• Know how to compute exponential probabilities using R
• Know how to simulate exponential distribution in R
• Introduce the normal distribution
• Activity: The Galton Board and Normal Distribution

Exponential Distribution

Exponential R.V.

Let \(\lambda=\frac{1}{15}\) be the success probability.

\[ \begin{aligned} \text{R.V. } & \longrightarrow X \sim \text{Exp}\left(\frac{1}{15}\right) \\ \text{PDF } & \longrightarrow f(x) = \frac{1}{15} e^{- \frac{1}{15} x} \\ \text{for } & x \in [0,\infty) \end{aligned} \]

Exponential Distribution

Example:

What is the probability that “success” happens on 15 unit length or less, given \(\lambda=\frac{1}{15}\)? \[ \begin{aligned} P(X \le 15) & = \int_0^{15} f(x) \ dx \\ & = \int_0^{15} \frac{1}{15} e^{-\frac{1}{15} x} \ dx \\ P(X \le 15) & \approx 0.632 \end{aligned} \]

Using R:

lambda <- 1/15
pexp(15,lambda)

## [1] 0.6321206

Exponential Distribution with Expected Value

Exponential R.V.

Let \(\lambda=\frac{1}{15}\) be the success probability.

\[ \begin{aligned} \text{R.V. } & \longrightarrow X \sim \text{Exp}\left(\frac{1}{15}\right) \\ \text{PDF } & \longrightarrow f(x) = \frac{1}{15} e^{- \frac{1}{15} x} \\ \text{for } & x \in [0,\infty) \\ \text{expected value} & \longrightarrow \text{E}(X) = 15 \end{aligned} \]

In general, the expected value of the exponential r.v. is given by \[\text{E}(X) = \frac{1}{\lambda},\] which is the reciprocal of the “success” rate.

Random Sampling from the Exponential Distribution

Sample Mean vs the Expected Value

The sample mean of \(0.88\) is not exactly equal to the expected value of \(1\) due to sampling variability. As we increase the number of samples, the sample mean gets closer to the expectation.

Geometric Random Sampling using R

N <- 100 # number of simulations
mean <- 15
lambda <- 1/15 # set rate
rexp(N,lambda)

Binomial Distribution

Binomial R.V.

Let \(p=0.50\) be the success probability and \(n=10\) the number of trials.

\[ \begin{aligned} \text{R.V. } & \longrightarrow X \sim \text{Binom}(p) \\ \text{PMF } & \longrightarrow P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \\ \text{for } & k = 0,1,2,3, \cdots, n \end{aligned} \]

Example

• Each fair coin flip is an independent Bernoulli trial with:
  • \(P(H) = \frac{1}{2}\)
  • \(P(T) = 1 - \frac{1}{2}\)
• Suppose that \(n=2\).
  • We expect to have \(2 \times \frac{1}{2} = 1\) \(H\) outcome in \(2\) trials.

In general:

• Let \(p\) be the probability of “success” for each independent Bernoulli trial.
• Let \(n\) be the number of trials.
• The expected value of “success” outcomes is \(n \times p\).