Euler Letters AI logo

Exponential and Gamma Distributions

SD06 — Continuous Distributions

Author

Samer El Kababji

Course: Statistics and Probability for Data Science (DS14270).
Source deck: Samer El Kababji, SD06 — Continuous Distributions.

The Poisson distribution counts events; the exponential and gamma distributions measure how long we wait for those events.

Exponential waiting time

For rate \(\lambda>0\), \(X\sim\operatorname{Exp}(\lambda)\) has density and CDF

\[ f(x)=\lambda e^{-\lambda x},\qquad F(x)=1-e^{-\lambda x},\qquad x\geq0. \]

Its survival probability is

\[ P(X>x)=e^{-\lambda x}. \]

Code 
lambda <- 0.5
x <- c(0, 1, 2, 4)

data.frame(
  x,
  density = dexp(x, rate = lambda),
  CDF = pexp(x, rate = lambda),
  survival = pexp(x, rate = lambda, lower.tail = FALSE)
)
  x    density       CDF  survival
1 0 0.50000000 0.0000000 1.0000000
2 1 0.30326533 0.3934693 0.6065307
3 2 0.18393972 0.6321206 0.3678794
4 4 0.06766764 0.8646647 0.1353353

The exponential moments are

\[ E(X)=\frac1\lambda, \qquad \operatorname{Var}(X)=\frac1{\lambda^2}. \]

Code 
lambda <- 0.5
c(mean = 1 / lambda, variance = 1 / lambda^2, sd = 1 / lambda)
    mean variance       sd 
       2        4        2

Exponential examples

For calls arriving at rate 3 per hour, the probability the next call arrives within an hour is

\[ P(X<1)=1-e^{-3}\approx0.9502. \]

For a pipeline failure time with rate \(0.5\) per year,

\[ P(X>2)=e^{-0.5(2)}=e^{-1}\approx0.3679. \]

Code 
c(
  call_within_one_hour = pexp(1, rate = 3),
  pipeline_survives_two_years = pexp(2, rate = 0.5, lower.tail = FALSE)
)
       call_within_one_hour pipeline_survives_two_years 
                  0.9502129                   0.3678794

Gamma waiting time

The gamma distribution measures the total wait to a later event. In the shape-rate parameterization,

\[ f(x)=\frac{\lambda^\alpha}{\Gamma(\alpha)} x^{\alpha-1}e^{-\lambda x},\qquad x\geq0. \]

With scale \(\beta=1/\lambda\), the same density is

\[ f(x)=\frac{1}{\Gamma(\alpha)\beta^\alpha} x^{\alpha-1}e^{-x/\beta}. \]

The moments are

\[ E(X)=\frac\alpha\lambda=\alpha\beta, \qquad \operatorname{Var}(X)=\frac\alpha{\lambda^2}=\alpha\beta^2. \]

Code 
alpha <- 3
lambda <- 0.5
beta <- 1 / lambda
x <- seq(0, 15, length.out = 301)

max(abs(
  dgamma(x, shape = alpha, rate = lambda) -
    dgamma(x, shape = alpha, scale = beta)
))
[1] 0
Code 
c(mean = alpha / lambda, variance = alpha / lambda^2)
    mean variance 
       6       12

Do not supply both rate and scale to R. State which convention is being used whenever gamma parameters are reported.

Time to the third component failure

A machine fails after its third internal component failure. If failures arrive at rate \(0.5\) per hour, then \(X\sim\operatorname{Gamma}(3,\text{rate}=0.5)\). For integer shape,

\[ F(x)=1-e^{-\lambda x} \sum_{j=0}^{\alpha-1}\frac{(\lambda x)^j}{j!}. \]

At \(x=4\) hours,

\[ P(X<4)=1-e^{-2}\left(1+2+\frac{2^2}{2!}\right) \approx0.3233. \]

Code 
manual <- 1 - exp(-2) * sum(2^(0:2) / factorial(0:2))
built_in <- pgamma(4, shape = 3, rate = 0.5)

c(manual = manual, R = built_in)
   manual         R 
0.3233236 0.3233236

Relationship among the three distributions

  • Poisson: number of events in a fixed interval.
  • Exponential: wait to the first event.
  • Gamma with integer shape \(\alpha\): wait to the \(\alpha\)th event.
Code 
set.seed(14270)
wait_to_third <- rgamma(100000, shape = 3, rate = 0.5)

c(
  simulated_mean = mean(wait_to_third),
  theoretical_mean = 6,
  simulated_P_less_than_4 = mean(wait_to_third < 4),
  exact_P_less_than_4 = pgamma(4, 3, rate = 0.5)
)
         simulated_mean        theoretical_mean simulated_P_less_than_4 
              6.0036694               6.0000000               0.3248400 
    exact_P_less_than_4 
              0.3233236

Practice questions

  1. If \(X\sim\operatorname{Exp}(2)\), find \(P(X>1.5)\) and \(E(X)\).
  2. If \(Y\sim\operatorname{Gamma}(4,\text{rate}=2)\), find its mean and variance.
  3. Find \(P(Y\leq3)\).
  4. Express the same gamma distribution using a scale parameter.
  1. \(P(X>1.5)=e^{-3}\approx0.0498\) and \(E(X)=0.5\).
  2. The mean is \(4/2=2\) and the variance is \(4/2^2=1\).
  3. pgamma(3, shape = 4, rate = 2) gives approximately \(0.8488\).
  4. The scale is \(\beta=1/2=0.5\), so use Gamma(shape = 4, scale = 0.5).