October 15

Exponential Distribution

A continuous probability distribution that models the time between consecutive events in a Poisson process (where events occur at a constant average rate). It is commonly used to model waiting times or lifespans of electronic components.

• Notation: $X \sim \text{Exp}(\lambda)$
• Parameter:
  • $\lambda$ (lambda, the “rate parameter”): The average number of events per unit of time.

Key Formulas

Probability Density Function (PDF)

Defines the relative likelihood for the waiting time to be equal to $x$.

\[f(x) = \lambda e^{-\lambda x} \quad \text{for } x \ge 0\]

Cumulative Distribution Function (CDF)

Gives the probability that the event occurs before or at some time $x$. This is most commonly used for calculations.

\[F(x) = P(X \le x) = 1 - e^{-\lambda x} \quad \text{for } x \ge 0\]

Mean and Variance

• Mean (Expected Value): The average waiting time between events.
  • $E[X] = \mu = \frac{1}{\lambda}$
• Variance: The spread of the waiting times.
  • $\text{Var}(X) = \sigma^2 = \frac{1}{\lambda^2}$
• Standard Deviation:
  • $\sigma = \frac{1}{\lambda}$

Important Note: Notice that the mean and the standard deviation are equal. Also, be careful with the parameter. Sometimes, the exponential distribution is parameterized using the mean ($\mu$ or $\beta$) instead of the rate ($\lambda$). In that case, $\lambda = 1/\mu$, and the formulas change accordingly. We will use the rate parameter $\lambda$.

The Memoryless Property

This is the most defining characteristic of the exponential distribution. It states that the probability of an event occurring in the future is independent of how long we have already waited.

• Formula: $P(X > s + t \mid X > s) = P(X > t)$
• Interpretation: If you’ve already waited $s$ minutes for a bus (and it hasn’t arrived), the probability you’ll have to wait at least an additional $t$ minutes is the same as the probability you had to wait at least $t$ minutes from the very beginning. The process “forgets” the time $s$ that has already passed

Finding Probabilities

We almost always use the CDF to find probabilities.

• Probability of waiting less than $a$:
  • $P(X \le a) = 1 - e^{-\lambda a}$
• Probability of waiting more than $a$:
  • $P(X > a) = 1 - P(X \le a) = 1 - (1 - e^{-\lambda a}) = e^{-\lambda a}$
• Probability of waiting between $a$ and $b$:
  • $P(a \le X \le b) = P(X \le b) - P(X \le a)$
  • $P(a \le X \le b) = (1 - e^{-\lambda b}) - (1 - e^{-\lambda a}) = e^{-\lambda a} - e^{-\lambda b}$

Examples

Problem: On average, a customer arrives at a coffee shop every 2 minutes. The time between arrivals follows an exponential distribution.

First, find the rate parameter $\lambda$.

• The mean waiting time is $\mu = 2$ minutes.
• The rate is $\lambda = 1 / \mu = 1/2 = 0.5$ customers per minute.
• So, $X \sim \text{Exp}(0.5)$.

1. What is the probability that the next customer arrives within 1 minute?

• We want $P(X \le 1)$.
• Use the CDF: $P(X \le 1) = 1 - e^{-\lambda x} = 1 - e^{-0.5 \times 1}$
• $P(X \le 1) = 1 - e^{-0.5} \approx 1 - 0.6065 = 0.3935$
• Answer: There’s a 39.35% chance.

2. What is the probability that we must wait more than 5 minutes for the next customer?

• We want $P(X > 5)$.
• Use the “more than” formula: $P(X > 5) = e^{-\lambda x} = e^{-0.5 \times 5}$
• $P(X > 5) = e^{-2.5} \approx 0.0821$
• Answer: There’s an 8.21% chance.

3. What is the probability the next customer arrives between 1 and 3 minutes?

• We want $P(1 \le X \le 3)$.
• Use the “between” formula: $P(1 \le X \le 3) = e^{-\lambda a} - e^{-\lambda b}$
• $P(1 \le X \le 3) = e^{-0.5 \times 1} - e^{-0.5 \times 3} = e^{-0.5} - e^{-1.5}$
• $P(1 \le X \le 3) \approx 0.6065 - 0.2231 = 0.3834$
• Answer: There’s a 38.34% chance.