Continuous uniform distribution

If \(X\) is a continuous random quantity that is “equally likely” to take on any value in a bounded interval \([a,\, b]\), then we can model it with the continuous uniform distribution.

Basic properties

Notation \(X\sim\text{Unif}(a,\,b)\)
Range \([a,\,b]\)
Parameter space \(-\infty < a < b < \infty\)
PDF \(f(x)=\begin{cases}\frac{1}{b-a} & a\leq x\leq b\\0&\text{else}\end{cases}\)
CDF \(F(x)=\begin{cases}0 & x < a\\\frac{x-a}{b-a} & a\leq x\leq b\\1&b<x\end{cases}\)
Expectation \(\frac{a+b}{2}\)
Variance \((b-a)^2/12\)

Fun facts:

  • Unif(0, 1) is called the standard uniform distribution.

R commands

Here is the documentation for the suite of commands that let you work with the normal distribution in R:

dunif(x, min, max) # PDF
punif(q, min, max) # CDF: P(X <= q)
qunif(p, min, max) # quantile function
runif(n, min, max) # random numbers

Play around!

#| '!! shinylive warning !!': |
#|   shinylive does not work in self-contained HTML documents.
#|   Please set `embed-resources: false` in your metadata.
#| standalone: true
#| viewerHeight: 600
#| fig-asp: 1

library(shiny)

# Define UI for application that draws a histogram
ui <- fluidPage(
  
  # Application title
  titlePanel("Uniform distribution CDF and PDF"),
  
  # Sidebar with a slider input for number of bins 
  sidebarLayout(
    sidebarPanel(
      sliderInput("a",
                  "Lower bound (a):",
                  min = 0,
                  max = 2,
                  value = 1,
                  step = 0.1),
      sliderInput("b",
                  "Upper bound (b):",
                  min = 3,
                  max = 5,
                  value = 4,
                  step = 0.1)
    ),
    
    # Show a plot of the generated distribution
    mainPanel(
      plotOutput("distPlot")
    )
  )
)

# Define server logic required to draw a histogram
server <- function(input, output) {
  
  output$distPlot <- renderPlot({
    
    a <- input$a
    b <- input$b

    par(mfrow = c(2, 1), mar = c(4, 4, 2, 2))
    
    curve(punif(x, min = a, max = b), from = -1, to = 6, n = 1000,
          ylab = "", col = "red", xlim = c(0, 5))
    curve(0*x, from = -1, to = a, n = 1000, 
          ylim = c(0, 1), ylab = "", col = "red", xlim = c(0, 5))
    curve(dunif(x, min = a, max = b), from = a, to = b, n = 1000, 
          ylim = c(0, 1), ylab = "", col = "red", add = TRUE)
    curve(0*x, from = b, to = 6, n = 1000, 
          ylim = c(0, 1), ylab = "", col = "red", add = TRUE)
  })
}

# Run the application 
shinyApp(ui = ui, server = server)

Derivations

\[ \begin{aligned} \int_{-\infty}^\infty f_X(x)\,\text{d} x &= \int_{-\infty}^a f_X(x)\,\text{d} x + \int_{a}^b f_X(x)\,\text{d} x + \int_{b}^\infty f_X(x)\,\text{d} x \\ &= {\int_{-\infty}^a 0\,\text{d} x} + \int_{a}^b \frac{1}{b-a}\,\text{d} x + {\int_{b}^\infty 0\,\text{d} x} \\ &= \frac{1}{b-a}\int_a^b1\,\text{d} x \\ &= \frac{1}{b-a} [x]_a^b \\ &= \frac{1}{b-a}(b-a) \\ &= 1 . \end{aligned} \]

The cdf is \(F_X(x)=P(X\leq x)=\int_{-\infty}^x f_X(t)\,\text{d} t\). If \(x< a\), then

\[ F_X(x)=\int_{-\infty}^xf_X(t)\,\text{d} t=\int_{-\infty}^x0\,\text{d} t=0. \]

If \(a\leq x\leq b\), then

\[ \begin{aligned} F_X(x) &= \int_{-\infty}^xf_X(t)\,\text{d} t \\ &= \int_{-\infty}^af_X(t)\,\text{d} t + \int_{a}^xf_X(t)\,\text{d} t \\ &= {\int_{-\infty}^a0\,\text{d} t} + \int_{a}^x\frac{1}{b-a}\,\text{d} t \\ &= \frac{1}{b-a}[t]_a^x \\ &= \frac{x-a}{b-a} . \end{aligned} \]

If \(b < x\), then

\[ \begin{aligned} F_X(x) &= \int_{-\infty}^xf_X(t)\,\text{d} t \\ &= \int_{-\infty}^af_X(t)\,\text{d} t + \int_{a}^bf_X(t)\,\text{d} t + \int_{b}^xf_X(t)\,\text{d} t \\ &= {\int_{-\infty}^a0\,\text{d} t} + \int_{a}^b\frac{1}{b-a}\,\text{d} t + {\int_{b}^x0\,\text{d} t} \\ &=1. \end{aligned} \]

So the cdf of \(X\sim\text{Unif}(a,\, b)\) is

\[ F_X(x) = \begin{cases} 0 & x<a \\ \frac{x-a}{b-a} & a\leq x\leq b \\ 1 & b<x. \end{cases} \]

The expected value is the midpoint of the interval:

\[ \begin{align*} E(X) &= \int_{-\infty}^\infty xf_X(x)\,\text{d} x \\ &= \int_{-\infty}^a xf_X(x)\,\text{d} x + \int_{a}^b xf_X(x)\,\text{d} x + \int_{b}^\infty xf_X(x)\,\text{d} x \\ &= {\int_{-\infty}^a x\cdot0\,\text{d} x} + \int_{a}^b x\frac{1}{b-a}\,\text{d} x + {\int_{b}^\infty x\cdot0\,\text{d} x} \\ &= 0+\frac{1}{b-a}\int_a^b x\,\text{d} x+0 \\ &= \frac{1}{b-a}\left[\frac{1}{2}x^2\right]_a^b \\ &= \frac{1}{b-a}\left[\frac{1}{2}b^2-\frac{1}{2}a^2\right] \\ &= \frac{1}{2(b-a)}\left[b^2-a^2\right] \\ &= \frac{1}{2{(b-a)}}{(b-a)}(b+a) \\ &=\frac{b+a}{2}. \end{align*} \]

That’s the proper derivation, but without the math, we could have reasoned our way to this answer by noting that the range is bounded, so the mean is guaranteed to exist, and the distribution is symmetric, so the mean matches the point of symmetry.