Lesson 18: Some Important Continuous RVs

Meike Niederhausen and Nicky Wakim

2025-11-19

Learning Objectives

Distinguish between Uniform, Exponential, Gamma, and Normal distributions when reading a story.
Identify the variable and the parameters in a story, and state what the variable and parameters mean.
Use the formulas for the pdf/CDF, expected value, and variance to answer questions and find probabilities.

Where are we?

Chapter 31: Uniform RVs

Properties of continuous uniform RVs

Scenario: Events are equally likely to happen anywhere or anytime in an interval of values
Shorthand: \(X \sim \text{U}[a,b]\)

\[ f_X(x) = \dfrac{1}{b-a}, \text{ for }a \leq x \leq b \]

\[ F_X(x) = \left\{ \begin{array}{ll} 0 & \quad x<a \quad \\ \dfrac{x-a}{b-a} & \quad a \leq x \leq b\quad \\ 1 & \quad x>b \quad \end{array} \right. \]

\[\text{E}(X) = \dfrac{a+b}{2} \text{, } \text{ Var}(X) = \dfrac{(b-a)^2}{12}\]

Identifying continuous uniform RV from word problems

Look for some indication that all events are equally likely
- Could also say “uniformly distributed”
Look for an interval
- Time example: Costumer in your store will approach the cash register in next 30 minutes. Approaching the register throughout the 30 minutes is equally likely.
- Length example: You have a 12 inch string that you need to cut. You are equally likely to cut anywhere on the string.
Different than the discrete uniform
- Discrete usually includes a countable number of events that are equally likely
- Continuous is not countable
  - Exact time and length can be measured with infinite decimal places

Helpful R code

Let’s say we’re looking at equally likely arrival times between 10 am and 11 am.

If we want to know the probability that someone arrives at 10:30am or earlier:
```
punif(q = 30, min = 0, max = 60)
```
```
[1] 0.5
```
If we want to know the time, say \(t\), where the probability of arriving at \(t\) or earlier is 0.35:
```
qunif(p = 0.35, min = 0, max = 60)
```
```
[1] 21
```

If we want to know the probability that someone arrives between 10:14 and 10:16 am:

punif(q = 16, min = 0, max = 60) - punif(q = 14, min = 0, max = 60)

[1] 0.03333333

If we want to sample 20 arrival times from the distribution:

runif(n = 20, min = 0, max = 60)

 [1] 36.1518032 56.6547857 45.1115403 51.8594366 34.8732771 12.8750386
 [7] 42.5877139 44.3007376 56.7688498 30.0011219 54.6299942 12.6420043
[13] 44.1304846 19.9011956 28.5120543  5.4834140 31.0400753 46.8370057
[19] 35.1252265  0.9729805

Bird on a wire (TB 31.5)

Example 1

A bird lands at a location that is Uniformly distributed along an electrical wire of length 150 feet. The wire is stretched tightly between two poles. What is the probability that the bird is 20 feet or less from one or the other of the poles?

Chapter 32: Exponential RVs

Properties of exponential RVs

Scenario: Modeling the time until the next (first) event
Continuous analog to the geometric distribution!
Shorthand: \(X \sim \text{Exp}(\lambda)\)

\[ f_X(x) = \lambda e^{-\lambda x}\text{ for } x>0, \lambda>0 \]

\[ F_X(x) = \left\{ \begin{array}{ll} 0 & \quad x<0 \quad \\ 1 - e^{-\lambda x} & \quad x\geq0 \\ \end{array} \right. \]

\[\text{E}(X) = \dfrac{1}{\lambda}\] \[\text{Var}(X) = \dfrac{1}{\lambda^2}\]

Memoryless Property

If \(b>0\),

\[P(X > a +b | X> a) = P(X > b)\]

This can be interpreted as:
- If you have waited \(a\) seconds (or any other measure of time) without a success
- Then the probability that you have to wait \(b\) more seconds is the same as as the probability of waiting \(b\) seconds initially.

Identifying exponential RV from word problems

Look for time between events/successes
Look for a rate of the events over time period
How does it differ from the geometric distribution?
- Geometric is number of trials until first success
- Exponential is time until first success
Relation to the Poisson distribution?
- When the time between arrivals is exponential, the number of arrivals in a fixed time interval is Poisson with the mean \(\lambda\)

Helpful R code

Let’s say we’re sitting at the bus stop, measuring the time until our bus arrives. We know the bus comes every 10 minutes on average.

If we want to know the probability that the bus arrives in the next 5 minutes:
```
pexp(q = 5, rate = 1/10)
```
```
[1] 0.3934693
```
If we want to know the time, say \(t\), where the probability of the bus arriving at \(t\) or earlier is 0.35:
```
qexp(p = 0.35, rate = 1/10)
```
```
[1] 4.307829
```
If we want to know the probability that the bus arrives between 3 and 5 minutes:
```
pexp(q = 5, rate = 1/10) - pexp(q = 3, rate = 1/10)
```
```
[1] 0.1342876
```

If we want to sample 20 bus arrival times from the distribution:

rexp(n = 20, rate = 1/10)

 [1] 11.3663960 18.3833292 11.9529412  8.0393384  2.9907828  1.8604949
 [7] 16.9187460  9.8558311  0.1888765 10.6110339  7.0503523  9.8269015
[13]  3.4872349 11.3613407 11.9240648 19.0331605 14.0223053  2.1621659
[19] 20.1604482  0.2328047

Transformation of independent exponential RVs

Revisit after joint notes:

Example 1

Let \(X_i \sim \textrm{Exp}(\lambda_i)\) be independent RVs, for \(i=1 \ldots n\). Find the pdf for the first of the arrival times.

Chapter 33: Gamma RVs

Properties of gamma RVs

Scenario: Modeling the time until the \(r^{th}\) event.
Continuous analog to the Negative Binomial distribution
Shorthand: \(X \sim \text{Gamma}(r, \lambda)\)

\[ f_X(x) = \dfrac{\lambda^r}{\Gamma(r)}x^{r-1} e^{-\lambda x}\text{ for } x>0, \lambda>0, \Gamma(r) = (r-1)! \]

\[ F_X(x) = \left\{ \begin{array}{ll} 0 & \quad x<0 \quad \\ 1 - e^{-\lambda x}\displaystyle\sum_{j=0}^{r-1}\dfrac{(\lambda x)^j}{j!} & \quad x\geq0 \\ \end{array} \right. \]

\[\text{E}(X) = \dfrac{r}{\lambda}\text{, }\text{ Var}(X) = \dfrac{r}{\lambda^2}\]

Common to see \(\alpha = r\) and \(\beta = \lambda\)

Identifying gamma RV from word problems

Gamma distribution with \(r=1\) is same as exponential
- Just like Negative Binomial with \(r=1\) is same as the geometric distribution
Similar to exponential
- Look for time between or until events/successes
  - BUT now we are measuring time until more than 1 success
- Look for a rate of the events over time period

Helpful R code

Let’s say we’re sitting at the bus stop, measuring the time until 4 buses arrive. We know the bus comes every 10 minutes on average.

If we want to know the probability that the 4 buses arrive in the next 50 minutes:

pgamma(q = 50, rate = 1/10, shape = 4)

[1] 0.7349741

pgamma(q = 50, scale = 10, shape = 4)

[1] 0.7349741

If we want to know the time, say \(t\), where the probability of the 4 buses arriving at \(t\) or earlier is 0.35:
```
qgamma(p = 0.35, rate = 1/10, shape = 4)
```
```
[1] 29.87645
```

If we want to know the probability that the 4 buses arrives between 30 and 50 minutes:

pgamma(q = 50, scale = 10, shape = 4) - pgamma(q = 30, scale = 10, shape = 4)

[1] 0.382206

If we want to sample 20 arrival times for the 4 buses:

rgamma(n = 20, scale = 10, shape = 4)

 [1] 24.17767 15.75058 16.75578 60.37217 36.67145 29.17471 61.12207 33.22881
 [9] 25.03580 39.81007 60.30513 45.66314 26.98178 56.14269 21.37449 36.87710
[17] 44.48559 25.77701 90.81206 14.69105

Remarks

The parameter \(r\) in a Gamma(\(r\),\(\lambda\)) distribution does NOT need to be a positive integer
- \(r\) is usually a positive integer
When \(r\) is a positive integer, the distribution is sometimes called an Erlang(\(r\),\(\lambda\)) distribution
When \(r\) is any positive real number, we have a general gamma distribution that is usually instead parameterized by \(\alpha>0\) and \(\beta>0\), where:
- \(\alpha = \text{shape parameter}\) : same as \(k\), the total number of events we must witness
  - In R code example: 4 buses to wait for
- \(\beta = \text{scale parameter}\) : same as \(\lambda\), the rate parameter
  - In R code example: 1 bus per 10 minutes (1/10)

Sending money orders

Example 1

On average, someone sends a money order once per 15 minutes. What is the probability someone sends 10 money orders in less than 3 hours?

Additional Resource

Another helpful site with R code: https://rpubs.com/mpfoley73/459051

Chapter 35: Normal RVs

Properties of Normal RVs

No scenario description here because the Normal distribution is so universal
- Central Limit Theorem (next class) makes it applicable to many types of events
Shorthand: \(X \sim \text{Normal}(\mu, \sigma^2)\)

\[ f_X(x) = \dfrac{1}{\sqrt{2\pi \sigma^2}}e^{-(x-\mu)^2/(2\sigma^2)} \text{, for} -inf < x < inf \]

\[\text{E}(X) = \mu \] \[\text{Var}(X) = \sigma^2\]

Helpful R code

Let’s say we’re measuring the high temperature today. The average high temperature on this day across many, many years is 50 degrees with a standard deviation of 4 degrees.

If we want to know the probability that the high temperature is below 45 degrees:
```
pnorm(q = 45, mean = 50, sd = 4)
```
```
[1] 0.1056498
```
If we want to know the temoerature, say \(t\), where the probability of that the temperature is at \(t\) or lower is 0.35:
```
qnorm(p = 0.35, mean = 50, sd = 4)
```
```
[1] 48.45872
```

If we want to know the probability that the temperature is between 45 and 50 degrees:

pnorm(q = 50, mean = 50, sd = 4) - pnorm(q = 45, mean = 50, sd = 4)

[1] 0.3943502

If we want to sample 20 days’ temperature (over the years) from the distribution:

rnorm(n = 20, mean = 50, sd = 4)

 [1] 44.33876 50.08607 50.70593 55.97296 49.52620 51.15660 51.42931 53.65328
 [9] 50.27942 40.35001 49.48731 64.33081 50.43379 54.01803 54.65349 45.49719
[17] 52.28530 47.64023 52.54549 50.48006

Movie night while studying

Example 1

Children’s movies run an average of 98 minutes with a standard deviation of 10 minutes. You check out a random movie from the library to entertain your kids so you can study for your test. Assume that your kids will be occupied for the entire length of the movie.

What is the probability that your kids will be occupied for at least the 2 hours you would like to study?
What is range for the bottom quartile (lowest 25%) of time they will be occupied?

Standard Normal Distribution

\[ Z \sim \text{Normal}(\mu = 0, \sigma^2 = 1)\]

Used to be more helpful when computing was not as advanced
- Use tables of the standard normal
- You can convert any normal distribution to a standard normal through transformation
\(Z = \dfrac{X - \mu_X}{\sigma_X}\)
- Comes from \(X = \sigma_X Z + \mu_X\)
- Since \(\sigma_X\) and \(\mu_X\) are constants, then \(E(X) = \mu_X\) and \(SD(X) = \sigma_X SD(Z) = \sigma_X\)