• The continuous analog of a probability mass function (pmf) is a probability density function (pdf).
• However, while pmfs and pdfs play analogous roles, they are different in one fundamental way; namely, a pmf outputs probabilities directly, while a pdf does not.
• A pdf of a continuous random variable must be integrated to find probabilities of related events.
• The probability density function (pdf) (a.k.a. density) of a continuous RV $$X$$, defined on a probability space with probability measure $$\text{P}$$, is a function $$f_X:\mathbb{R}\mapsto[0,\infty)$$ which satisfies \begin{align*} \text{P}(a \le X \le b) & =\int_a^b f_X(x) dx, \qquad \text{for all } -\infty \le a \le b \le \infty \end{align*}
• For a continuous random variable $$X$$ with pdf $$f_X$$, the probability that $$X$$ takes a value in the interval $$[a, b]$$ is the area under the pdf over the region $$[a,b]$$.
• The axioms of probability imply that a valid pdf must satisfy \begin{align*} f_X(x) & \ge 0 \qquad \text{for all } x,\\ \int_{-\infty}^\infty f_X(x) dx & = 1 \end{align*}

Example 15.1 Two resistors are randomly selected and connected in series, so that the system resistance $$X$$ (ohms) is the sum of the two resistances. Assume that the first resistor has a marginal Uniform(135, 165) distribution, and the second has a marginal Uniform(162, 198) distribution (corresponding to nominal 150 and 180 ohm resistors with 10% tolerance). Also assume that the two resistances are independent.

Simulation suggests that the pdf of $$X$$ is

$f_X(x) = \begin{cases} \frac{1}{33}\left(1-\frac{1}{33}|x - 330|\right), & 297 < x < 363\\ 0, & \text{otherwise.} \end{cases}$

1. Sketch a plot of the pdf. What does this say about the distribution of the system resistance?

2. Verify that $$f_X$$ is a valid pdf.

3. Find $$\text{P}(X > 340)$$.

4. Find $$\text{P}(330 < X< 340)$$.

## 15.1 Density is not probability

Example 15.2 Continuing Example 15.1, where the series system resistance $$X$$ (ohms) has pdf $f_X(x) = \begin{cases} \frac{1}{33}\left(1-\frac{1}{33}|x - 330|\right), & 297 < x < 363\\ 0, & \text{otherwise.} \end{cases}$

1. Compute $$\text{P}(X = 340)$$.

2. Compute the probability that $$X$$, rounded to one decimal place, is 340.0.

3. How is the probability in the previous part related to $$f_X(340)$$?

4. How many times more likely is $$X$$, rounded to one decimal place, to be 340.0 than to be 350.0?

• The probability that a continuous random variable $$X$$ equals any particular value is 0. That is, if $$X$$ is continuous then $$\text{P}(X=x)=0$$ for all $$x$$.
• For a continuous random variable $$X$$, $$\text{P}(X \le x) = \text{P}(X < x)$$, etc.
• Careful: this is NOT true for discrete random variables; for a discrete random variable $$\text{P}(X \le x) = \text{P}(X < x) + \text{P}(X = x)$$.
• For continuous random variables, it doesn’t really make sense to talk about the probability that the random value is equal to a particular value. However, we can consider the probability that a random variable is close to a particular value.
• The density $$f_X(x)$$ at value $$x$$ is not a probability.
• Rather, the density $$f_X(x)$$ at value $$x$$ is related to the probability that the RV $$X$$ takes a value “close to $$x$$” in the following sense $\text{P}\left(x-\frac{\epsilon}{2} \le X \le x+\frac{\epsilon}{2}\right) \approx \epsilon f_X(x), \qquad \text{for small \epsilon}$
• The quantity $$\epsilon$$ is a small number that represents the desired degree of precision. For example, rounding to two decimal places corresponds to $$\epsilon=0.01$$.
• What’s important about a pdf is relative heights. For example, if $$f_X(x_2)/ f_X(x_1) = 2$$ then $$X$$ is roughly “twice as likely to be near $$x_2$$ than to be near $$x_1$$” in the above sense. $\frac{f_X(x_2)}{f_X(x_1)} = \frac{\epsilon f_X(x_2)}{\epsilon f_X(x_1)} \approx \frac{\text{P}\left(x_2-\frac{\epsilon}{2} \le X \le x_2+\frac{\epsilon}{2}\right)}{\text{P}\left(x_1-\frac{\epsilon}{2} \le X \le x_1+\frac{\epsilon}{2}\right)}$

## 15.2 Exponential distributions

Example 15.3 Suppose that we model the waiting time, measured continuously in hours, from now until the next earthquake (of any magnitude) occurs in southern CA as a continuous random variable $$X$$ with pdf $f_X(x) = 2 e^{-2x}, \; x \ge0.$ This is the pdf of the “Exponential(2)” distribution.

1. Sketch the pdf of $$X$$. What does this tell you about waiting times?

2. Compute $$\text{P}(X = 0.5)$$.

3. Without doing any integration, approximate the probability that $$X$$ rounded to the nearest minute is 0.5 hours.

4. Without doing any integration determine how much more likely is it for $$X$$ rounded to the nearest minute to be 0.5 than 1.5.

5. Compute and interpret $$\text{P}(X > 0.25)$$.

6. Compute and interpret $$\text{P}(X \le 3)$$.

7. Compute and interpret the 25th percentile of $$X$$.

8. Compute and interpret the 50th percentile of $$X$$.

9. Compute and interpret the 75th percentile of $$X$$.

10. Start to construct a spinner corresponding to this Exponential(2) distribution.

11. Use simulation to approximate the long run average value of $$X$$. Interpret this value. At what rate do earthquakes tend to occur?

12. Use simulation to approximate the standard deviation of $$X$$. What do you notice?

• Exponential distributions are often used to model the waiting times between events in a random process that occurs continuously over time.
• A continuous random variable $$X$$ has an Exponential distribution with rate parameter $$\lambda>0$$ if its pdf is $f_X(x) = \begin{cases}\lambda e^{-\lambda x}, & x \ge 0,\\ 0, & \text{otherwise} \end{cases}$
• If $$X$$ has an Exponential($$\lambda$$) distribution then \begin{align*} \text{P}(X>x) & = e^{-\lambda x}, \quad x\ge 0\\ \text{Long run average of X} & = \frac{1}{\lambda}\\ \text{Standard deviation of X} & = \frac{1}{\lambda} \end{align*}
• Exponential distributions are often used to model the waiting time in a random process until some event occurs.
• $$\lambda$$ is the average rate at which events occur over time (e.g., 2 per hour)
• $$1/\lambda$$ is the mean time between events (e.g., 1/2 hour)
• The “standard” Exponential distribution is the Exponential(1) distribution, with rate parameter 1 and mean 1.
• If $$X$$ has an Exponential(1) distribution and $$\lambda>0$$ is a constant then $$X/\lambda$$ has an Exponential($$\lambda$$) distribution.
• A continuous random variable $$X$$ has a Uniform distribution with parameters $$a$$ and $$b$$, with $$a<b$$, if its probability density function $$f_X$$ satisfies \begin{align*} f_X(x) & \propto \text{constant}, \quad & & a<x<b\\ & = \frac{1}{b-a}, \quad & & a<x<b. \end{align*}
• If $$X$$ has a Uniform($$a$$, $$b$$) distribution then \begin{align*} \text{Long run average value of X} & = \frac{a+b}{2}\\ \text{Variance of X} & = \frac{|b-a|^2}{12}\\ \text{SD of X} & = \frac{|b-a|}{\sqrt{12}} \end{align*}
• If $$U$$ has a Uniform(0, 1) distribution then $$X = a + (b-a)U$$ has a Uniform($$a$$, $$b$$) distribution.