Visualizing The Exponential Distribution (1/2)

Exponential Probability Density Function (PDF)

\(\star\) The exponential r.v. has infinitely sized sample space. The plot of the exponential PDF only shows the range \(0 \le x \le 20\).

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

Visualizing The Exponential Distribution (2/2)

Exponential Cumulative Distribution Function (CDF)

\(\star\) The CDF, in general, computes the cumulative probability of a given PDF.

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 Probabilities (1/2)

Exponential Distribution

\(\star\) Note that the pexp function computes the probability \(P(X \le x)\), meaning it computes the area under the PDF from \(X=0\) to \(X=x\). The dexp function computes the density, not probability because \(P(X = x) = 0\) at any \(x\).

Example:

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

Using R:

pexp(2.5,1/3)

## [1] 0.5654018

Exponential Probabilities (2/2)

Exponential Distribution

\(\star\) Use the complement rule when computing the probability \(P(X \ge x) = 1 - P(X \le x)\).

Example:

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

Using R:

1-pexp(2.5,1/3)

## [1] 0.4345982

Exponential Expected Value

Exponential Distribution with Expected Value

\(\star\) The law of large numbers still can be used to interpret the expected value, which in this case, you would expect to have \(3\) unit length until a “success” occurs in the long run.

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.

Visualizing the Normal Distribution (1/2)

Normal Probability Density Function (PDF)

\(\star\) The normal r.v. has infinitely sized sample space. The plot of the exponential PDF only shows the range \(0 \le x \le 20\).

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

Visualizing the Normal Distribution (2/2)

Normal Cumulative Density Function (CDF)

\(\star\) The CDF of the Normal distribution is also known as the sigmoid function.

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 Probabilities

Normal Distribution

Example:

What is \(P(X \le 13)\) for \(X \sim \text{N}\left(10,2.24^2\right)\)? \[ \begin{aligned} P(X \le 13) & = \int_0^{13} f(x) \ dx \\ & = \int_0^{13} \frac{1}{\sqrt{2 \pi (2.24)^2}} e^{-\frac{(x-10)^2}{2(2.24)^2}} \ dx \\ P(X \le 13) & \approx 0.9098 \end{aligned} \]

Using R:

pnorm(13,10,2.24)

## [1] 0.9097612

\(\star\) Note that the pnorm() function computes the probability \(P(X \le x)\), meaning it computes the area under \(f(x)\) from \(X=0\) to \(X=x\) using the Normal PDF. The dnorm() function computes the density, not probability because \(P(X = x)=0\) at any \(x\).

Normal Expected Value

Normal Distribution with Expected Value

In general, the expected value of the normal r.v. is given by \[\text{E}(X) = \mu,\] which is the center of the normal distribution.

Exponential vs Normal Distribution Summary

R.V. \(X\)	Exponential	Normal
Description	Unit length until a “success” event happens	Approximation to the Binomial with sufficiently large number of \(n\) independent trials
Sampling	With replacement	With replacement
Parameters	\(\lambda \longrightarrow\) rate of “success”	\(\mu \longrightarrow\) mean \(\sigma^2 \longrightarrow\) variance
PDF	\(f(x) = \lambda e^{- \lambda x}\) \(x \in [0,\infty)\)	\(f(x) = \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}\) \(x \in (-\infty,\infty)\)
Expected Value \(\text{E}(X)\)	\(\frac{1}{\lambda}\)	\(\mu\)
PDF	`dexp`	`dnorm`
CDF	`pexp`	`pnorm`
Simulations	`rexp`	`rnorm`

Exponential and Normal Distributions

Applied Statistics

Objectives

Visualizing The Exponential Distribution (1/2)

Visualizing The Exponential Distribution (2/2)

Exponential Probabilities (1/2)

Exponential Probabilities (2/2)

Exponential Expected Value

Visualizing the Normal Distribution (1/2)

Visualizing the Normal Distribution (2/2)

Normal Probabilities

Normal Expected Value

Exponential vs Normal Distribution Summary