## 1.2 Discrete Distributions

These notes rely heavily on PSU STATS 504 course notes.

The most important discrete distributions are the Binomial, Poisson, and Multinomial. Sometimes useful are the related Bernoulli, negative binomial, geometric, and hypergeometric distributions.

A discrete random variable $$X$$ is described by its probability mass function $$f(x) = P(X = x)$$. The set of $$x$$ values for which $$f(x) > 0$$ is called the support. If the distribution depends on unknown parameter(s) $$\theta$$ we write it as $$f(x; \theta)$$ (frequentist) or $$f(x | \theta)$$ (Bayesian).

### 1.2.1 Bernoulli

If $$X$$ is the result of a trial with two outcomes of probability $$P(X = 1) = \pi$$ and $$P(X = 0) = 1 - \pi$$, then $$X$$ is a random variable with a Bernoulli distribution

$f(x) = \pi^x (1 - \pi)^{1 - x}, \hspace{1cm} x \in (0, 1)$

with $$E(X) = \pi$$ and $$Var(X) = \pi(1 - \pi)$$.

### 1.2.2 Binomial

If $$X$$ is the count of successful events in $$n$$ identical and independent Bernoulli trials of success probability $$\pi$$, then $$X$$ is a random variable with a binomial distribution $$X \sim Bin(n,\pi)$$

$f(x;n, \pi) = \frac{n!}{x!(n-x)!} \pi^x (1-\pi)^{n-x} \hspace{1cm} x \in (0, 1, ..., n), \hspace{2mm} \pi \in [0, 1]$

with $$E(X)=n\pi$$ and $$Var(X) = n\pi(1-\pi)$$.

Binomial sampling is used to model counts of one level of a categorical variable over a fixed sample size. Here is a simple analysis of data from a Binomial process. Data set dat contains frequencies of high-risk drinkers vs non-high-risk drinkers in a college survey.

##
##  No Yes
## 685 630

The MLE of $$\pi$$ from the Binomial distribution is the sample mean.

x <- sum(dathigh_risk == "Yes") n <- nrow(dat) p <- x / n print(p) ##  0.4790875 Here is the binomial distribution $$f(x; \pi), \hspace{5mm} x \in [550, 700]$$. events <- round(seq(from = 550, to = 700, length = 20), 0) density <- dbinom(x = events, prob = p, size = n) prob <- pbinom(q = events, prob = p, size = n, lower.tail = TRUE) df <- data.frame(events, density, prob) ggplot(df, aes(x = factor(events))) + # geom_col(aes(y = density)) + geom_col(aes(y = density), fill = mf_pal()(1), alpha = 0.8) + geom_text( aes(label = round(density, 3), y = density + 0.001), position = position_dodge(0.9), size = 3, vjust = 0 ) + geom_line( data = df, aes(x = as.numeric(factor(events)), y = prob/40), color = mf_pal()(1), size = 1) + scale_y_continuous(sec.axis = sec_axis(~.*40, name = "Cum Prob")) + theme_mf() + labs(title = "PMF and CDF of Binomial Distribution", subtitle = "Bin(1315, 0.479).", x = "Events (x)", y = "Density") There are several ways to calculate a confidence interval for $$\pi$$. One method is the normal approximation (Wald) interval. $\pi = p \pm z_{\alpha /2} \sqrt{\frac{p (1 - p)}{n}}$ alpha <- .05 z <- qnorm(1 - alpha / 2) se <- sqrt(p * (1 - p) / n) p + c(-z*se, z*se) ##  0.4520868 0.5060882 This method is easy to understand and calculate by hand, but its accuracy suffers when $$np<5$$ or $$n(1-p)<5$$ and it does not work at all when $$p = 0$$ or $$p = 1$$. Option two is the Wilson method. $\frac{p + \frac{z^2}{2n}}{1 + \frac{z^2}{n}} \pm \frac{z}{1 + \frac{z^2}{n}} \sqrt{\frac{p(1 - p)}{n} + \frac{z^2}{4n^2}}$ est <- (p + (z^2)/(2*n)) / (1 + (z^2) / n) pm <- z / (1 + (z^2)/n) * sqrt(p*(1-p)/n + (z^2) / (4*(n^2))) est + c(-pm, pm) ##  0.4521869 0.5061098 This is what prop.test() does when you set correct = FALSE. prop.test(x = x, n = n, correct = FALSE) ## ## 1-sample proportions test without continuity correction ## ## data: x out of n, null probability 0.5 ## X-squared = 2.3004, df = 1, p-value = 0.1293 ## alternative hypothesis: true p is not equal to 0.5 ## 95 percent confidence interval: ## 0.4521869 0.5061098 ## sample estimates: ## p ## 0.4790875 There is a second version of the Wilson interval that applies a “continuity correction” that aligns the “minimum coverage probability”, rather than the “average probability”, with the nominal value. I’ll need to learn what’s inside those quotations at some point. prop.test(x = x, n = n) ## ## 1-sample proportions test with continuity correction ## ## data: x out of n, null probability 0.5 ## X-squared = 2.2175, df = 1, p-value = 0.1365 ## alternative hypothesis: true p is not equal to 0.5 ## 95 percent confidence interval: ## 0.4518087 0.5064898 ## sample estimates: ## p ## 0.4790875 Finally, there is the Clopper-Pearson exact confidence interval. Clopper-Pearson inverts two single-tailed binomial tests at the desired alpha. This is a non-trivial calculation, so there is no easy formula to crank through. Just use the binom.test() function and pray no one asks for an explanation. binom.test(x = x, n = n) ## ## Exact binomial test ## ## data: x and n ## number of successes = 630, number of trials = 1315, p-value = 0.1364 ## alternative hypothesis: true probability of success is not equal to 0.5 ## 95 percent confidence interval: ## 0.4517790 0.5064896 ## sample estimates: ## probability of success ## 0.4790875 The expected probability of no one being a high-risk drinker is $$f(0;0.479) = \frac{1315!}{0!(1315-0)!} 0.479^0 (1-0.479)^{1315-0} = 0$$. dbinom(x = 0, size = n, p = p) ##  0 The expected probability of half the population being a high-risk drinker, $$f(658, 0.479)$$, is impossible to write out, and slow to calculate. pbinom(q = .5*n, size = n, prob = p, lower.tail = FALSE) ##  0.06455096 As n increases for fixed $$\pi$$, the binomial distribution approaches normal distribution $$N(n\pi, n\pi(1−\pi))$$. The normal distribution is a good approximation when $$n$$ is large. pnorm(q = 0.5, mean = p, sd = se, lower.tail = FALSE) ##  0.06450357 Here are some more examples using smaller sample sizes. The probability 2 out of 10 coin flips are heads if the probability of heads is 0.3: dbinom(x = 2, size = 10, prob = 0.3) ##  0.2334744 Here is a simulation from n = 10,000 random samples of size 10. rbinom() generates a random sample of numbers from the binomial distribution. data.frame(cnt = rbinom(n = 10000, size = 10, prob = 0.3)) %>% count(cnt) %>% ungroup() %>% mutate(pct = n / sum(n), X_eq_x = cnt == 2) %>% ggplot(aes(x = as.factor(cnt), y = n, fill = X_eq_x, label = pct)) + geom_col(alpha = 0.8) + scale_fill_mf() + geom_label(aes(label = round(pct, 2)), size = 3, alpha = .6) + theme_mf() + theme(legend.position = "none") + labs(title = "Binomial Distribution", subtitle = paste0( "P(X=2) successes in 10 trials when p = 0.3 is ", round(dbinom(2, 10, 0.3), 4), "." ), x = "Successes", y = "Count", caption = "Simulation from n = 10,000 binomial random samples.") What is the probability of <=2 heads in 10 coin flips where probability of heads is 0.3? The cumulative probability is the sum of the first three bars in the simulation above. Function pbinom() calculates the cumulative binomial probability. pbinom(q = 2, size = 10, prob = 0.3, lower.tail = TRUE) ##  0.3827828 What is the expected number of heads in 25 coin flips if the probability of heads is 0.3? The expected value, $$\mu = np$$, is 7.5. Here’s an empirical test from 10,000 samples. mean(rbinom(n = 10000, size = 25, prob = .3)) ##  7.5191 The variance, $$\sigma^2 = np (1 - p)$$, is 5.25. Here’s an empirical test. var(rbinom(n = 10000, size = 25, prob = .3)) ##  5.248123 Suppose X and Y are independent random variables distributed $$X \sim Bin(10, .6)$$ and $$Y \sim Bin(10, .7)$$. What is the probability that either variable is <=4? Let $$P(A) = P(X<=4)$$ and $$P(B) = P(Y<=4)$$. Then $$P(A|B) = P(A) + P(B) - P(AB)$$, and because the events are independent, $$P(AB) = P(A)P(B)$$. p_a <- pbinom(q = 4, size = 10, prob = 0.6, lower.tail = TRUE) p_b <- pbinom(q = 4, size = 10, prob = 0.7, lower.tail = TRUE) p_a + p_b - (p_a * p_b) ##  0.2057164 Here’s an empirical test. df <- data.frame( x = rbinom(10000, 10, 0.6), y = rbinom(10000, 10, 0.7) ) mean(if_else(dfx <= 4 | df$y <= 4, 1, 0)) ##  0.2116 A couple other points to remember: • The Bernoulli distribution is a special case of the binomial with $$n = 1$$. • The binomial distribution assumes independent trials. If you sample without replacement from a finite population, use the hypergeometric distribution. ### 1.2.3 Poission If $$X$$ is the number of successes in $$n$$ (many) trials when the probability of success $$\lambda / n$$ is small, then $$X$$ is a random variable with a Poisson distribution $$X \sim Poisson(\lambda)$$ $f(x;\lambda) = \frac{e^{-\lambda} \lambda^x}{x!} \hspace{1cm} x \in (0, 1, ...), \hspace{2mm} \lambda > 0$ with $$E(X)=\lambda$$ and $$Var(X) = \lambda$$. The Poisson likelihood function is $L(\lambda; x) = \prod_{i=1}^N f(x_i; \lambda) = \prod_{i=1}^N \frac{e^{-\lambda} \lambda^x_i}{x_i !} = \frac{e^{-n \lambda} \lambda^{\sum x_i}}{\prod x_i}.$ The Poisson loglikelihood function is $l(\lambda; x) = \sum_{i=1}^N x_i \log \lambda - n \lambda.$ One can show that the loglikelihood function is maximized at $\hat{\lambda} = \sum_{i=1}^N x_i / n.$ Thus, for a Poisson sample, the MLE for $$\lambda$$ is just the sample mean. Poisson sampling is used to model counts of events that occur randomly over a fixed period of time. Here is a simple analysis of data from a Poisson process. Data set dat contains frequencies of goal counts during the first round matches of the 2002 World Cup. ## goals freq ## 1 0 23 ## 2 1 37 ## 3 2 20 ## 4 3 11 ## 5 4 2 ## 6 5 1 ## 7 6 0 ## 8 7 0 ## 9 8 1 The MLE of $$\lambda$$ from the Poisson distribution is the sample mean. lambda <- weighted.mean(dat$goals, dat$freq) print(lambda) ##  1.378947 The 0.95 CI is $$\lambda \pm z_{.05/2} \sqrt{\lambda / n}$$ n <- sum(dat$freq)
z <- qnorm(0.975)
se <- sqrt(lambda / n)
paste0("[", round(lambda - z*se, 2), ", ", round(lambda + z*se, 2),"]")
##  "[1.14, 1.62]"

The expected probability of scoring 2 goals in a match is $$\frac{e^{-1.38} 1.38^2}{2!} = 0.239$$.

dpois(x = 2, lambda = lambda)
##  0.2394397
events <- 0:10
density <- dpois(x = events, lambda = 3)
prob <- ppois(q = events, lambda = 3, lower.tail = TRUE)
df <- data.frame(events, density, prob)
ggplot(df, aes(x = factor(events), y = density)) +
geom_col() +
geom_text(
aes(label = round(density, 3), y = density + 0.01),
position = position_dodge(0.9),
size = 3,
vjust = 0
) +
geom_line(
data = df,
aes(x = events, y = prob/4),
size = 1) +
scale_y_continuous(sec.axis = sec_axis(~.*4, name = "Cum Prob")) +
theme_mf() +
scale_fill_mf() +
labs(title = "PMF and CDF of Poisson Distribution",
subtitle = "Poisson(3).",
x = "Events (x)",
y = "Density") The expected probability of scoring 2 to 4 goals in a match is

sum(dpois(x = c(2:4), lambda = lambda))
##  0.3874391

Or, using the cumulative probability distribution,

ppois(q = 4, lambda = lambda) - ppois(q = 1, lambda = lambda)
##  0.3874391

How well does the Poisson distribution fit the 2002 World Cup data?

dat %>%
mutate(pred = n * dpois(x = goals, lambda = lambda)) %>%
rename(obs = freq) %>%
pivot_longer(cols = -goals) %>%
ggplot(aes(x = goals, y = value, color = name)) +
geom_point() +
theme_mf() +
scale_color_mf() +
geom_smooth(se = FALSE) +
labs(
title = "Poisson Dist: Observed vs Expected",
color = "",
y = "frequencey"
) It fits the data pretty good!

$$Poison(\lambda) \rightarrow Bin(n, \pi)$$ when $$n\pi = \lambda$$ and $$n \rightarrow \infty$$ and $$\pi \rightarrow 0$$. Because the Poisson is limit of the $$Bin(n, \pi)$$, it is useful as an approximation to the binomial when $$n$$ is large ($$n>=20$$) and $$\pi$$ small ($$p<=0.05$$).

For example, suppose a baseball player has a p=.03 chance of hitting a homerun. What is the probability of X>=20 homeruns in 500 at-bats? This is a binomial process because the sample size is fixed.

pbinom(q = 20, size = 500, prob = 0.03, lower.tail = FALSE)
##  0.07979678

But $$n$$ is large and $$\pi$$ is small, so the Poission distribution will work well too.

ppois(q = 20, lambda = 0.03 * 500, lower.tail = FALSE)
##  0.08297091

What is the distribution of successes from a sample of n = 50 when the probability of success is p = .03?

n = 500
p = 0.03
x = 0:30
data.frame(
events = x,
Poisson = dpois(x = x, lambda = p * n),
Binomial = dbinom(x = x, size = n, p = p)
) %>%
pivot_longer(cols = -events) %>%
ggplot(aes(x = events, y = value, color = name)) +
geom_point() +
theme_mf() +
scale_color_mf() +
labs(title = "Poisson(15) vs. Bin(500, .03)",
subtitle = "Poisson approximation to binomial.",
x = "Events",
y = "Density",
color = "") When the observed variance is greater than $$\lambda$$ (overdispersion), the Negative Binomial distribution can be used instead of Poisson.

Suppose the probability that a drug produces a certain side effect is p = = 0.1% and n = 1,000 patients in a clinical trial receive the drug. What is the probability 0 people experience the side effect?

The expected value is np, 1. The probability of measuring 0 when the expected value is 1 is dpois(x = 0, lambda = 1000 * .001) = 0.3678794. ### 1.2.4 Multinomial

If $$X = (X_1, X_2, \cdots, X_k)$$ are the counts of successful events in $$n$$ identical and independent trials of success probabilities $$\pi = (\pi_1, \pi_2, \cdots, \pi_k)$$, then $$X$$ is a random variable with a multinomial distribution $$X \sim Mult(n,\pi)$$

$f(x; n, \pi) = \frac{n!}{x_{1}! x_{2}! \cdots x_{k}!} \pi^{x_1} \pi^{x_2} \cdots \pi^{x_k} \hspace{1cm} x \in \{0, 1, ..., n \}, \hspace{2mm} \pi \in [0, 1]$

with expected values vector $$E(X_j) = n\pi_j$$ and covariance matrix

$Var(X) = \begin{bmatrix}n\pi_{1}(1-\pi_{1}) & -n\pi_{1}\pi_{2} & \cdots & -n\pi_{1}\pi_{k}\\ -n\pi_{1}\pi_{2} & n\pi_{2}(1-\pi_{2}) & \cdots & -n\pi_{2}\pi_{k}\\ \vdots & \vdots & \ddots & \vdots\\ -n\pi_{1}\pi_{k} & -n\pi_{2}\pi_{k} & \cdots & n\pi_{k}(1-\pi_{k}) \end{bmatrix}$

so $$Var(X_j) = n \pi_j (1 - \pi_j)$$ and $$cov(X_j, X_k) = -n \pi_j \pi_k$$.

The individual components of a multinomial random vector are binomial and have a binomial distribution, $$X_i = Bin(n, \pi_i)$$. Binomial is a special case of multinomial for k = 2.

Suppose a city population is 20% black, 15% Hispanic, and 65% other. From a random sample of $$n = 12$$ persons, what is the probability of 4 black and 8 other?

$f(x;\pi) = \frac{12!}{4! 0! 8!} (0.20)^4 (0.15)^0 (0.65)^8 = 0.0252$

Function dmultinom() calculates the multinomial probability.

dmultinom(x = c(4, 0, 8), prob = c(0.20, 0.15, 0.65))
##  0.025

To calculate the probability of <= 1 black, combine Hispanic and other, then sum the probability of black = 1 and black = 2.

$f(x;\pi) = \frac{12!}{0! 12!} (0.20)^0 (0.80)^{12} + \frac{12!}{1! 11!} (0.20)^1 (0.80)^{11} = 0.2748$

dmultinom(x = c(0, 12), prob = c(0.20, 0.80)) +
dmultinom(x = c(1, 11), prob = c(0.20, 0.80))
##  0.27

### 1.2.5 Negative-Binomial

If $$X$$ is the count of failure events ocurring prior to reaching $$r$$ successful events in a sequence of Bernouli trias of success probability $$p$$, then $$X$$ is a random variable with a negative-binomial distribution $$X \sim NB(r, p)$$. The probability of $$X = x$$ failures prior to $$r$$ successes is

$f(x;r, p) = {{x + r - 1} \choose {r - 1}} p^r (1-p)^{x}.$

with $$E(X) = r (1 - p) / p$$ and $$Var(X) = r (1-p) / p^2$$.

When the data has overdispersion, model the data with the negative-binomial distribution instead of Poission.

#### Examples

An oil company has a $$p = 0.20$$ chance of striking oil when drilling a well. What is the probability the company drills $$x + r = 7$$ wells to strike oil $$r = 3$$ times? Note that the question is formulated as counting total events, $$x + r = 7$$, so translate it to total failed events, $$x = 4$$.

$f(x;r, p) = {{4 + 3 - 1} \choose {3 - 1}} (0.20)^3 (1 - 0.20)^4 = 0.049.$

Function dnbinom() calculates the negative-binomial probability. Parameter x equals the number of failures, $$x - r$$.

dnbinom(x = 4, size = 3, prob = 0.2)
##  0.049

The expected number of failures prior to 3 successes is $$E(X) = 3 (1 - 0.20) / 0.20 = 12$$ with variance $$Var(X) = 3 (1 - 0.20) / 0.20^2 = 60$$. Confirm this with a simulation from n = 10,000 random samples using rnbinom().

my_dat <- rnbinom(n = 10000, size = 3, prob = 0.20)
mean(my_dat)
##  12
var(my_dat)
##  60 ### 1.2.6 Geometric

If $$X$$ is the count of Bernoulli trials of success probability $$p$$ required to achieve the first successful event, then $$X$$ is a random variable with a geometric distribution $$X \sim G(p)$$. The probability of $$X = x$$ trials is

$f(x; p) = p(1-p)^{x-1}.$

with $$E(X)=\frac{{n}}{{p}}$$ and $$Var(X) = \frac{(1-p)}{p^2}$$. The probability of $$X<=n$$ trials is

$F(X=n) = 1 - (1-p)^n.$

#### Examples

What is the probability a marketer encounters x = 3 people on the street who did not attend a sporting event before the first success if the population probability is p = 0.20?

$f(4; 0.20) = 0.20(1-0.20)^{4-1} = 0.102.$

Function dgeom() calculates the geometric distribution probability. Parameter x is the number of failures, not the number of trials.

dgeom(x = 3, prob = 0.20)
##  0.1
data.frame(cnt = rgeom(n = 10000, prob = 0.20)) %>%
count(cnt) %>%
top_n(n = 15, wt = n) %>%
ungroup() %>%
mutate(pct = round(n / sum(n), 3),
X_eq_x = cnt == 3) %>%
ggplot(aes(x = as.factor(cnt), y = n, fill = X_eq_x, label = pct)) +
geom_col(alpha = 0.8) +
scale_fill_mf() +
geom_text(size = 3) +
theme_mf() +
theme(legend.position = "none") +
labs(title = "Distribution of trials prior to first success",
subtitle = paste("P(X = 3) | X ~ G(.2) = ", round(dgeom(2, .2), 3)),
x = "Unsuccessful trials",
y = "Count",
caption = "simulation of n = 10,000 samples from geometric dist.") What is the probability the marketer fails to find someone who attended a game in x <= 5 trials before finding someone who attended a game on the sixth trial when the population probability is p = 0.20?

p = 0.20
n = 5
# exact
pgeom(q = n, prob = p, lower.tail = TRUE)
##  0.74
# simulated
mean(rgeom(n = 10000, prob = p) <= n)
##  0.74 What is the probability the marketer fails to find someone who attended a game on x >= 5 trials before finding someone who attended a game on the next trial?

p = 0.20
n = 5
# exact
pgeom(q = n, prob = p, lower.tail = FALSE)
##  0.26
# simulated
mean(rgeom(n = 10000, prob = p) > n)
##  0.26 The expected number of trials to achieve the first success is 1 / 0.20 = 5, Var(X) = (1 - 0.20) / 0.20^2 = 20?

p = 0.20
# mean
# exact
1 / p
##  5
# simulated
mean(rgeom(n = 10000, prob = p)) + 1
##  5
# Variance
# exact
(1 - p) / p^2
##  20
# simulated
var(rgeom(n = 100000, prob = p))
##  20

### 1.2.7 Hypergeometric

If $$X$$ is the count of successful events in a sample of size $$n$$ without replacement from a population of size $$N$$ containing $$K$$ successes and $$N-K$$ non-successes, then $$X$$ is a random variable with a hypergeometric distribution

$f(x|N,K,n) = \frac{{{K}\choose{k}}{{N-K}\choose{n-k}}}{{N}\choose{n}}.$

with $$E(X) = n\frac{K}{N}$$ and $$Var(X) = n \frac{K}{N} \cdot \frac{N-n}{N} \cdot \frac{N-K}{N-1}$$.

The formula follows from the frequency table of the possible outcomes.

Sampled Not Sampled Total
success k K-k K
non-success n-k (N-K)-(n-k) N-K
Total n N-n N

If $$X$$ is the count of successful events in a sample of size $$k$$ without replacement from a population containing $$M$$ successes and $$N$$ non-successes, then $$X$$ is a random variable with a hypergeometric distribution

$f(x|m,n,k) = \frac{{{m}\choose{x}}{{n}\choose{k-x}}}{{m+n}\choose{k}}.$

with $$E(X)=k\frac{m}{m+n}$$ and $$Var(X) = k\frac{m}{m+n}\cdot\frac{m+n-k}{m+n}\cdot\frac{n}{m+n-1}$$.

phyper returns the cumulative probability (percentile) p at the specified value (quantile) q. qhyper returns the value (quantile) q at the specified cumulative probability (percentile) p.

#### Example

What is the probability of selecting $$X = 14$$ red marbles from a sample of $$k = 20$$ taken from an urn containing $$m = 70$$ red marbles and $$n = 30$$ green marbles?

Function dhyper() calculates the hypergeometric probability.

x = 14
m = 70
n = 30
k = 20

dhyper(x = x, m = m, n = n, k = k)
##  0.21

The expected value is 14 and variance is 3.39. The hypergeometric random variable is similar to the binomial random variable except that it applies to situations of sampling without replacement from a small population. As the population size increases, sampling without replacement converges to sampling with replacement, and the hypergeometric distribution converges to the binomial. What if the total population size is 250? 500? 1000? ### 1.2.8 Gamma

If $$X$$ is the interval until the $$\alpha^{th}$$ successful event when the average interval is $$\theta$$, then $$X$$ is a random variable with a gamma distribution $$X \sim \Gamma(\alpha, \theta)$$. The probability of an interval of $$X = x$$ is

$f(x; \alpha, \theta) = \frac{1}{\Gamma(\alpha)\theta^\alpha}x^{\alpha-1}e^{-x/\theta}.$

where $$\Gamma(\alpha) = (1 - \alpha)!$$ with $$E(X) = \alpha \theta$$ and $$Var(X) = \alpha \theta^2$$.

#### Examples

On average, someone sends a money order once per 15 minutes ($$\theta = .25$$). What is the probability someone sends $$\alpha = 10$$ money orders in less than $$x = 3$$ hours?*

theta = 0.25
alpha = 10
pgamma(q = 3, shape = alpha, scale = 0.25)
##  0.76
data.frame(x = 0:1000 / 100, prob = pgamma(q = 0:1000 / 100, shape = alpha, scale = theta, lower.tail = TRUE)) %>%
mutate(Interval = ifelse(x >= 0 & x <= 3, "0 to 3", "other")) %>%
ggplot(aes(x = x, y = prob, fill = Interval)) +
geom_area(alpha = 0.9) +
theme_mf() +
scale_fill_mf() +
labs(title = "X ~ Gam(alpha = 10, theta = .25)",
subtitle = "Probability of 10 events in X hours when the mean time to an event is .25 hours.",
x = "Interval (x)",
y = "pgamma") 