Names of Distributions

R has an abbreviated name for every probability distribution. This name is used to identify the functions associated with the distribution. For example, the name of the Normal distribution is “norm”, which is the root of these function names:

Table 8-1 describes some common discrete distributions, and Table 8-2 describes several common continuous distributions.

Getting Help on Probability Distributions

To see the R functions related to a particular probability distribution, use the help command and the full name of the distribution. For example, this will show the functions related to the Normal distribution:

?Normal

Some distributions have names that don’t work well with the help command, such as “Student’s t”. They have special help names, as noted in Tables Table 8-1 and Table 8-2: NegBinomial, Chisquare, Lognormal, and TDist. Thus, to get help on the Student’s t distribution, use this:

?TDist

See Also

There are many other distributions implemented in downloadable packages; see the CRAN task view devoted to probability distributions. The SuppDists package is part of the R base, and it includes ten supplemental distributions. The MASS package, which is also part of the base, provides additional support for distributions, such as maximum-likelihood fitting for some common distributions as well as sampling from a multivariate Normal distribution.

Problem

You want to calculate the number of combinations of n items taken k at a time.

Solution

Use the choose function:

n <- 10
k <- 2
choose(n, k)
#> [1] 45

Discussion

A common problem in computing probabilities of discrete variables is counting combinations: the number of distinct subsets of size k that can be created from n items. The number is given by n!/r!(n − r)!, but it’s much more convenient to use the choose function—especially as n and k grow larger:

choose(5, 3)   # How many ways can we select 3 items from 5 items?
#> [1] 10
choose(50, 3)  # How many ways can we select 3 items from 50 items?
#> [1] 19600
choose(50, 30) # How many ways can we select 30 items from 50 items?
#> [1] 4.71e+13

These numbers are also known as binomial coefficients.

See Also

This recipe merely counts the combinations; see “Generating Combinations” to actually generate them.

Problem

You want to generate all combinations of n items taken k at a time.

Solution

Use the combn function:

items <- 2:5
k <- 2
combn(items, k)
#>      [,1] [,2] [,3] [,4] [,5] [,6]
#> [1,]    2    2    2    3    3    4
#> [2,]    3    4    5    4    5    5

Discussion

We can use combn(1:5,3) to generate all combinations of the numbers 1 through 5 taken three at a time:

combn(1:5, 3)
#>      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
#> [1,]    1    1    1    1    1    1    2    2    2     3
#> [2,]    2    2    2    3    3    4    3    3    4     4
#> [3,]    3    4    5    4    5    5    4    5    5     5

The function is not restricted to numbers. We can generate combinations of strings, too. Here are all combinations of five treatments taken three at a time:

combn(c("T1", "T2", "T3", "T4", "T5"), 3)
#>      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
#> [1,] "T1" "T1" "T1" "T1" "T1" "T1" "T2" "T2" "T2" "T3"
#> [2,] "T2" "T2" "T2" "T3" "T3" "T4" "T3" "T3" "T4" "T4"
#> [3,] "T3" "T4" "T5" "T4" "T5" "T5" "T4" "T5" "T5" "T5"

See Also

See “Counting the Number of Combinations” to count the number of possible combinations before you generate a huge set.

Problem

You want to generate random numbers.

Solution

The simple case of generating a uniform random number between 0 and 1 is handled by the runif function. This example generates one uniform random number:

runif(1)
#> [1] 0.915

R can generate random variates from other distributions as well. For a given distribution, the name of the random number generator is “r” prefixed to the distribution’s abbreviated name (e.g., rnorm for the Normal distribution’s random number generator). This example generates one random value from the standard normal distribution:

rnorm(1)
#> [1] 1.53

Discussion

Most programming languages have a wimpy random number generator that generates one random number, uniformly distributed between 0.0 and 1.0, and that’s all. Not R.

R can generate random numbers from many probability distributions other than the uniform distribution. The simple case of generating uniform random numbers between 0 and 1 is handled by the runif function:

runif(1)
#> [1] 0.83

The argument of runif is the number of random values to be generated. Generating a vector of 10 such values is as easy as generating one:

runif(10)
#>  [1] 0.642 0.519 0.737 0.135 0.657 0.705 0.458 0.719 0.935 0.255

There are random number generators for all built-in distributions. Simply prefix the distribution name with “r” and you have the name of the corresponding random number generator. Here are some common ones:

set.seed(42)
runif(1, min = -3, max = 3)      # One uniform variate between -3 and +3
#> [1] 2.49
rnorm(1)                         # One standard Normal variate
#> [1] 1.53
rnorm(1, mean = 100, sd = 15)    # One Normal variate, mean 100 and SD 15
#> [1] 114
rbinom(1, size = 10, prob = 0.5) # One binomial variate
#> [1] 5
rpois(1, lambda = 10)            # One Poisson variate
#> [1] 12
rexp(1, rate = 0.1)              # One exponential variate
#> [1] 3.14
rgamma(1, shape = 2, rate = 0.1) # One gamma variate
#> [1] 22.3

As with runif, the first argument is the number of random values to be generated. Subsequent arguments are the parameters of the distribution, such as mean and sd for the Normal distribution or size and prob for the binomial. See the function’s R help page for details.

The examples given so far use simple scalars for distributional parameters. Yet the parameters can also be vectors, in which case R will cycle through the vector while generating random values. The following example generates three normal random values drawn from distributions with means of −10, 0, and +10, respectively (all distributions have a standard deviation of 1.0):

rnorm(3, mean = c(-10, 0, +10), sd = 1)
#> [1] -9.420 -0.658 11.555

That is a powerful capability in such cases as hierarchical models, where the parameters are themselves random. The next example calculates 30 draws of a normal variate whose mean is itself randomly distributed and with hyperparameters of μ = 0 and σ = 0.2:

means <- rnorm(30, mean = 0, sd = 0.2)
rnorm(30, mean = means, sd = 1)
#>  [1] -0.5549 -2.9232 -1.2203  0.6962  0.1673 -1.0779 -0.3138 -3.3165
#>  [9]  1.5952  0.8184 -0.1251  0.3601 -0.8142  0.1050  2.1264  0.6943
#> [17] -2.7771  0.9026  0.0389  0.2280 -0.5599  0.9572  0.1972  0.2602
#> [25] -0.4423  1.9707  0.4553  0.0467  1.5229  0.3176

If you are generating many random values and the vector of parameters is too short, R will apply the Recycling Rule to the parameter vector.

See Also

See the “Introduction” to this chapter.

Problem

You want to generate a sequence of random numbers, but you want to reproduce the same sequence every time your program runs.

Solution

Before running your R code, call the set.seed function to initialize the random number generator to a known state:

set.seed(42) # Or use any other positive integer...

Discussion

After generating random numbers, you may often want to reproduce the same sequence of “random” numbers every time your program executes. That way, you get the same results from run to run. One of the authors (Paul) once supported a complicated Monte Carlo analysis of a huge portfolio of securities. The users complained about getting slightly different results each time the program ran. No kidding! The analysis was driven entirely by random numbers, so of course there was randomness in the output. The solution was to set the random number generator to a known state at the beginning of the program. That way, it would generate the same (quasi-)random numbers each time and thus yield consistent, reproducible results.

In R, the set.seed function sets the random number generator to a known state. The function takes one argument, an integer. Any positive integer will work, but you must use the same one in order to get the same initial state.

The function returns nothing. It works behind the scenes, initializing (or reinitializing) the random number generator. The key here is that using the same seed restarts the random number generator back at the same place:

set.seed(165)   # Initialize generator to known state
runif(10)       # Generate ten random numbers
#>  [1] 0.116 0.450 0.996 0.611 0.616 0.426 0.666 0.168 0.788 0.442

set.seed(165)   # Reinitialize to the same known state
runif(10)       # Generate the same ten "random" numbers
#>  [1] 0.116 0.450 0.996 0.611 0.616 0.426 0.666 0.168 0.788 0.442

See Also

See “Generating Random Numbers” for more about generating random numbers.

Problem

You want to sample a dataset randomly.

Solution

The sample function will randomly select n items from a set:

sample(set, n)

Discussion

Suppose your World Series data contains a vector of years when the Series was played. You can select 10 years at random using sample:

world_series <- read_csv("./data/world_series.csv")
sample(world_series$year, 10)
#>  [1] 2010 1961 1906 1992 1982 1948 1910 1973 1967 1931

The items are randomly selected, so running sample again (usually) produces a different result:

sample(world_series$year, 10)
#>  [1] 1941 1973 1921 1958 1979 1946 1932 1919 1971 1974

The sample function normally samples without replacement, meaning it will not select the same item twice. Some statistical procedures (especially the bootstrap) require sampling with replacement, which means that one item can appear multiple times in the sample. Specify replace=TRUE to sample with replacement.

It’s easy to implement a simple bootstrap using sampling with replacement. Suppose we have a vector, x, of 1,000 random numbers, drawn from a normal distribution with mean 4 and standard deviation 10.

set.seed(42)
x <- rnorm(1000, 4, 10)

This code fragment samples 1,000 times from x and calculates the median of each sample:

medians <- numeric(1000)   # empty vector of 1000 numbers
for (i in 1:1000) {
  medians[i] <- median(sample(x, replace = TRUE))
}

From the bootstrap estimates, we can estimate the confidence interval for the median:

ci <- quantile(medians, c(0.025, 0.975))
cat("95% confidence interval is (", ci, ")\n")
#> 95% confidence interval is ( 3.16 4.49 )

We know that x was created from a normal distribution with a mean of 4 and, hence, the sample median should be 4 also. (In a symetrical distribution like the normal, the mean and the median are the same.) Our confidence interval easily contains the value.

See Also

See “Randomly Permuting a Vector” for randomly permuting a vector and Recipe X-X for more about bootstrapping. “Generating Reproducible Random Numbers” discusses setting seeds for quasi-random numbers.

Problem

You want to generate a random sequence, such as a series of simulated coin tosses or a simulated sequence of Bernoulli trials.

Solution

Use the sample function. Sample n draws from the set of possible values, and set replace=TRUE:

sample(set, n, replace = TRUE)

Discussion

The sample function randomly selects items from a set. It normally samples without replacement, which means that it will not select the same item twice and will return an error if you try to sample more items than exist in the set. With replace=TRUE, however, sample can select items over and over; this allows you to generate long, random sequences of items.

The following example generates a random sequence of 10 simulated flips of a coin:

sample(c("H", "T"), 10, replace = TRUE)
#>  [1] "H" "T" "H" "T" "T" "T" "H" "T" "T" "H"

The next example generates a sequence of 20 Bernoulli trials—random successes or failures. We use TRUE to signify a success:

sample(c(FALSE, TRUE), 20, replace = TRUE)
#>  [1]  TRUE FALSE  TRUE  TRUE FALSE  TRUE FALSE FALSE  TRUE  TRUE FALSE
#> [12]  TRUE  TRUE FALSE  TRUE  TRUE FALSE FALSE FALSE FALSE

By default, sample will choose equally among the set elements and so the probability of selecting either TRUE or FALSE is 0.5. With a Bernoulli trial, the probability p of success is not necessarily 0.5. You can bias the sample by using the prob argument of sample; this argument is a vector of probabilities, one for each set element. Suppose we want to generate 20 Bernoulli trials with a probability of success p = 0.8. We set the probability of FALSE to be 0.2 and the probability of TRUE to 0.8:

sample(c(FALSE, TRUE), 20, replace = TRUE, prob = c(0.2, 0.8))
#>  [1]  TRUE  TRUE FALSE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE
#> [12]  TRUE  TRUE  TRUE  TRUE  TRUE FALSE FALSE  TRUE  TRUE

The resulting sequence is clearly biased toward TRUE. I chose this example because it’s a simple demonstration of a general technique. For the special case of a binary-valued sequence you can use rbinom, the random generator for binomial variates:

rbinom(10, 1, 0.8)
#>  [1] 1 0 1 1 1 1 1 0 1 1

Problem

You want to generate a random permutation of a vector.

Solution

If v is your vector, then sample(v) returns a random permutation.

Discussion

We typically think of the sample function for sampling from large datasets. However, the default parameters enable you to create a random rearrangement of the dataset. The function call sample(v) is equivalent to:

sample(v, size = length(v), replace = FALSE)

which means “select all the elements of v in random order while using each element exactly once.” That is a random permutation. Here is a random permutation of 1, …, 10:

sample(1:10)
#>  [1]  7  3  6  1  5  2  4  8 10  9

See Also

See “Generating a Random Sample” for more about sample.

Problem

You want to calculate either the simple or the cumulative probability associated with a discrete random variable.

Solution

For a simple probability, P(X = x), use the density function. All built-in probability distributions have a density function whose name is “d” prefixed to the distribution name. For example, dbinom for the binomial distribution.

For a cumulative probability, P(X ≤ x), use the distribution function. All built-in probability distributions have a distribution function whose name is “p” prefixed to the distribution name; thus, pbinom is the distribution function for the binomial distribution.

Discussion

Suppose we have a binomial random variable X over 10 trials, where each trial has a success probability of 1/2. Then we can calculate the probability of observing x = 7 by calling dbinom:

dbinom(7, size = 10, prob = 0.5)
#> [1] 0.117

That calculates a probability of about 0.117. R calls dbinom the density function. Some textbooks call it the probability mass function or the probability function. Calling it a density function keeps the terminology consistent between discrete and continuous distributions (“Calculating Probabilities for Continuous Distributions”).

The cumulative probability, P(X ≤ x), is given by the distribution function, which is sometimes called the cumulative probability function. The distribution function for the binomial distribution is pbinom. Here is the cumulative probability for x = 7 (i.e., P(X ≤ 7)):

pbinom(7, size = 10, prob = 0.5)
#> [1] 0.945

It appears the probability of observing X ≤ 7 is about 0.945.

The density functions and distribution functions for some common discrete distributions are shown in Table @ref(tab:distributions).

The complement of the cumulative probability is the survival function, P(X > x). All of the distribution functions let you find this right-tail probability simply by specifying lower.tail=FALSE:

pbinom(7, size = 10, prob = 0.5, lower.tail = FALSE)
#> [1] 0.0547

Thus we see that the probability of observing X > 7 is about 0.055.

The interval probability, P(x₁ < X ≤ x₂), is the probability of observing X between the limits x₁ and x₂. It is calculated as the difference between two cumulative probabilities: P(X ≤ x₂) − P(X ≤ x₁). Here is P(3 < X ≤ 7) for our binomial variable:

pbinom(7, size = 10, prob = 0.5) - pbinom(3, size = 10, prob = 0.5)
#> [1] 0.773

R lets you specify multiple values of x for these functions and will return a vector of the corresponding probabilities. Here we calculate two cumulative probabilities, P(X ≤ 3) and P(X ≤ 7), in one call to pbinom:

pbinom(c(3, 7), size = 10, prob = 0.5)
#> [1] 0.172 0.945

This leads to a one-liner for calculating interval probabilities. The diff function calculates the difference between successive elements of a vector. We apply it to the output of pbinom to obtain the difference in cumulative probabilities—in other words, the interval probability:

diff(pbinom(c(3, 7), size = 10, prob = 0.5))
#> [1] 0.773

See Also

See this chapter’s “Introduction” for more about the built-in probability distributions.

Problem

You want to calculate the distribution function (DF) or cumulative distribution function (CDF) for a continuous random variable.

Solution

Use the distribution function, which calculates P(X ≤ x). All built-in probability distributions have a distribution function whose name is “p” prefixed to the distribution’s abbreviated name—for instance, pnorm for the Normal distribution.

Example: what’s the probability of a draw being below .8 for a draw from a random standard normal distribution?

pnorm(q = .8, mean = 0, sd = 1)
#> [1] 0.788

Discussion

The R functions for probability distributions follow a consistent pattern, so the solution to this recipe is essentially identical to the solution for discrete random variables (“Calculating Probabilities for Discrete Distributions”). The significant difference is that continuous variables have no “probability” at a single point, P(X = x). Instead, they have a density at a point.

Given that consistency, the discussion of distribution functions in “Calculating Probabilities for Discrete Distributions” is applicable here, too. Table @ref(tab:continuous) gives the distribution functions for several continuous distributions.

We can use pnorm to calculate the probability that a man is shorter than 66 inches, assuming that men’s heights are normally distributed with a mean of 70 inches and a standard deviation of 3 inches. Mathematically speaking, we want P(X ≤ 66) given that X ~ N(70, 3):

pnorm(66, mean = 70, sd = 3)
#> [1] 0.0912

Likewise, we can use pexp to calculate the probability that an exponential variable with a mean of 40 could be less than 20:

pexp(20, rate = 1 / 40)
#> [1] 0.393

Just as for discrete probabilities, the functions for continuous probabilities use lower.tail=FALSE to specify the survival function, P(X > x). This call to pexp gives the probability that the same exponential variable could be greater than 50:

pexp(50, rate = 1 / 40, lower.tail = FALSE)
#> [1] 0.287

Also like discrete probabilities, the interval probability for a continuous variable, P(x₁ < X < x₂), is computed as the difference between two cumulative probabilities, P(X < x₂) − P(X < x₁). For the same exponential variable, here is P(20 < X < 50), the probability that it could fall between 20 and 50:

pexp(50, rate = 1 / 40) - pexp(20, rate = 1 / 40)
#> [1] 0.32

See Also

See this chapter’s “Introduction” for more about the built-in probability distributions.

Problem

Given a probability p and a distribution, you want to determine the corresponding quantile for p: the value x such that P(X ≤ x) = p.

Solution

Every built-in distribution includes a quantile function that converts probabilities to quantiles. The function’s name is “q” prefixed to the distribution name; thus, for instance, qnorm is the quantile function for the Normal distribution.

The first argument of the quantile function is the probability. The remaining arguments are the distribution’s parameters, such as mean, shape, or rate:

qnorm(0.05, mean = 100, sd = 15)
#> [1] 75.3

Discussion

A common example of computing quantiles is when we compute the limits of a confidence interval. If we want to know the 95% confidence interval (α = 0.05) of a standard normal variable, then we need the quantiles with probabilities of α/2 = 0.025 and (1 − α)/2 = 0.975:

qnorm(0.025)
#> [1] -1.96
qnorm(0.975)
#> [1] 1.96

In the true spirit of R, the first argument of the quantile functions can be a vector of probabilities, in which case we get a vector of quantiles. We can simplify this example into a one-liner:

qnorm(c(0.025, 0.975))
#> [1] -1.96  1.96

All the built-in probability distributions provide a quantile function. Table @ref(tab:discrete-quant-dist) shows the quantile functions for some common discrete distributions.

Table @ref(tab:cont-quant-dist) shows the quantile functions for common continuous distributions.

See Also

Determining the quantiles of a data set is different from determining the quantiles of a distribution—see “Calculating Quantiles (and Quartiles) of a Dataset”.

Problem

You want to plot the density function of a probability distribution.

Solution

Define a vector x over the domain. Apply the distribution’s density function to x and then plot the result. If x is a vector of points over the domain we care about plotting, we then calculate the density using one of the d_____ density functions like dlnorm for lognormal or dnorm for normal.

dens <- data.frame(x = x,
                   y = d_____(x))
ggplot(dens, aes(x, y)) + geom_line()

Here is a specific example that plots the standard normal distribution for the interval -3 to +3:

library(ggplot2)

x <- seq(-3, +3, 0.1)
dens <- data.frame(x = x, y = dnorm(x))

ggplot(dens, aes(x, y)) + geom_line()

Figure 8-1. Standard Normal

Figure 8-1 shows the smooth density function.

Discussion

All the built-in probability distributions include a density function. For a particular density, the function name is “d” prepended to the density name. The density function for the Normal distribution is dnorm, the density for the gamma distribution is dgamma, and so forth.

If the first argument of the density function is a vector, then the function calculates the density at each point and returns the vector of densities.

The following code creates a 2 × 2 plot of four densities:

x <- seq(from = 0, to = 6, length.out = 100) # Define the density domains
ylim <- c(0, 0.6)

# Make a data.frame with densities of several distributions
df <- rbind(
  data.frame(x = x, dist_name = "Uniform"    , y = dunif(x, min   = 2, max = 4)),
  data.frame(x = x, dist_name = "Normal"     , y = dnorm(x, mean  = 3, sd = 1)),
  data.frame(x = x, dist_name = "Exponential", y = dexp(x, rate  = 1 / 2)),
  data.frame(x = x, dist_name = "Gamma"      , y = dgamma(x, shape = 2, rate = 1)) )

# Make a line plot like before, but use facet_wrap to create the grid
ggplot(data = df, aes(x = x, y = y)) +
  geom_line() +
  facet_wrap(~dist_name)   # facet and wrap by the variable dist_name

Figure 8-2. Multiple Density Plots

Figure 8-2 shows four density plots. However, a raw density plot is rarely useful or interesting by itself, and we often shade a region of interest.

Figure 8-3. Standard Normal with Shading

Figure 8-3 is a normal distribution with shading from the 75^th percentile to the 95^th percentile.

We create the plot by first plotting the density and then creating a shaded region with the geom_ribbon function from ggplot2.

First, we create some data and draw the density curve shown in Figure 8-4

x <- seq(from = -3, to = 3, length.out = 100)
df <- data.frame(x = x, y = dnorm(x, mean = 0, sd = 1))

p <- ggplot(df, aes(x, y)) +
  geom_line() +
  labs(
    title = "Standard Normal Distribution",
    y = "Density",
    x = "Quantile"
  )
p

Figure 8-4. Density Plot

Next, we define the region of interest by calculating the x value for the quantiles we’re interested in. Then finally we add a geom_ribbon to add a subset of our original data as a colored region. The resulting plot is shown here:

q75 <- quantile(df$x, .75)
q95 <- quantile(df$x, .95)

p +
  geom_ribbon(
    data = subset(df, x > q75 & x < q95),
    aes(ymax = y),
    ymin = 0,
    fill = "blue",
    colour = NA,
    alpha = 0.5
  )