9 Continuous Random Variables

We now turn our attention to continuous random variables.

9.1 Distribution Functions

Definition 9.1 (Distribution Function of a Random Variable) Let $X$ be a random variable. The distribution function of $X,$ denoted $F(x),$ is the function defined on all real numbers $x$ such that $$F(x) = P(X \leq x).$$

Example 9.1 Suppose $X$ is the discrete random variable given by density function $$\begin{array}{c|c|c|c|c} x & 0 & 1 & 2 & 3 \\ \hline p(x) & 1/8 & 3/8 & 3/8 & 1/8 \end{array}$$ Note $F(-2.7) = P(X \leq -2.7) = 0,$ and $F(1.3) = P(X \leq 1.3) = 1/8 + 3/8 = 4/8$ (since the only $X$ values less than or equal to 1.3 with positive probability are $X=0$ or $X=1$).

The distribution function for $X$ is the following step function:

$$F(x)= \begin{cases} 0 &\text{ if }x < 0 \\ 1/8 &\text{ if } 0 \leq x < 1 \\ 4/8 &\text{ if } 1 \leq x < 2 \\ 7/8 &\text{ if } 2 \leq x < 3 \\ 1 &\text{ if } x \geq 3 \end{cases}$$

Figure 9.1: Distribution function for X

Observe that this function $F(x)$ is defined for all $-\infty < x < \infty$ (check out those arrows :)). The jumps in the graph indicate that the function $F$ is not continuous, and the points of discontinuity occur exactly at the values of $X$ in the probability table for $X$.

Theorem 9.1 (Properties of any distribution function) If $F(x)$ is a distribution function, then

1. $\displaystyle \lim_{x \to -\infty} F(x) = 0$;
2. $F$ is non-decreasing. That is, if $x_1 \leq x_2$ then $F(x_1) \leq F(x_2)$; and
3. $\displaystyle \lim_{x \to \infty} F(x) = 1$.

Definition 9.2 (Continuous Random Variable) A random variable is called a continuous random variable if its distribution function $F$ is continuous for all $x$.

So the distribution function for any continuous random variable has the following sort of look, descriptively (as in Figure 9.2):

• it is continuous,
• its domain is $(-\infty,\infty)$
• as $x$ progresses away from $-\infty$ toward $\infty,$ the values of $F(x)$ rise from 0 to 1, never decreasing along the way.

Definition 9.3 Let $F$ be the distribution for a continuous random variable $X$. Then the derivative of $F,$ wherever it exists is called the probability density function for $X$. When continuous $X$ has a probability density function, we usually denote it as $f(x)$.

The density function $f(x)$ is a theoretical curve for the frequency distribution of a population of measurements. We’ll look at examples shortly.

Theorem 9.2 (Properties of a density function) If $f(x)$ is a density function for a continuous random variable $X,$ then

1. $f(x) \geq 0$ for all $x,$ and
2. $\displaystyle \int_{-\infty}^{\infty} f(x)~dx = 1$.

Sketch of Proof:

For a) Recall $f(x) = F^\prime(x)$. One feature of any distribution function is that it is never decreasing, so its slope (derivative) is never negative. Since $f(x)$ gives the slope of $F,$ $f(x) \geq 0$.

For b) $F$ is the antiderivative of $f,$ which is useful to know when we integrate $f$. Also, $\displaystyle \int_{-\infty}^{\infty} f(x)~dx$ is an improper integral, which we can tackle by splitting it into two integrals, assuming each of these new integrals converges:

$$\begin{align*} \int_{-\infty}^{\infty} f(x)~dx &= \int_{-\infty}^{0} f(x)~dx + \int_{0}^{\infty} f(x)~dx \\ &= \lim_{a \to -\infty} \int_a^0 f(x)~dx + \lim_{b \to \infty} \int_0^b f(x)~dx \\ &= \lim_{a \to -\infty} \left[F(0)-F(a)\right] + \lim_{b \to \infty} \left[F(b)-F(0)\right] \\ &= (F(0)-0) + (1 - F(0)) & \text{ by limit properties of } F \\ &= 1. \end{align*}$$

Example 9.2 Consider distribution function $F$ pictured below, where $c > 0$ is a fixed constant.

Figure 9.2: Piece-wise linear distribution function

This function is piece-wise linear, continuous, and it is differentiable everywhere except the sharp corners at $x = 0$ and $x = c$. At any other point, $f(x) = F^\prime(x)$ equals the slope of the segment running through the point $(x,F(X))$.

So the probability density function for this random variable is $$f(x)= \begin{cases} 0 &\text{ if }x < 0 \\ 1/c &\text{ if } 0 < x < c \\ 0 &\text{ if } x > c, \end{cases}$$ and the graph of $f$ looks like this:

Figure 9.3: probability density function for X

Note that $f(x) \geq 0$ for all $x$. Also, $$\int_{-\infty}^{\infty} f(x)~dx = \int_0^c f(x)~dx$$ (we only have to integrate over intervals in which $f(x) > 0$), and this later integral is the area of a rectangle of width $c$ and height $1/c,$ so it has area 1. Thus, we have a valid pdf!

Example 9.3 Find the value of $k$ that makes the following function a valid pdf. $$f(x)= \begin{cases} kx^8 &\text{ if }0 \leq x \leq 1 \\ 0 &\text{ else.} \end{cases}$$ We need $k \geq 0$ os that $f(x) \geq 0$ for all $x$. We also need $$1 = \int_{-\infty}^\infty f(x)~dx = \int_0^1 kx^8~dx = \frac{k}{9}x^9 ~\biggr|_0^1.$$ It follows that $k = 9$.

Definition 9.4 (Quantiles) Let $X$ denote a random variable. If $0 < p < 1,$ the $p$th quantile of $X,$ denoted $\phi_p,$ is the smallest value such that $F(\phi_p) \geq p$. If $X$ is continuous, $\phi_p$ is the smallest value such that $F(\phi_p) = p$.

Some special quantiles:

• $\phi_.25,$ denoted $Q1,$ is called the first quartile,
• $\phi_.5,$ denoted $M,$ is called the median of the random variable,
• $\phi_.75,$ denoted $Q3,$ is called the third quartile

Theorem 9.3 If $X$ is a continuous random variable with density function $f,$ then for any real numbers $a < b,$ $$P(a \leq X \leq b) = \int_a^b f(x)~dx.$$

Proof Idea: The distribution function $F$ is an antiderivative of the density function $f,$ so using the Fundamental Theorem of Calculus,

$$\begin{align*} \int_a^b f(x)~dx &= F(b) - F(a) \\ &= P(X\leq b) - P(X \leq a) \\ &= P(a < X \leq b) &\text{ since } a < b\\ &= P(a \leq X \leq b) &\text{ since } P(X = a) = 0 \end{align*}$$

Note: For any continuous random variable $X,$ and $a < b,$

$$P(a < X < b) = P(a \leq X < b) = P(a < X \leq b) = P(a \leq X \leq B)$$ since $P(X = c) = 0$ for any real number $c$.

Example 9.4 (A quadratic density function) Suppose $X$ has density function $$f(x)= \begin{cases} \frac{3}{8}x^2 &\text{ if }0 \leq x \leq 2 \\ 0 &\text{ else.} \end{cases}$$

Wait! Is this actually a valid density function?

1. Ok, yes, $f(x) \geq 0$ for all $x$.
2. And…

$$\begin{align*} \int_{-\infty}^{\infty} f(x) ~dx &= \int_0^2 \frac{3}{8} x^2~dx \\ &= \frac{1}{8} x^3 \Big|_0^2 \\ &= 1. \end{align*}$$

Ok, now to the question: Find $P(1 \leq X \leq 2)$:

$$\begin{align*} P(1 \leq X \leq 2) &= \int_1^2 \frac{3}{8} x^2~dx \\ &= \frac{1}{8} x^3 ~\biggr|_1^2 \\ &= 1 - \frac{1}{8} \\ &= \frac{7}{8}. \end{align*}$$

Even though $X$ can take any value between 0 and 2, the probability is 7/8 that $X$ will be between 1 and 2. Most of the area under the density curve is at the high end of the $X$ range:

Figure 9.4: A quadratic density function

Example 9.5 Suppose $X$ has density function $$f(x)= \begin{cases} 0 &\text{ if } x < 1 \\ \frac{1}{x^2} &\text{ if } x \geq 1. \end{cases}$$

a) Check that this gives a valid density function:

$$\begin{align*} \int_{-\infty}^{\infty} f(x)~dx &= \int_1^\infty x^{-2}~dx \\ &= \lim_{b \to \infty} \left[\int_1^b x^{-2}~dx \right] \\ &= \lim_{b \to \infty} \left[ -\frac{1}{x} \biggr|_1^b \right]\\ &= \lim_{b \to \infty} \left[ -\frac{1}{b}+1\right]\\ &= 1. \end{align*}$$ The limit equals 1 in the end since $1/b \to 0$ as $b \to \infty$.

b) Find $F(x),$ the cumulative probability distribution function.

By definition, for any real number $x,$ $$F(x) = \int_{-\infty}^x f(t)~dt,$$ which, of course, gives the area under $f$ over the interval $(-\infty, x]$. Since $f$ is piece-wise defined, the integrand used in the integral to evaluate $F$ depends on the bounds of the integral.

$$F(x)= \begin{cases} \displaystyle\int_{-\infty}^x 0 ~dt &\text{ if } x < 1 \\ \displaystyle\int_{-\infty}^1 0 ~dt + \displaystyle\int_{1}^x \frac{1}{t^2} ~dt&\text{ if } x \geq 1. \end{cases}$$ We leave it to the reader to integrate these expressions to obtain $$F(x)= \begin{cases} 0 &\text{ if } x < 1 \\ \displaystyle 1 - \frac{1}{x} &\text{ if } x \geq 1. \end{cases}$$

Figure 9.5: Distribution function for X

c) Find $P(1 < X < 3).$

Well, $$P(1 < X < 3) = \int_1^3 f(x)~dx = F(3)-F(1),$$ by the Fundamental Theorem of Calculus (FTC), so $$P(1 < X < 3) = (1 - 1/3) - (1 - 1/1) = 2/3.$$

9.2 Expected Value for Continuous Random Variables

Definition 9.5 If $X$ is a continuous random variable with probability density function $f(x),$ then the expected value of $X$, denoted $E(X),$ is $$E(X) = \int_{-\infty}^\infty x \cdot f(x)~dx,$$ provided this integral exists. The expected value $E(X)$ is also called the mean of $X$, and is often denoted as $\mu_X,$ or $\mu$ if the random variable $X$ is understood.

The expected value of the function $g(X)$ of $X$ is $$E(g(X)) = \int_{-\infty}^\infty g(x) \cdot f(x)~dx,$$ provided this integral exists.

The variance of $X$ is $$V(X) = E((X-\mu_X)^2),$$ provided this integral exists.

As in the discrete case, one can show $V(X) = E(X^2)-E(X)^2,$ a working formula for variance which is sometimes easier to use to calculate variance.

Example 9.6

Find $E(X)$ and $V(X)$ where $X$ is the continuous random variable from Example 9.4.

Recall $X$ has density function $\displaystyle f(x) = 3x^2/8$ for $0 \leq x \leq 2$.

Expected Value: $$\begin{align*} E(X) &= \int_0^2 x \cdot 3x^2/8~dx \\ &= \frac{3}{8} \int_0^2 x^3~dx \\ &= \frac{3}{8}\frac{1}{4}x^4 ~\biggr|_0^2 \\ &= \frac{3}{2}. \end{align*}$$

Variance: We first find $E(X^2)$: $$\begin{align*} E(X^2) &= \int_0^2 x^2 \cdot 3x^2/8~dx \\ &= \frac{3}{8} \int_0^2 x^4~dx \\ &= \frac{3}{8}\frac{1}{5}x^5 ~\biggr|_0^2 \\ &= \frac{12}{5}. \end{align*}$$

Then, $$\begin{align*} V(X) &= E(X^2) - E(X)^2 \\ &= (12/5) - (3/2)^2\\ &= 0.15. \end{align*}$$

The properties of expected value that held for discrete random variables also hold for continuous random variables.

Theorem 9.4 Suppose $X$ is a continuous random variable, $c \in \mathbb{R}$ is a constant, and $g,$ $g_1,$ and $g_2$ are functions of $X$.

1. $E(c) = c$.
2. $E(c\cdot g(X))= cE(g(X))$.
3. $E(g_1(X) \pm g_2(X)) = E(g_1(X)\pm g_2(X))$.

These results follow immediately from properties of integration. For instance, to prove property 1 we observe that for constant $c,$ $$E(c) = \int_{-\infty}^\infty c\cdot f(x)~ dx = c \int_{-\infty}^\infty f(x)~ dx,$$ and the integral in the last expression equals 1 by definition of a valid probability density function.

Theorem 9.5 Let $X$ be a random variable (discrete or continuous) with $E(X) = \mu$ and $V(X) = \sigma^2,$ and let $a, b$ be constants. Then

1. $\displaystyle E(aX + b) = aE(X) + b = a \mu + b.$
2. $\displaystyle V(aX + b) = a^2V(X) = a^2 \sigma^2.$

Proof.

1. This result follows immediately from properties of expected value (Theorems 9.4 and 9.4).

2. Let $Y = aX + b$. Then (a) says that $E(Y) = a \mu + b,$ so $$\begin{align*} V(Y) &= E((Y-(a\mu + b))^2) \\ &= E\left(((aX+b)-(a\mu + b))^2\right)\\ &= E\left((aX-a\mu)^2\right)\\ &= a^2 E\left((X-\mu)^2\right) \end{align*}$$ But $E\left((X-\mu)^2\right)=V(X)$ by the definition of variance, so the result follows.

Example 9.7 (Ore Sample Impurities) For certain ore samples, the proportion $X$ of impurities per sample is a random variable with density function $$f(x)= \begin{cases} 1.5x^2 + x &\text{ if }0 \leq x \leq 1 \\ 0 &\text{ else. } \end{cases}$$ The dollar value of each sample is $W = 5 - 0.5X$.

Find the mean, variance, and standard deviation of $W$.

First, let’s consider the variable $X$ itself.

$$\begin{align*} E(X) &= \int_0^1 x \cdot (1.5x^2 + x)~dx \\ &= \int_0^1 1.5x^3 + x^2~dx \\ &= \frac{1.5}{4}x^4 + \frac{1}{3} x^3 ~\biggr|_0^1 \\ &= \frac{17}{24}. \end{align*}$$

Also, $$\begin{align*} E(X^2) &= \int_0^1 x^2 \cdot (1.5x^2 + x)~dx \\ &= \int_0^1 1.5x^4 + x^3~dx \\ &= \frac{1.5}{5}x^5 + \frac{1}{4} x^4 ~\biggr|_0^1 \\ &= \frac{11}{20}. \end{align*}$$

Thus, $V(X) = (11/20)-(17/24)^2 \approx 0.0483.$

Then, by Theorem 9.5,

$$E(W) = E(5 - 0.5X) = 5 - 0.5E(X) = 5 - 0.5\cdot (17/24) = 4.65,$$ and $$V(W) = V(5 - 0.5X) = 0.25V(X) \approx 0.012,$$ so that the standard deviation is $\sigma = \sqrt{V(W)} ~\approx 0.11$ (about 11 cents).

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