11.2.1 Continuous distribution functions

So how do we describe the randomness of continuous random variables? In the case of discrete random variables, the probability mass function (pmf) and the cumulative distribution function (cdf) are used to describe randomness. However, recall that the pmf is a function that returns the probability that the random variable takes the inputted value. Due to the nature of continuous random variables, the probability that a continuous random variable takes on any one individual value is technically 0. Thus, a pmf cannot apply to a continuous random variable.

Rather, we describe the randomness of continuous random variables with the probability density function (pdf) and the cumulative distribution function (cdf). Note that the cdf has the same interpretation and application as in the discrete case.

11.2.2 Probability density function

Let \(X\) be a continuous random variable. The probability density function (pdf) of \(X\), given by \(f_X(x)\) is a function that describes the behavior of \(X\). It is important to note that in the continuous case, \(f_X(x)\neq \mbox{P}(X=x)\), as the probability of \(X\) taking any one individual value is 0.

The pdf is a function. The input of a pdf is any real number. The output is known as the density. The pdf has three main properties:

1. \(f_X(x)\geq 0\)

2. \(\int_{S_X} f_X(x)\mathop{}\!\mathrm{d}x = 1\)

3. \(\mbox{P}(X\in A)=\int_{x\in A} f_X(x)\mathop{}\!\mathrm{d}x\) or another way to write this \(\mbox{P}(a \leq X \leq b)=\int_{a}^{b} f_X(x)\mathop{}\!\mathrm{d}x\)

Properties 2) and 3) imply that the area underneath a pdf represents probability. The pdf is a non-negative function, it cannot have negative values.

11.2.3 Cumulative distribution function

The cumulative distribution function (cdf) of a continuous random variable has the same interpretation as it does for a discrete random variable. It is a function. The input of a cdf is any real number, and the output is the probability that the random variable takes a value less than or equal to the inputted value. It is denoted as \(F\) and is given by: \[ F_X(x)=\mbox{P}(X\leq x)=\int_{-\infty}^x f_x(t) \mathop{}\!\mathrm{d}t \]

Example:
Let \(X\) be a continuous random variable with \(f_X(x)=2x\) where \(0 \leq x \leq 1\). Verify that \(f\) is a valid pdf. Find the cdf of \(X\). Also, find the following probabilities: \(\mbox{P}(X<0.5)\), \(\mbox{P}(X>0.5)\), and \(\mbox{P}(0.1\leq X < 0.75)\). Finally, find the median of \(X\).

To verify that \(f\) is a valid pdf, we simply note that \(f_X(x) \geq 0\) on the range \(0 \leq x \leq 1\). Also, we note that \(\int_0^1 2x \mathop{}\!\mathrm{d}x = x^2\bigg|_0^1 = 1\).

Using R, we find

integrate(function(x)2*x,0,1)

## 1 with absolute error < 1.1e-14

Or we can use the mosaicCalc package to find the anti-derivative. If the package is not installed, you can use the Packages tab in RStudio or type install.packages("mosaicCalc") at the command prompt. Load the library.

library(mosaicCalc)

(Fx<-antiD(2*x~x))

## function (x, C = 0) 
## 1 * x^2 + C

Fx(1)-Fx(0)

## [1] 1

Figure 11.1: pdf of \(X\)

The cdf of \(X\) is found by \[ \int_0^x 2t \mathop{}\!\mathrm{d}t = t^2\bigg|_0^x = x^2 \] This is antiD found from the calculations above.

So, \[ F_X(x)=\left\{ \begin{array}{ll} 0, & x<0 \\ x^2, & 0\leq x \leq 1 \\ 1, & x>1 \end{array}\right. \]

The plot of the cdf of \(X\) is shown in Figure 11.2.

Figure 11.2: cdf of \(X\)

Probabilities are found either by integrating the pdf or using the cdf:

\(\mbox{P}(X < 0.5)=\mbox{P}(X\leq 0.5)=F_X(0.5)=0.5^2=0.25\). See Figure 11.3.

Figure 11.3: Probability represented by shaded area

\(\mbox{P}(X > 0.5) = 1-\mbox{P}(X\leq 0.5)=1-0.25 = 0.75\) See Figure 11.4.

Figure 11.4: Probability represented by shaded area

\(\mbox{P}(0.1\leq X < 0.75) = \int_{0.1}^{0.75}2x\mathop{}\!\mathrm{d}x = 0.75^2 - 0.1^2 = 0.5525\) See Figure 11.5.

integrate(function(x)2*x,.1,.75)

## 0.5525 with absolute error < 6.1e-15

Alternatively, \(\mbox{P}(0.1\leq X < 0.75) = \mbox{P}(X < 0.75) -\mbox{P}(x \leq 0.1) = F(0.75)-F(0.1)=0.75^2-0.1^2 =0.5525\)

Fx(0.75)-Fx(0.1)

## [1] 0.5525

Notice for a continuous random variable, we are loose with the use of the = sign. This is because for a continuous random variable \(\mbox{P}(X=x)=0\). Do not get sloppy when working with discrete random variables.

Figure 11.5: Probability represented by shaded area

The median of \(X\) is the value \(x\) such that \(\mbox{P}(X\leq x)=0.5\), the area under a single point is 0. So we simply solve \(x^2=0.5\) for \(x\). Thus, the median of \(X\) is \(\sqrt{0.5}=0.707\).

Or using R

uniroot(function(x)(Fx(x)-.5),c(0,1))$root

## [1] 0.7071067

11.2.4 Simulation

As in the case of the discrete random variable, we can simulate a continuous random variable if we have an inverse for the cdf. The range of the cdf is \([0,1]\), so we generate a random number in this interval and then apply the inverse cdf to obtain a random variable. In a similar manner, for a continuous random variable, we use the following pseudo code:
1. Generate a random number in the interval \([0,1]\), \(U\).
2. Find the random variable \(X\) from \(F_{X}^{-1}(U)\).
In R for our example, this looks like the following.

sqrt(runif(1))

## [1] 0.3227367

results <- do(10000)*sqrt(runif(1))

inspect(results)

## 
## quantitative variables:  
##      name   class         min        Q1    median        Q3       max      mean
## ...1 sqrt numeric 0.004462668 0.5017449 0.7070588 0.8664043 0.9999366 0.6660716
##             sd     n missing
## ...1 0.2361588 10000       0

Figure 11.6 is a density plot of the simulated density function.

results %>%
  gf_density(~sqrt,xlab="X") %>%
  gf_theme(theme_bw()) %>%
  gf_labs(x="X",y="")

Figure 11.6: Density plot of the simulated random variable.

11.3.1 Mean and variance

Let \(X\) be a continuous random variable. The mean of \(X\), or \(\mu_X\), is simply \(\mbox{E}(X)\). Thus, \[ \mbox{E}(X)=\int_{S_X}x\cdot f_X(x)\mathop{}\!\mathrm{d}x \]

As in the discrete case, the variance of \(X\) is the expected squared difference from the mean, or \(\mbox{E}[(X-\mu_X)^2]\). Thus, \[ \sigma^2_X = \mbox{Var}(X)=\mbox{E}[(X-\mu_X)^2]= \int_{S_X} (x-\mu_X)^2\cdot f_X(x) \mathop{}\!\mathrm{d}x \]

Recall homework problem 6 from the last chapter. In this problem, you showed that \(\mbox{Var}(X)=\mbox{E}(X^2)-\mbox{E}(X)^2\). Thus, \[ \mbox{Var}(X)=\mbox{E}(X^2)-\mbox{E}(X)^2 = \int_{S_X} x^2\cdot f_X(x)\mathop{}\!\mathrm{d}x - \mu_X^2 \]

Example:
Consider the random variable \(X\) from above. Find the mean and variance of \(X\). \[ \mu_X= \mbox{E}(X)=\int_0^1 x\cdot 2x\mathop{}\!\mathrm{d}x = \frac{2x^3}{3}\bigg|_0^1 = \frac{2}{3}=0.667 \]

Side note: Since the mean of \(X\) is smaller than the median of \(X\), we say that \(X\) is skewed to the left, or negatively skewed.

Using R.

integrate(function(x)x*2*x,0,1)

## 0.6666667 with absolute error < 7.4e-15

Or using antiD()

Ex<-antiD(2*x^2~x)
Ex(1)-Ex(0)

## [1] 0.6666667

Using our simulation.

mean(~sqrt,data=results)

## [1] 0.6660716

\[ \sigma^2_X = \mbox{Var}(X)= \mbox{E}(X^2)-\mbox{E}(X)^2 = \int_0^1 x^2\cdot 2x\mathop{}\!\mathrm{d}x - \left(\frac{2}{3}\right)^2 = \frac{2x^4}{4}\bigg|_0^1-\frac{4}{9}=\frac{1}{2}-\frac{4}{9}=\frac{1}{18}=0.056 \]

integrate(function(x)x^2*2*x,0,1)$value-(2/3)^2

## [1] 0.05555556

or

Vx<-antiD(x^2*2*x~x)
Vx(1)-Vx(0)-(2/3)^2

## [1] 0.05555556

var(~sqrt,data=results)*9999/10000

## [1] 0.05576538

And finally, the standard deviation of \(X\) is \(\sigma_X = \sqrt{\sigma^2_X}=\sqrt{1/18}=0.236\).