given a mean number of events that happen within unit time (𝜆):

• the time between events has Exponential Distribution
• the number of events occurring within a unit time has Poisson Distribution
• the total time of 𝛼 events has Gamma Distribution

Probability Density Function

𝑓(𝑋=𝑥) = 𝜆𝑒^−𝜆𝑥for 𝑥 > 0

where:

• 𝜆 = number of events per unit time (i.e. inverse of average time between events)
• 𝑒 = 2.71828
• 𝑥 = is the time between events in question
• 𝑋 = is random variable with exponential distribution

see Deriving Exponential Distribution from Poisson Distribution

Expectation

𝐄[𝑋] = 1/𝜆

proof

• 𝐄[𝑋] = ₀∫^∞𝑥𝑓(𝑥)𝑑𝑥
• 𝐄[𝑋] = ₀∫^∞𝑥𝜆𝑒^−𝜆𝑥𝑑𝑥
• 𝐄[𝑋] = 𝜆 ₀∫^∞𝑥𝑒^−𝜆𝑥𝑑𝑥
• by Integration by Parts
  • 𝑢 = 𝑥
  • 𝑑𝑣 = 𝑒^−𝜆𝑥
  • 𝑑𝑢 = 𝑑𝑥
  • 𝑣 = -(1/𝜆)𝑒^−𝜆𝑥
• 𝐄[𝑋] = 𝜆 [𝑢𝑣₀|^∞- ₀∫^∞𝑣𝑑𝑢]
• 𝐄[𝑋] = 𝜆 [-(𝑥/𝜆)𝑒^−𝜆𝑥₀|^∞- ₀∫^∞-(1/𝜆)𝑒^−𝜆𝑥𝑑𝑥]
• 𝐄[𝑋] = 𝜆 [0 - ₀∫^∞-(1/𝜆)𝑒^−𝜆𝑥𝑑𝑥]
• 𝐄[𝑋] = 𝜆 [0 + (1/𝜆) ₀∫^∞𝑒^−𝜆𝑥𝑑𝑥]
• 𝐄[𝑋] = ₀∫^∞𝑒^−𝜆𝑥𝑑𝑥
• 𝐄[𝑋] = -(1/𝜆)𝑒^−𝜆𝑥₀|^∞
• 𝐄[𝑋] = -(1/𝜆)𝑒^−𝜆∞- (-(1/𝜆)𝑒^−𝜆0)
• 𝐄[𝑋] = -(1/𝜆)𝑒^−𝜆∞+ (1/𝜆)
• 𝐄[𝑋] = 0+ (1/𝜆)
• 𝐄[𝑋] = 1/𝜆

Variance

𝑉𝑎𝑟(𝑋) = 1/𝜆²

proof

• 𝑉𝑎𝑟(𝑋) = ₀∫^∞𝑥²𝑓(𝑥)𝑑𝑥 - (1/𝜆)²
• 𝑉𝑎𝑟(𝑋) = ₀∫^∞𝑥²𝜆𝑒^−𝜆𝑥𝑑𝑥 - (1/𝜆)²
• 𝑉𝑎𝑟(𝑋) = 𝜆 ₀∫^∞𝑥²𝑒^−𝜆𝑥𝑑𝑥 - (1/𝜆)²
• by Integration by Parts
  • 𝑢 = 𝑥²
  • 𝑑𝑣 = 𝑒^−𝜆𝑥
  • 𝑑𝑢 = 2𝑥𝑑𝑥
  • 𝑣 = -(1/𝜆)𝑒^−𝜆𝑥
• 𝑉𝑎𝑟(𝑋) = 𝜆 [𝑢𝑣₀|^∞- ₀∫^∞𝑣𝑑𝑢] - (1/𝜆)²
• 𝑉𝑎𝑟(𝑋) = 𝜆 [-(𝑥²/𝜆)𝑒^−𝜆𝑥₀|^∞- ₀∫^∞-(2/𝜆)𝑥𝑒^−𝜆𝑥𝑑𝑥] - (1/𝜆)²
• 𝑉𝑎𝑟(𝑋) = 𝜆 [0 - ₀∫^∞-(2/𝜆)𝑥𝑒^−𝜆𝑥𝑑𝑥] - (1/𝜆)²
• 𝑉𝑎𝑟(𝑋) = 𝜆 [(2/𝜆) ₀∫^∞𝑥𝑒^−𝜆𝑥𝑑𝑥] - (1/𝜆)²
• 𝑉𝑎𝑟(𝑋) = 2₀∫^∞𝑥𝑒^−𝜆𝑥𝑑𝑥 - (1/𝜆)²
• by Integration by Parts
  • 𝑢 = 𝑥
  • 𝑑𝑣 = 𝑒^−𝜆𝑥
  • 𝑑𝑢 = 𝑑𝑥
  • 𝑣 = -(1/𝜆)𝑒^−𝜆𝑥
• 𝑉𝑎𝑟(𝑋) = 2 [𝑢𝑣₀|^∞- ₀∫^∞𝑣𝑑𝑢] - (1/𝜆)²
• 𝑉𝑎𝑟(𝑋) = 2 [-(𝑥/𝜆)𝑒^−𝜆𝑥₀|^∞- ₀∫^∞-(1/𝜆)𝑒^−𝜆𝑥𝑑𝑥] - (1/𝜆)²
• 𝑉𝑎𝑟(𝑋) = 2 [0 - ₀∫^∞-(1/𝜆)𝑒^−𝜆𝑥𝑑𝑥] - (1/𝜆)²
• 𝑉𝑎𝑟(𝑋) = 2 [0 + (1/𝜆) ₀∫^∞𝑒^−𝜆𝑥𝑑𝑥] - (1/𝜆)²
• 𝑉𝑎𝑟(𝑋) = 2/𝜆₀∫^∞𝑒^−𝜆𝑥𝑑𝑥 - (1/𝜆)²
• 𝑉𝑎𝑟(𝑋) = 2/𝜆₀∫^∞𝑒^−𝜆𝑥𝑑𝑥 - (1/𝜆)²
• 𝑉𝑎𝑟(𝑋) = 2/𝜆 [-(1/𝜆)𝑒^−𝜆𝑥₀|^∞] - (1/𝜆)²
• 𝑉𝑎𝑟(𝑋) = 2/𝜆 [-(1/𝜆)𝑒^−𝜆∞- (-(1/𝜆)𝑒^−𝜆0)] - (1/𝜆)²
• 𝑉𝑎𝑟(𝑋) = 2/𝜆 [-(1/𝜆)𝑒^−𝜆∞+ (1/𝜆)] - (1/𝜆)²
• 𝑉𝑎𝑟(𝑋) = 2/𝜆 [0 + (1/𝜆)] - (1/𝜆)²
• 𝑉𝑎𝑟(𝑋) = 2/𝜆 (1/𝜆) - (1/𝜆)²
• 𝑉𝑎𝑟(𝑋) = (2/𝜆²) - (1/𝜆²)
• 𝑉𝑎𝑟(𝑋) = 1/𝜆²

Cumulative Distribution Function (𝐶𝐷𝐹)

𝐶𝐷𝐹(𝑥) = 1 - 𝑒^−𝜆𝑥

proof

definite integration from 0 to 𝑥

• 𝐶𝐷𝐹(𝑥) = ₀∫^ˣ𝑓(𝑥)𝑑𝑥
• 𝐶𝐷𝐹(𝑥) = -𝑒^−𝜆𝑥+ constant ₀|^ˣ
• 𝐶𝐷𝐹(𝑥) = (-𝑒^−𝜆𝑥+ constant) - (-𝑒^−𝜆0+ constant)
• 𝐶𝐷𝐹(𝑥) = (-𝑒^−𝜆𝑥+ constant) - (-1+ constant)
• 𝐶𝐷𝐹(𝑥) = -𝑒^−𝜆𝑥+ constant + 1 - constant
• 𝐶𝐷𝐹(𝑥) = 1 - 𝑒^−𝜆𝑥

definite integration from 0 to INF

• ₀∫^∞𝑓(𝑥)𝑑𝑥 = ₀∫^∞𝜆𝑒^−𝜆𝑥𝑑𝑥
• ₀∫^∞𝑓(𝑥)𝑑𝑥 = -𝑒^−𝜆𝑥+ constant ₀|^∞
• ₀∫^∞𝑓(𝑥)𝑑𝑥 = (-𝑒^−𝜆∞+ constant) - (-𝑒^−𝜆0+ constant)
• ₀∫^∞𝑓(𝑥)𝑑𝑥 = (0+ constant) - (-1+ constant)
• ₀∫^∞𝑓(𝑥)𝑑𝑥 = 0+ constant + 1 - constant
• ₀∫^∞𝑓(𝑥)𝑑𝑥 = 1

indefinite integration of probability density function

• ∫𝑓(𝑥)𝑑𝑥 = ∫𝜆𝑒^−𝜆𝑥𝑑𝑥
• ∫𝑓(𝑥)𝑑𝑥 = -𝑒^−𝜆𝑥+ constant

Example 1

Let 𝑋 = amount number of minutes a postal clerk spends with his or her customer. The time is known to have an exponential distribution with the average amount of time equal to 4 minutes

𝜆 = 0.25 customers per minute, therefore the probability density function is:

𝑓(𝑋=𝑥) = 0.25 𝑒^−0.25𝑥for 𝑥 > 0

Example 2

Jobs are sent to a printer at an average rate of 3 jobs per hour.

What is the expected time between jobs?

• 𝜆 = 3
• 𝐄[𝑇] = 1/𝜆 = 1/3 hours = 20 minutes

What is the probability that the next job is sent within 5 minutes?

• 𝑷 {𝑇 < 5 minutes} = 𝑷 {𝑇 < 1/12 hours}
• 𝑷 {𝑇 < 5 minutes} = 𝐶𝐷𝐹(𝑇 = 1/12)
• 𝑷 {𝑇 < 5 minutes} = 1 - 𝑒^−3(1/12)
• 𝑷 {𝑇 < 5 minutes} = 1 - 𝑒^−1/4
• 𝑷 {𝑇 < 5 minutes} = 0.221199..

Subpages

• Poisson vs Exponential
• Exponential Distribution is the only Continuous Distribution with the Memoryless Property (similar to its discrete counterpart Geometric Distribution)