Definition: A random variable is continuous if possible values comprise either a single interval on the number line or a union of disjoint intervals.

Examples:

  • In the study of the ecology of a lake, a rv $X$ could be the depth measurements at a random chosen locations

    $X\in[0,\text{maximum depth of lake}]$.

  • In a study of a chemical reaction, $Y$ could be the concentration level of a particular chemical in solution.

  • In a study of customer service, $W$ could be the time a customer waits for service.

Note: If $X$ is continuous, $P(X=x)=0$ for any $x$!

Motivation example: Suppose a train is scheduled to arrive at 1 pm. Let $X$ be the minutes past the hour that it arrives and $X\in{0,1,2,3,4,5}$.

$x$ 0 1 2 3 4 5
$p(x)$ 0.1 0.15 0.3 0.25 0.15 0.05
\[X\in[0,5]\\ P(1.5\lt X\lt 2.5)=\int_{1.5}^{2.5}f(x)d(x)\]

$y = f(x)\leftarrow$ the probability density function for the continuous rv $X$.

\[P(a\le X\le b)=\int_a^b f(x)dx\]
Properties of the probability density function

For any continuous rv $X$ with probability density function (pdf) $f$ we have:

  • The probability density function $f:(-\infty,\infty)\rightarrow[0,\infty)$ so $f(x)\ge0$.

  • $P(-\infty\lt X\lt\infty)=\int_{-\infty}^{\infty}f(x)dx=1\quad(=P(S))$.

  • $P(a\le X\le b)=\int_{a}^{b}f(x)dx$

Note: $P(X=a)=\int_a^a f(x)dx = 0$ for all real numbers $a$.

Cumulative Distribution Function

Definition: The cumulative distribution function (cdf) for a continuous rv $X$ is given by

\[F(x) = P(X\le x)=\int_{-\infty}^{x}f(t)dt\]

  • $0\le F(x)\le 1$

  • $\lim_{x \to -\infty}F(x)=0$ and $\lim_{x \to \infty}F(x)=1$

  • $F’(x)=f(x)$ by fundamental theorem of Calculus.

  • $F(x)$ is always increasing.

Uniform Random Variable

Definition: A random variable $X$ has the uniform distribution on the interval $[a,b]$ if its density function is given by

\[f(x)=\begin{cases} \begin{align} &\frac{1}{b-a}&&\text{if }a\le x \le b\\ &0&&\text{else} \end{align} \end{cases}\]

Notation: $X\sim U[a,b]$

Cumulative Distribution Function:

\[\begin{align} F(x)&=P(X\le x)\\ &=\int_{-\infty}^{x}f(t)dt\\ &=\int_{a}^{x}\frac{1}{b-a}dt,&&a\lt x\lt b \end{align}\]

Solve integral, we have

\[P(X\le x)= \begin{cases} \begin{align} &0&&\text{if }x\lt a\\ &\frac{x-a}{b-a}&&\text{if }a\le x\le b\\ &1&&\text{if }b\lt x \end{align} \end{cases}\]

Example 1: Random number generators select numbers uniformly from a specific interval, usually $[0,1]$.

Example 2: Suppose the diameter of aerosol particles in a particular application is uniformly distributed between 2 and 6 nanometers. Find the probability that a random measured particle has diameter greater that 3 nanometers.

$X\sim U[2,6]$

\[f(x)=\begin{cases} \begin{align} &\frac{1}{4}&&\text{for }2\le x\le6\\ &0&&\text{else} \end{align} \end{cases}\] \[\begin{align} P(X\ge3)= 1-P(X\lt3)&=\int_2^3\frac{1}{4}dt\\ &=\frac{3}{4}\\ \end{align}\]

Or

\[P(X\ge3)=\int_3^6\frac{1}{4}dt=\frac{3}{4}\]

Example 3: You throw a dart at a dashboard. The radial distance of the dart from the x-axis can be modeled by a uniform random variable.

\[Y\sim U[0,360]\]
Expectation and Variance for a continuous rv $X$

Recall: $Y$ is discrete

\[E(Y)=\sum_k kP(Y=k)\quad V(Y)=\sum_k(k-\mu_Y)^2P(Y=k)\]

If $X$ is continuous RV with pdf $f(x)$

Expected Value

\[\color{red}{E(X) = \int_{-\infty}^{\infty}xf(x)dx}\]

Variance

\[\begin{align} \color{red}{V(X)}&=\int_{-\infty}^{\infty}(x-\mu_X)^2f(x)dx\\ &=\int_{-\infty}^{\infty}(x^2-2\mu_X x+\mu_X^2)f(x)dx\\ &=\underbrace{\int_{-\infty}^{\infty}x^2f(x)dx}_{E(X^2)}\\ &\qquad-2\mu_X x\underbrace{\int_{-\infty}^{\infty}xf(x)dx}_{E(X)}\\ &\qquad+\mu_X^2\underbrace{\int_{-\infty}^{\infty}f(x)dx}_{1}\\ &=\color{red}{E(X^2)-(E(X))^2} \end{align}\]

Compute expectation and variance for $X\sim U[a,b]$

\[f(x)=\begin{cases} \begin{align} &\frac{1}{b-a}&&\text{if }a\le x \le b\\ &0&&\text{else} \end{align} \end{cases}\]

Expectation

\[\begin{align} \color{red}{E(X)}&=\int_a^b x.\frac{1}{b-a}dx\\ &=\frac{1}{b-a}\frac{x^2}{2}\bigg\rvert_{a}^{b}\\ &=\frac{b^2-a^2}{2(b-a)}\\ &=\color{red}{\frac{b+a}{2}} \end{align}\]

Variance

We have second moment

\[\begin{align} E(X^2)&=\int_a^b x^2.\frac{1}{b-a}dx\\ &=\frac{1}{b-a}\frac{x^3}{3}\bigg\rvert_{a}^{b}\\ &=\frac{b^3-a^3}{3(b-a)}\\ &=\frac{b^2+ab+a^2}{3} \end{align}\]

Variance

\[\begin{align} \color{red}{V(X)}&=E(X^2)-(E(X))^2\\ &=\frac{b^2+ab+a^2}{3}-\bigg(\frac{b+a}{2}\bigg)^2\\ &=\color{red}{\frac{(b-a)^2}{12}} \end{align}\]