• Develop and understanding of the exponential distribution
• Develop and understanding of the normal distribution
• Know how to compute exponential and normal probabilities using R
• Introduce how to simulate exponential and normal random sampling in R

Exponential Probability Density Function (PDF)

Exponential R.V. and PDF:

Let \(\lambda\) be the rate parameter.

• \(\displaystyle X \sim \text{Exp}\left(\lambda\right)\)
• \(\displaystyle f(x) = \lambda e^{-\lambda x}, \quad x \ge 0\)

Relevant R Functions:

• dexp \(\leftarrow\) PDF
• rexp \(\leftarrow\) random sampling simulation

Exponential Cumulative Density Function (CDF)

Exponential PDF and CDF:

Let \(\lambda\) be the rate parameter.

• \(\displaystyle f(x) = \lambda e^{-\lambda x}, \quad x \ge 0\)
• \(\displaystyle P(X \le x) = \int_{0}^{x} f(t) \ dt, \quad x \ge 0\)

Relevant R Functions:

• pexp \(\leftarrow\) CDF
• qexp \(\leftarrow\) Inverse CDF

Exponential Distribution:

Example:

What is the probability that “success” happens on 2.5 unit length or less, given \(\lambda=\frac{1}{3}\)? \[ \begin{aligned} P(X \le 2.5) & = \int_0^{2.5} f(t) \ dt \\ & = \int_0^{2.5} \frac{1}{3} e^{-\frac{1}{3} t} \ dt \\ P(X \le 2.5) & \approx 0.565 \end{aligned} \]

Using R:

pexp(2.5,1/3)

## [1] 0.5654018

Exponential Distribution:

Example:

What is the probability that “success” happens on at least 2.5 unit length, given \(\lambda=\frac{1}{3}\)? \[ \begin{aligned} P(X \ge 2.5) & = 1 - P(X \le 2.5) \\ & = 1 - \int_0^{2.5} f(t) \ dt \\ & = 1 - \int_0^{2.5} \frac{1}{3} e^{-\frac{1}{3} t} \ dt \\ & = \int_{2.5}^{\infty} \frac{1}{3} e^{-\frac{1}{3} t} \ dt \\ P(X \ge 2.5) & \approx 0.435 \end{aligned} \]

Using R:

1-pexp(2.5,1/3)

## [1] 0.4345982

Exponential Distribution:

Example:

What is the probability that “success” happens between 1 to 5 unit length, given \(\lambda=\frac{1}{3}\)? \[ \begin{aligned} P(1 \le X \le 5) & = P(X \le 5) - P(X \le 1) \\ & = \int_0^{5} f(t) \ dt - \int_0^{1} f(t) \ dt \\ & = \int_1^{5} f(t) \ dt \\ & = \int_1^{5} \frac{1}{3} e^{-\frac{1}{3} t} \ dt \\ P(1 \le X \le 5) & \approx 0.528 \end{aligned} \]

Using R:

pexp(5,1/3)-pexp(1,1/3)

## [1] 0.5276557

Exponential Distribution with Expected Value

Exponential R.V.

Let \(\lambda=\frac{1}{3}\) be the rate parameter.

• \(\displaystyle X \sim \text{Exp}\left(\frac{1}{3}\right)\)
• \(\displaystyle f(x) = \lambda e^{-\frac{1}{3} x}, \quad x \ge 0\)
• \(\displaystyle \text{E}(X) = 3\)

In general, \[ \begin{aligned} \text{E}(X) & = \int_0^{\infty} x f(x) \ dx \\ & = \int_0^{\infty} x \lambda e^{-\lambda x} \ dx \\ \text{E}(X) & = \frac{1}{\lambda} \end{aligned} \] which is the reciprocal of the rate parameter.

Normal Probability Density Function (PDF)

Normal R.V. and PDF:

Let \(\mu\) be the mean and \(\sigma^2\) be the variance.

• \(\displaystyle X \sim \text{N}\left(\mu,\sigma^2\right)\)
• \(\displaystyle f(x) = \frac{1}{\sigma \sqrt{2 \pi}} \exp{\left(-\frac{1}{2} \left( \frac{x - \mu}{\sigma} \right)^2\right)}, \ -\infty < x < \infty\)

Relevant R Functions:

• dnorm \(\leftarrow\) PDF
• rnorm \(\leftarrow\) random sampling simulation

Normal Cumulative Density Function (CDF)

Normal PDF and CDF:

Let \(\mu\) be the mean and \(\sigma^2\) be the variance.

• \(\displaystyle f(x) = \frac{1}{\sigma \sqrt{2 \pi}} \exp{\left(-\frac{1}{2} \left( \frac{x - \mu}{\sigma} \right)^2\right)}, \ -\infty < x < \infty\)
• \(\displaystyle P(X \le x) = \int_{\infty}^{x} f(t) \ dt\)

Relevant R Functions:

• pnorm \(\leftarrow\) CDF
• qnorm \(\leftarrow\) Inverse CDF

Normal Distribution:

Example:

What is the probability that a randomly selected value is at most \(13\)?

\[ \begin{aligned} P(X \le 13) & = \int_{-\infty}^{13} f(t) \ dt \\ & = \int_{-\infty}^{13} \frac{1}{2.24 \sqrt{2 \pi}} \exp{\left(-\frac{1}{2} \left( \frac{t - 10}{2.24} \right)^2\right)} \ dt \\ P(X \le 13) & \approx 0.910 \end{aligned} \]

Using R:

pnorm(13,10,2.24)

## [1] 0.9097612

Normal Distribution:

Example:

What is the probability that a randomly selected value is at least \(13\)?

\[ \begin{aligned} P(X \ge 13) & = 1 - P(X \le 13) \\ & = 1 - \int_{-\infty}^{13} f(t) \ dt \\ & = 1 - \int_{-\infty}^{13} \frac{1}{2.24 \sqrt{2 \pi}} \exp{\left(-\frac{1}{2} \left( \frac{t - 10}{2.24} \right)^2\right)} \ dt \\ & = \int_{13}^{\infty} \frac{1}{2.24 \sqrt{2 \pi}} \exp{\left(-\frac{1}{2} \left( \frac{t - 10}{2.24} \right)^2\right)} \ dt \\ P(X \le 13) & \approx 0.090 \end{aligned} \]

Using R:

1-pnorm(13,10,2.24)

## [1] 0.09023884

Normal Distribution:

Example:

What is the probability that a randomly selected value is between \(7\) and \(13\)?

\[ \begin{aligned} P(7 \le X \le 13) & = P(X \le 13)-P(X \le 7) \\ & = \int_{-\infty}^{13} f(t) \ dt - \int_{-\infty}^{7} f(t) \ dt \\ & = \int_{7}^{13} f(t) \ dt \\ & = \int_{7}^{13} \frac{1}{2.24 \sqrt{2 \pi}} \exp{\left(-\frac{1}{2} \left( \frac{t - 10}{2.24} \right)^2\right)} \ dt \\ P(7 \le X \le 13) & \approx 0.820 \end{aligned} \]

Using R:

pnorm(13,10,2.24)-pnorm(7,10,2.24)

## [1] 0.8195223

Normal Distribution with Expected Value:

Exponential R.V.

Let \(\mu = 10\) be the mean and \(\sigma^2 = 2.24^2\) be the variance.

• \(\displaystyle X \sim \text{N}\left(10,2.24^2\right)\)
• \(\displaystyle f(x) = \frac{1}{\sigma \sqrt{2 \pi}} \exp{\left(-\frac{1}{2} \left( \frac{x - \mu}{\sigma} \right)^2\right)}, \ -\infty < x < \infty\)
• \(\displaystyle \text{E}(X) = 10\)

In general, the expected value of the normal r.v. is given by \[ \begin{aligned} \text{E}(X) & = \int_{-\infty}^{\infty} x f(x) \ dx \\ & = \int_{-\infty}^{\infty} \frac{x}{\sigma \sqrt{2 \pi}} \exp{\left(-\frac{1}{2} \left( \frac{x - \mu}{\sigma} \right)^2\right)} \ dx \\ \text{E}(X) & = \mu \end{aligned} \] which is the center of the normal distribution.

The z-score formula:

\[z = \frac{x - \mu}{\sigma}\]

Standard Normal R.V. and PDF:

Let \(\mu = 0\) be the mean and \(\sigma^2 = 1\) be the variance.

• \(\displaystyle Z \sim \text{N}\left(0,1\right)\)
• \(\displaystyle f(z) = \frac{1}{\sqrt{2 \pi}} \exp{\left(-\frac{z^2}{2} \right)}; \\ \quad \text{ for } -\infty < z < \infty\)

How the transformation works:

• Let \(x = \mu\) and \(\sigma = 1\).
• Then, \(z = 0\), meaning the distance from the mean to itself is \(0\) standard deviations.
• A positive z-score indicates the value is above the mean, while a negative z-score indicates it is below.