Continuous Random Variables and Their Probability Distributions

Introduction

• The first few lectures come from Mathematical Statistics with Applications, by Wackerly.
  • We must understand the underlying probability and random variable theory before moving into the Bayesian world.
• We will be covering the following chapters:
  • Chapter 2: probability theory
  • Chapter 3: discrete random variables
  • Chapter 4: continuous random variables

4.2: Probability Distributions for Continuous RV

• Distribution function:

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

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

• Theorem: Properties of a Distribution Function
  • 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).
• Note: To be rigorous, if F(y) is a valid distribution function, then F(y) also must be right continuous.

4.2: 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.
    • Note: To be mathematically precise, we also need the first derivative of F(y) to exist and be continuous.
• 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.

4.2: 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: Properties of a Density Function
  • 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.

4.2: 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}.

4.2: Probability Distributions for Continuous RV

• Suppose that F(x) = \begin{cases} 0, & y < 0 \\ y, & 0 \leq y \leq 1 \\ 1, & y > 1 \end{cases}

• Find the probability density function for Y.

4.2: 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.

4.2: Probability Distributions for Continuous RV

• Given f(y) = cy^2, where 0 \le y \le 2 and f(y)=0 elsewhere,

  1. Find the value of c for which f(y) is a valid density function.
     
     
     
     
     
     
     
     

  2. Find P[1 \le Y \le 2].

4.3: 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 mean for a discrete random variable,

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

4.3: 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_2(Y)+...+g_k(Y)\right] = E\left[ g_1(Y) \right] + E\left[ g_2(Y) \right] + ... + E\left[ g_k(Y) \right]

4.3: Expected Values for Continuous RV

• In a previous example, we determined that f(y) = \frac{3}{8}y^2 for 0 \le y \le 2 and f(y) = 0 elsewhere is a valid density function.

• If the random variable Y has this density function, find \mu = E[y] and \sigma^2 = V[Y]

4.4: Uniform Probability Distribution

• Uniform Distribution

4.4: Uniform Probability Distribution

• Uniform Distribution

  • If \theta_1 < \theta_2, a random variable Y is said to have a uniform distribution on the interval (\theta_1, \theta_2) iff

f(y) = \begin{cases} \frac{1}{\theta_2 - \theta_1}, & \theta_1 \le y \le \theta_2 \\ 0, & \text{elsewhere} \end{cases}

• Theorem:

  • If \theta_1 < \theta_2 and Y is a random variable uniformly distributed 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.

4.4: Uniform Probability Distribution

• The change in depth of a river from one day to the next, measured in feet, at a specific location is a random variable, Y, with the following density function:

f(y) = \begin{cases} k, & -2 \le y \le 2 \\ 0, & \text{elsewhere} \end{cases}

1. Determine the value of k.
   
   
   
   
   
   

2. Find the distribution function for Y.

4.4: Uniform Probability Distribution

• The change in depth of a river from one day to the next, measured in feet, at a specific location is a random variable, Y, with the following density function:

f(y) = \begin{cases} k, & -2 \le y \le 2 \\ 0, & \text{elsewhere} \end{cases}

3. What is the mean of the distribution?
   
   
   
   
   
   

4. What is the variance of the distribution?

4.5: Normal Probability Distribution

• Normal Distribution

4.5: Normal Probability Distribution

• Normal 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)}

• Theorem:

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

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

4.5: Normal Probability Distribution

• The weekly amount of money spent on maintenance and repairs by a company was observed over a long period of time to be approximately normally distributed with mean $400 and standard deviation $20.

1. If $450 is budgeted for next week, what is the probability that the actual costs will exceed the budgeted amount?
   
   
   
   
   
   

2. How much should be budgeted for weekly repairs and maintenance to provide that the probability the budgeted amount will be exceeded in a given week is only 0.10?

4.6: Gamma Probability Distribution

• Gamma Distribution

4.6: Gamma Probability Distribution

• Gamma Distribution

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

f(y) = \begin{cases} \frac{y^{\alpha-1} e^{-y/\beta}}{\beta^{\alpha} \Gamma(\alpha)}, & 0 \le y < \infty \\ 0, & \text{elsewhere} \end{cases}

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

• Theorem:

  • 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.

4.6: Gamma Probability Distribution

• Annual incomes for heads of household in a section of a city have approximately a gamma distribution with \alpha=20 and \beta=1000.

  1. What is f(y)?
     
     
     
     

  2. What is the mean of Y?
     
     
     
     

  3. What is the variance of Y?
     
     
     
     

  4. What proportion of heads of households in this section of the city have incomes in excess of $30,000?

4.7: Beta Probability Distribution

• Beta Distribution

4.7: Beta Probability Distribution

• Beta Distribution

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

f(y) = \begin{cases} \frac{y^{\alpha-1}(1-y)^{\beta-1}}{B(\alpha,\beta)}, & 0 \le y \le 1 \\ 0, & \text{elsewhere} \end{cases}

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

• Theorem:

  • 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)}

4.7: Beta Probability Distribution

• The percentage of impurities per batch in a chemical product is a random variable Y with density function

f(y) = \begin{cases} 12y^2(1-y), & 0 \le y \le 1 \\ 0, & \text{elsewhere} \end{cases}

• A batch with more than 40% impurities cannot be sold.

1. Use integration to determine the probability that a randomly selected batch cannot be sold because of excessive impurities.
   
   
   
   
   

2. Use R to find the answer to part (a).

4.7: Beta Probability Distribution

• The percentage of impurities per batch in a chemical product is a random variable Y with density function

f(y) = \begin{cases} 12y^2(1-y), & 0 \le y \le 1 \\ 0, & \text{elsewhere} \end{cases}

• A batch with more than 40% impurities cannot be sold.

3. Find the mean of the percentage of impurities in a randomly selected batch of the chemical.
   
   
   
   
   

4. Find the variance of the percentage of impurities in a randomly selected batch of the chemical.

Homework