B|A] \to P[B|A], we also need P[A] and P[B].

Bayes’ Theorem

• Using Bayes Theorem and existing probabilities, find the probability that an employee works in the Grocery department given that they are in favor of extended hours.

Bayes’ Theorem

• Using Bayes Theorem and existing probabilities, find the probability that an employee is against extended hours given that they work in the Electronics department.

Basic Definitions

• Discrete random variable: a variable that can assume only a finite or countably infinite number of distinct values.

• Probability distribution of a random variable: collection of probabilities for each value of the random variable.

• Notation:

  • Uppercase letter (e.g., Y) denotes a random variable.
  • Lowercase letter (e.g., y) denotes a particular value that the random variable may assume.
    • The specific observed value, y, is not random.

Probability Distributions for Discrete RV

• Probability function for \boldsymbol Y: sum of the the probabilities of all sample points in S that are assigned the value y

  • P[Y = y] = p(y): the probability that Y takes on the value y.

• Probability distribution for \boldsymbol Y: a formula, table, or graph that provides p(y) \ \forall \ y.

• Theorem: For any discrete probability distribution, the following must be true:

  • 0 \le p(y) \le 1 \ \forall \ y
  • \sum_y p(y) = 1 \ \forall \ p(y) > 0.

Expected Values for Discrete RV

• Expected value: Let Y be a discrete random variable with probability function p(y). Then the expected value of Y, E[Y], is defined to be

E(Y) = \sum_{y} y p(y)

• When p(y) is an accurate characterization of the population frequency distribution, then the expected value is the population mean.

E[Y] = \mu

Expected Values for Discrete RV

• Variance: if Y is a random variable with mean E[Y] = \mu, the variance of a random variable Y is defined to be the expected value of (Y-\mu)^2.

V[Y] = E\left[ (Y-\mu)^2 \right]

• If p(y) is an accurate characterization of the population frequency distribution, then V(Y) is the population variance,

V[Y] = \sigma^2

• Standard deviation: the positive square root of V[Y].

Expected Values for Discrete RV

• There is an alternative (and easier) way to calculate the variance manually,

• Theorem: Let Y be a discrete random variable with probability function p(y) and mean E[Y] = \mu. Then,

V[Y] = \sigma^2 = E\left[(Y-\mu)^2\right] = E\left[Y^2\right] - \mu^2

Expected Values for Discrete RV

• The probability distribution for a random variable Y is given below.

• Find the mean of Y.

Expected Values for Discrete RV

• The probability distribution for a random variable Y is given below.

• Find the variance and standard deviation of Y.

Binomial Probability Distribution

• Binomial experiment:

  1. The experiment consists of a fixed number, n, of identical trials.

  2. Each trial results in one of two outcomes: success (S) or failure (F).

  3. The probability of success on a single trial is equal to some value p and remains the same from trial to trial.

     • The probability of failure is equal to q = (1-p).

  4. The trials are independent.

  5. The random variable of interest is Y, the number of successes observed during the n trials.

Binomial Probability Distribution

• A random variable Y is said to have a binomial distribution based on n trials with success probability p iff

p(y) = {n \choose y}p^y q^{n-y}, \text{ where } y = 0, 1, 2, ..., n, \text{ and } 0 \le p \le1

• Let Y be a binomial random variable based on n trials and success probability p. Then

E[Y] = \mu = np \ \ \ \text{and} \ \ \ V[Y] = \sigma^2 = npq

• See Wackerly pg. 107 for derivation.

Binomial Probability Distribution

• We can use R to find information related to the binomial distribution.
  • P[X = x]: dbinom(x, size, prob)
  • P[X \le x]: pbinom(q, size, prob)
  • P[X > x]: pbinom(q, size, prob, lower.tail = FALSE)
• In the functions,
  • x or q is the value of X we are interested in
  • size is the sample size (n)
  • prob is the probability of success, \pi
  • lower.tail has two options:
    • TRUE (default) returns P[X \le x]
    • FALSE returns P[X > x]

Binomial Probability Distribution

• The manufacturer of a dairy drink wishes to compare a new formula (B) with that of the standard formula (A). Each of four judges perform a blinded taste test and report which glass he or she most enjoyed. Suppose that the two formulas are equally attractive.
  1. What is the probability distribution?
     
     
     
     
     
  2. What is the mean of the distribution?
     
     
     
  3. What is the variance of the distribution?

Binomial Probability Distribution

• The manufacturer of a dairy drink wishes to compare a new formula (B) with that of the standard formula (A). Each of four judges perform a blinded taste test and report which glass he or she most enjoyed. Suppose that the two formulas are equally attractive.

• Use R to find:

  1. P[X = 2]
  2. P[X > 2]
  3. P[X < 4]

Binomial Probability Distribution

• The manufacturer of a dairy drink wishes to compare a new formula (B) with that of the standard formula (A). Each of four judges perform a blinded taste test and report which glass he or she most enjoyed. Suppose that the two formulas are equally attractive.

• Use R to find:

  1. P[X = 2]

dbinom(x = 2, size = 4, prob = 0.5)

[1] 0.375

Binomial Probability Distribution

• The manufacturer of a dairy drink wishes to compare a new formula (B) with that of the standard formula (A). Each of four judges perform a blinded taste test and report which glass he or she most enjoyed. Suppose that the two formulas are equally attractive.

• Use R to find:

  2. P[X > 2]

pbinom(q = 2, size = 4, prob = 0.5, lower.tail = FALSE)

[1] 0.3125

Binomial Probability Distribution

• The manufacturer of a dairy drink wishes to compare a new formula (B) with that of the standard formula (A). Each of four judges perform a blinded taste test and report which glass he or she most enjoyed. Suppose that the two formulas are equally attractive.

• Use R to find:

  3. P[X < 4] = P[X \le 3]

pbinom(q = 3, size = 4, prob = 0.5)

[1] 0.9375

Poisson Probability Distribution

• We often use the Poisson distribution to model count data.

• A random variable Y is said to have a Poisson probability distribution iff

p(y) = \frac{\lambda^y}{y!}e^{-\lambda}, \text{ where } y=0,1,2,..., \text{ and } \lambda > 0

• If Y is a random variable with a Poisson distribution with parameter \lambda, then

E[Y] = \mu = \lambda \text{ and } V[Y] = \sigma^2 = \lambda

• See Wackerly pg. 134 for derivation.

Poisson Probability Distribution

• We can use R to find information related to the Poisson distribution.
  • P[X = x]: dpois(x, lambda)
    
  • P[X \le x]: ppois(q, lambda)
    
  • P[X > x]: ppois(q, lambda, lower.tail = FALSE)
• In the functions:
  • x or q is the value of X we are interested in
    
  • lambda is the rate of occurrence
  • lower.tail has two options:
    • TRUE (default) returns P[X \le x]
      
    • FALSE returns P[X > x]

Poisson Probability Distribution

• Customers arrive at a checkout counter in a department store according to a Poisson distribution at an average of seven per hour.
  1. What is the probability distribution?
     
     
     
     
     
  2. What is the mean of the distribution?
     
     
     
  3. What is the variance of the distribution?

Poisson Probability Distribution

• Customers arrive at a checkout counter in a department store according to a Poisson distribution at an average of seven per hour. Use R to find the following probabilities.
  1. No more than three customers arrive.
  2. At least two customers arrive.
  3. Exactly five customers arrive.

Poisson Probability Distribution

• Customers arrive at a checkout counter in a department store according to a Poisson distribution at an average of seven per hour. Use R to find the following probabilities.
  1. No more than three customers arrive.

ppois(q = 3, lambda = 7)

[1] 0.08176542

Poisson Probability Distribution

• Customers arrive at a checkout counter in a department store according to a Poisson distribution at an average of seven per hour. Use R to find the following probabilities.
  2. At least two customers arrive.

ppois(q = 1, lambda = 7, lower.tail = FALSE)

[1] 0.9927049

Poisson Probability Distribution

• Customers arrive at a checkout counter in a department store according to a Poisson distribution at an average of seven per hour. Use R to find the following probabilities.
  3. Exactly five customers arrive.

dpois(x = 5, lambda = 7)

[1] 0.1277167

Probability Distributions for Continuous RV

• The distribution function of Y (any random variable), denoted by F(y), is such that

F(y) = P[Y \le y] \text{ for } -\infty < y < \infty

• Theorem: If F(y) is a distribution function, then
  • F(-\infty) \equiv \underset{y \to -\infty}{\lim} F(y) = 0
  • F(\infty) \equiv \underset{y \to \infty}{\lim} F(y) = 1
  • F(y) is a nondecreasing function of y.
    • If y_1 and y_2 are any values such that y_1 < y_2, then F(y_1) \le F(y_2).

Probability Distributions for Continuous RV

• Continuous random variable:
  • A random variable Y with distribution function F(y) is said to be continuous if F(y) is continuous for -\infty < y < \infty.
• If Y is a continuous random variable, then for any real number y, P[Y = y] = 0.
  • i.e., we must remember to find the probability of an interval.

Probability Distributions for Continuous RV

• Probability density function:
  • Let F(y) be the cumulative density function for a continuous random variable, Y. Then

p[Y = y] = f(y) = \frac{dF(y)}{dy} = F'(y).

• Theorem: If f(y) is a density function for a continuous random variable, then
  • f(y) \ge 0 \ \forall y, \ -\infty < y < \infty.
  • \int_{-\infty}^{\infty} f(y) dy = 1.

Probability Distributions for Continuous RV

• Cumulative density function:

  • Let f(y) be the probability density function for a continuous random variable, Y. Then

P[Y \le y] = F(y) = \int_{-\infty}^y f(t) dt, \text{ for } y \in {\rm I\!R}.

Probability Distributions for Continuous RV

• Theorem: If the random variable Y has density function f(y) and a < b, then the probability that Y falls in the interval [a, b] is

P[a \le Y \le b] = \int_a^b f(y) dy.

Expected Values for Continuous RV

• Expected value: The expected value of a continuous variable Y is

E[Y] = \int_{-\infty}^{\infty} y f(y) \ dy

• This is the continuous version of the expected value for a discrete random variable,

E[Y] = \sum_y y p(y)

Expected Values for Continuous RV

• Theorem: Let g(Y) be a function of Y; then the expected value of g(Y) is given by

E\left[ g(Y) \right] = \int_{-\infty}^{\infty} g(y) f(y) \ dy

• Theorem: Let c be a constant and let g(Y), g_1(Y), g_2(Y), …, g_k(Y) be functions of a continuous random variable, Y. Then the following results hold:
  • E[c] = c
  • E\left[cg(Y)] = cE[g(Y)\right]
  • E\left[g_1(Y)+...+g_k(Y)\right] = E\left[ g_1(Y) \right] + ... + E\left[ g_k(Y) \right]

Uniform Probability Distribution

• Uniform Distribution

Uniform Probability Distribution

• A random variable Y is said to have a uniform distribution iff

f(y) = \frac{1}{\theta_2 - \theta_1}, \ \theta_1 \le y \le \theta_2

• If \theta_1 < \theta_2 and Y is a uniformly distributed r.v. on the interval (\theta_1, \theta_2), then

E[Y] = \mu = \frac{\theta_1+\theta_2}{2} \ \ \ \text{and} \ \ \ V[Y] = \sigma^2 = \frac{(\theta_2-\theta_1)^2}{12}

• See Wackerly pg. 176 for derivation.

Uniform Probability Distribution

• An industrial psychologist has determined that it takes a worker between 9 and 15 minutes to complete a task on an automobile assembly line. If the time to complete the task is uniformly distributed over the interval 9 \le y \le 15, then determine:
  1. The probability distribution.
     
     
     
     
  2. The mean of the distribution.
     
     
     
  3. The variance and standard deviation of the distribution.

Uniform Probability Distribution

• We can use R to find information related to the uniform distribution:
  • P[X \le x]: punif(q, min, max)
    
  • P[X \ge x]: punif(q, min, max, lower.tail = FALSE)
• In the functions:
  • q is the value of X we are interested in
    
  • min is the lower bound of the distribution
    
  • max is the upper bound of the distribution
    
  • lower.tail has two options:
    • TRUE (default) returns P[X \le x]
      
    • FALSE returns P[X \ge x]

Uniform Probability Distribution

• An industrial psychologist has determined that it takes a worker between 9 and 15 minutes to complete a task on an automobile assembly line. If the time to complete the task is uniformly distributed over the interval 9 \le y \le 15, then determine the following probabilities:
  1. A worker takes fewer than 13 minutes.
  2. A worker takes at least 11 minutes.
  3. A worker takes between 14 and 15 minutes.

Uniform Probability Distribution

• An industrial psychologist has determined that it takes a worker between 9 and 15 minutes to complete a task on an automobile assembly line. If the time to complete the task is uniformly distributed over the interval 9 \le y \le 15, then determine the following probabilities:
  1. A worker takes fewer than 13 minutes.

punif(13, 9, 15)

[1] 0.6666667

Uniform Probability Distribution

• An industrial psychologist has determined that it takes a worker between 9 and 15 minutes to complete a task on an automobile assembly line. If the time to complete the task is uniformly distributed over the interval 9 \le y \le 15, then determine the following probabilities:
  2. A worker takes at least 11 minutes.

punif(11, 9, 15, lower.tail = FALSE)

[1] 0.6666667

Uniform Probability Distribution

• An industrial psychologist has determined that it takes a worker between 9 and 15 minutes to complete a task on an automobile assembly line. If the time to complete the task is uniformly distributed over the interval 9 \le y \le 15, then determine the following probabilities:
  3. A worker takes between 14 and 15 minutes.

punif(15, 9, 15) - punif(14, 9, 15)

[1] 0.1666667

Normal Probability Distribution

• Normal Distribution

Normal Probability Distribution

• A random variable Y is said to have a normal distribution iff, for \sigma > 0 and -\infty < \mu < \infty,

f(y) = \frac{1}{\sigma \sqrt{2\pi}} e^{-(y-\mu)^2/(2\sigma^2)}

• If Y is a random variable normally distributed with parameters \mu and \sigma, then

E[Y] = \mu \ \ \ \text{and} \ \ \ V[Y] = \sigma^2

Normal Probability Distribution

• We can use R to find information related to the normal distribution.
  • P[X \le x]: pnorm(q, mean, sd)
    
  • P[X \ge x]: pnorm(q, mean, sd, lower.tail = FALSE)
• In the functions:
  • q is the value of X we are interested in
    
  • mean is the population mean \mu
    
  • sd is the standard deviation \sigma
    
  • lower.tail has two options:
    • TRUE (default) returns P[X \le x]
      
    • FALSE returns P[X \ge x]

Normal Probability Distribution

• A random variable Y is said to have a standard normal distribution iff

Y \sim N(\mu=0,\sigma=1)

• The normal distribution is then simplified to

f(y) = \frac{1}{\sqrt{2\pi}} e^{-y^2/2}

• Note that in all cases of the normal distribution, we assume -\infty < y < \infty.

Normal Probability Distribution

• When using pnorm(), the default values for mean and sd are 0 and 1.

• Thus, if we have the standard normal our R functions simplify to:

  • P[Z \le z]: pnorm(z)
    
  • P[Z \ge z]: pnorm(z, lower.tail = FALSE)

• In the functions:

  • q is the z-score value of interest
    
  • lower.tail = TRUE returns P[Z \le z]
    
  • lower.tail = FALSE returns P[Z \ge z]

Normal Probability Distribution

• A geneticist working for a seed company develops a new carrot for growing in heavy clay soil. After measuring 5000 of these carrots, it can be said that the carrot length, Y, is normally distributed with \mu = 11.5 cm and \sigma = 1.15 cm. Determine
  1. The probability distribution.
     
     
     
     
  2. The mean of the distribution.
     
     
     
  3. The variance and standard deviation of the distribution.

Normal Probability Distribution

• A geneticist working for a seed company develops a new carrot for growing in heavy clay soil. After measuring 5000 of these carrots, it can be said that the carrot length, Y, is normally distributed with \mu = 11.5 cm and \sigma = 1.15 cm.
  1. What is the probability that a carrot will be between 10 and 13 cm?
  2. What is the probability that a carrot will be less than 9 cm?
  3. What is the probability that a carrot will be 12 cm or larger?

Normal Probability Distribution

• A geneticist working for a seed company develops a new carrot for growing in heavy clay soil. After measuring 5000 of these carrots, it can be said that the carrot length, Y, is normally distributed with \mu = 11.5 cm and \sigma = 1.15 cm.
  1. What is the probability that a carrot will be between 10 and 13 cm?

pnorm(q = 13, mean = 11.5, sd = 1.15) - pnorm(q = 10, mean = 11.5, sd = 1.15)

[1] 0.807885

Normal Probability Distribution

• A geneticist working for a seed company develops a new carrot for growing in heavy clay soil. After measuring 5000 of these carrots, it can be said that the carrot length, Y, is normally distributed with \mu = 11.5 cm and \sigma = 1.15 cm.
  2. What is the probability that a carrot will be less than 9 cm?

pnorm(q = 9, mean = 11.5, sd = 1.15)

[1] 0.01485583

Normal Probability Distribution

• A geneticist working for a seed company develops a new carrot for growing in heavy clay soil. After measuring 5000 of these carrots, it can be said that the carrot length, Y, is normally distributed with \mu = 11.5 cm and \sigma = 1.15 cm.
  3. What is the probability that a carrot will be 12 cm or larger?

pnorm(q = 12, mean = 11.5, sd = 1.15, lower.tail = FALSE)

[1] 0.3318601

Gamma Probability Distribution

• Gamma Distribution

Gamma Probability Distribution

• A random variable Y is said to have a gamma distribution with parameters \alpha > 0 and \beta > 0 iff,

f(y) = \frac{y^{\alpha-1} e^{-y/\beta}}{\beta^{\alpha} \Gamma(\alpha)}, \ 0 \le y < \infty

• Note that \Gamma(\alpha) = \int_{0}^{\infty} y^{\alpha-1} e^{-y} \ dy.

• If Y has a gamma distribution with parameters \alpha and \beta, then

E[Y] = \mu = \alpha\beta \ \ \ \text{and} \ \ \ V[Y] = \sigma^2 = \alpha\beta^2

• See Wackerly pg. 187 for derivation.

Gamma Probability Distribution

• We can use R to find information related to the Gamma distribution.
  • P[X \le x]: pgamma(q, shape, rate)
    
  • P[X \ge x]: pgamma(q, shape, rate, lower.tail = FALSE)
• In the functions:
  • q is the value of X we are interested in
    
  • shape is the shape parameter, \alpha
    
  • scale is the scale parameter, \beta
    • Alternatively, can parameterize with rate = 1/\beta, rate = 1 / scale
  • lower.tail has two options:
    • TRUE (default) returns P[X \le x]
      
    • FALSE returns P[X \ge x]

Gamma Probability Distribution

• Annual incomes for heads of household in an affluent section of a city have approximately a gamma distribution with \alpha=32 and \beta=2500. Determine:
  1. The probability distribution.
     
     
     
     
  2. The mean of the distribution.
     
     
     
  3. The variance and standard deviation of the distribution.

Gamma Probability Distribution

• Annual incomes for heads of household in an affluent section of a city have approximately a gamma distribution with \alpha=32 and \beta=2500.
  1. What proportion have incomes in excess of $100,000?
  2. What proportion have incomes between $75,000 and $150,000?

Gamma Probability Distribution

• Annual incomes for heads of household in a section of a city have approximately a gamma distribution with \alpha=32 and \beta=2500.
  1. What proportion have incomes in excess of $30,000?

pgamma(q = 30000, shape = 32, scale = 2500, lower.tail = FALSE)

[1] 0.9999988

Gamma Probability Distribution

• Annual incomes for heads of household in an affluent section of a city have approximately a gamma distribution with \alpha=32 and \beta=2500.

  2. What proportion have incomes between $75,000 and $150,000?
     

  pgamma(q = 150000, shape = 32, scale = 2500) - pgamma(q = 75000, shape = 32, scale = 2500)

  [1] 0.6186147

Beta Probability Distribution

• Beta Distribution

Beta Probability Distribution

• A random variable Y is said to have a beta distribution with parameters \alpha > 0 and \beta > 0 iff,

f(y) = \frac{y^{\alpha-1}(1-y)^{\beta-1}}{B(\alpha,\beta)}, \ 0 \le y \le 1

• Note: B(\alpha,\beta) = \int_0^1 y^{\alpha-1}(1-y)^{\beta-1} \ dy = \frac{\Gamma(\alpha) \Gamma(\beta)}{\Gamma(\alpha+\beta)}.

• If Y has a beta distribution with parameters \alpha > 0 and \beta > 0, then

E[Y] = \mu = \frac{\alpha}{\alpha+\beta} \ \ \ \text{and} \ \ \ V[Y] = \sigma^2 = \frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}

Beta Probability Distribution

• We can use R to find information related to the Beta distribution.
  • P[X \le x]: pbeta(q, shape1, shape2)
    
  • P[X \ge x]: pbeta(q, shape1, shape2, lower.tail = FALSE)
• In the functions:
  • q is the value of X we are interested in – must be in [0, 1]!
  • shape1 is the first shape parameter, \alpha
    
  • shape2 is the second shape parameter, \beta
    
  • lower.tail has two options:
    • TRUE (default) returns P[X \le x]
      
    • FALSE returns P[X \ge x]

Beta Probability Distribution

• In a survey of cupcake preferences, 8 respondents liked the new cupcake flavor and 2 did not. We will model the proportion of all respondents who would like the cupcake flavor using a Beta distribution with \alpha = 8 and \beta = 2. Determine:
  1. The probability distribution.
     
     
     
     
  2. The mean of the distribution.
     
     
     
  3. The variance and standard deviation of the distribution.

Beta Probability Distribution

• In a survey of cupcake preferences, 8 respondents liked the new cupcake flavor and 2 did not. We will model the proportion of all respondents who would like the cupcake flavor using a Beta distribution with \alpha = 8 and \beta = 2. What is the probability that:
  1. fewer than 60% of respondents like the new flavor?
  2. more than 90% of respondents like the new flavor?
  3. somewhere between 70% and 90% of respondents like the new flavor?

Beta Probability Distribution

• In a survey of cupcake preferences, 8 respondents liked the new cupcake flavor and 2 did not. We will model the proportion of all respondents who would like the cupcake flavor using a Beta distribution with \alpha = 8 and \beta = 2. What is the probability that:
  1. fewer than 60% of respondents like the new flavor?

pbeta(q = 0.6, shape1 = 8, shape2 = 2)

[1] 0.07054387

Beta Probability Distribution

• In a survey of cupcake preferences, 8 respondents liked the new cupcake flavor and 2 did not. We will model the proportion of all respondents who would like the cupcake flavor using a Beta distribution with \alpha = 8 and \beta = 2. What is the probability that:
  2. more than 90% of respondents like the new flavor?

pbeta(q = 0.9, shape1 = 8, shape2 = 2, lower.tail = FALSE)

[1] 0.225159

Beta Probability Distribution

• In a survey of cupcake preferences, 8 respondents liked the new cupcake flavor and 2 did not. We will model the proportion of all respondents who would like the cupcake flavor using a Beta distribution with \alpha = 8 and \beta = 2. What is the probability that:
  3. somewhere between 70% and 90% of respondents like the new flavor?

pbeta(q = 0.9, shape1 = 8, shape2 = 2) - pbeta(q = 0.7, shape1 = 8, shape2 = 2)

[1] 0.5788377

Wrap Up

• We reviewed basic probability rules and distributions today.
  • We are setting up for our next lecture(s).
• You now are ready to work on Assignment 1: Probability Distributions.
  • .qmd file is available to download on Canvas.
  • Goals:
    • Identify the applicable distribution.
    • Find probabilities using said distributions.
• Wednesday: Thinking Like a Bayesian