given the mean number of events per unit time (𝜆):

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

Probability Density Function #1

• 𝑓(𝑥) = (𝜆^𝛼/𝛤(𝛼))·𝑥^𝛼-1·𝑒^-𝜆𝑥# for 𝑥 > 0

where:

• 𝛼 - number of events
• 𝜆 - number of events per unit time
• 𝛤(𝛼) = gamma function = (𝛼 - 1)!

special cases of a Gamma Distribution:

• Gamma(𝛼=1, 𝜆) = Exponential(𝜆)
• Gamma(𝛼>0, 𝜆=1/2) = Chi-Square(2𝛼)

Probability Density Function #2

• 𝑓(𝑥) = [1/(𝛤(𝛼)𝜆^𝛼)]·𝑥^𝛼-1·𝑒^-𝑥/𝜆# for 𝑥 > 0

Expectation

𝐄[𝑋] = 𝛼/𝜆

proof

via exponential distribution Exponential(𝜆) variables 𝑋₁, ..., 𝑋_𝛼

𝑋 = 𝑋₁ + … + 𝑋_𝛼

• 𝐄[𝑋] = 𝐄[𝑋₁ + … + 𝑋_𝛼]
• 𝐄[𝑋] = 𝐄[𝑋₁] + … + 𝐄[𝑋_𝛼]
• 𝐄[𝑋] = 1/𝜆 + … + 1/𝜆
• 𝐄[𝑋] = 𝛼/𝜆

representing a Gamma variable 𝑋 as a sum of independent

way 2 ^∞𝑓(𝑥)𝑑𝑥 = 1 for Gamma and all the other densities

• 1 = ₀∫^∞𝑓(𝑥)𝑑𝑥
• 1 = ₀∫^∞(𝜆^𝛼/𝛤(𝛼)) (𝑥^𝛼-1𝑒^-𝜆𝑥) 𝑑𝑥
• 1 = (𝜆^𝛼/𝛤(𝛼)) ₀∫^∞(𝑥^𝛼-1𝑒^-𝜆𝑥) 𝑑𝑥
• 𝛤(𝛼)/𝜆^𝛼 = ₀∫^∞(𝑥^𝛼-1𝑒^-𝜆𝑥) 𝑑𝑥
• 𝛤(𝛼+1)/𝜆^𝛼+1 = ₀∫^∞(𝑥^𝛼𝑒^-𝜆𝑥) 𝑑𝑥

now compute expectation:

• 𝐄[𝑋] = ₀∫^∞𝑥 𝑓(𝑥)𝑑𝑥
• 𝐄[𝑋] = ₀∫^∞𝑥 (𝜆^𝛼/𝛤(𝛼)) (𝑥^𝛼-1𝑒^-λ𝑥) 𝑑𝑥
• 𝐄[𝑋] = (𝜆^𝛼/𝛤(𝛼)) ₀∫^∞𝑥(𝑥^𝛼-1𝑒^-λ𝑥) 𝑑𝑥
• 𝐄[𝑋] = (𝜆^𝛼/𝛤(𝛼)) ₀∫^∞𝑥^𝛼𝑒^-λ𝑥 𝑑𝑥
• 𝐄[𝑋] = (𝜆^𝛼/𝛤(𝛼)) * (𝛤(𝛼+1)/𝜆^𝛼+1)
• 𝐄[𝑋] = (𝛤(𝛼+1)/𝛤(𝛼)) * (𝜆^𝛼/𝜆^𝛼+1)
• 𝐄[𝑋] = [(𝛼)! / (𝛼-1)!] * (1/𝜆)
• 𝐄[𝑋] = (𝛼) * (1/𝜆)
• 𝐄[𝑋] = 𝛼/𝜆

First, note that ₀∫

Variance

𝑉𝑎𝑟(𝑋) = 𝛼/𝜆²

proof

via exponential distribution Exponential(λ) variables 𝑋₁, ..., 𝑋_𝛼

𝑋 = 𝑋₁ + … + 𝑋_𝛼

• 𝑉𝑎𝑟(𝑋) = 𝑉𝑎𝑟(𝑋₁ + … + 𝑋_𝛼)
• 𝑉𝑎𝑟(𝑋) = 𝑉𝑎𝑟(𝑋₁) + … + 𝑉𝑎𝑟(𝑋_𝛼) # because of independence
• 𝑉𝑎𝑟(𝑋) = 1/𝜆² + … + 1/𝜆²
• 𝑉𝑎𝑟(𝑋) = 𝛼/𝜆²

representing a Gamma variable 𝑋 as a sum of INDEPENDENT

way 2 ^∞𝑓(𝑥)𝑑𝑥 = 1 for Gamma and all the other densities

• 1 = ₀∫^∞𝑓(𝑥)𝑑𝑥
• 1 = ₀∫^∞(𝜆^𝛼/𝛤(𝛼)) (𝑥^𝛼-1𝑒^-𝜆𝑥) 𝑑𝑥
• 1 = (𝜆^𝛼/𝛤(𝛼)) ₀∫^∞(𝑥^𝛼-1𝑒^-𝜆𝑥) 𝑑𝑥
• 𝛤(𝛼)/𝜆^𝛼 = ₀∫^∞(𝑥^𝛼-1𝑒^-𝜆𝑥) 𝑑𝑥
• 𝛤(𝛼+2)/𝜆^𝛼+2 = ₀∫^∞(𝑥^𝛼+1𝑒^-𝜆𝑥) 𝑑𝑥

now compute expectation:

• 𝑉𝑎𝑟(𝑋) = [₀∫^∞𝑥²𝑓(𝑥)𝑑𝑥] - 𝜇²
• 𝑉𝑎𝑟(𝑋) = [₀∫^∞𝑥²(𝜆^𝛼/𝛤(𝛼)) (𝑥^𝛼-1𝑒^-𝜆𝑥)𝑑𝑥] - (𝛼/𝜆)²
• 𝑉𝑎𝑟(𝑋) = [(𝜆^𝛼/𝛤(𝛼)) ₀∫^∞𝑥²(𝑥^𝛼-1𝑒^-𝜆𝑥)𝑑𝑥] - (𝛼/𝜆)²
• 𝑉𝑎𝑟(𝑋) = [(𝜆^𝛼/𝛤(𝛼)) ₀∫^∞𝑥^𝛼+1𝑒^-𝜆𝑥𝑑𝑥] - (𝛼/𝜆)²
• 𝑉𝑎𝑟(𝑋) = [(𝜆^𝛼/𝛤(𝛼)) (𝛤(𝛼+2)/𝜆^𝛼+2)] - (𝛼/𝜆)²
• 𝑉𝑎𝑟(𝑋) = [(𝛤(𝛼+2)/𝛤(𝛼)) * (𝜆^𝛼/𝜆^𝛼+2)] - (𝛼/𝜆)²
• 𝑉𝑎𝑟(𝑋) = [((𝛼+1)! / (𝛼-1)!) * (1/𝜆²)] - (𝛼/𝜆)²
• 𝑉𝑎𝑟(𝑋) = [(𝛼(𝛼+1)) * (1/𝜆²)] - (𝛼/𝜆)²
• 𝑉𝑎𝑟(𝑋) = [(𝛼²+ 𝛼) * (1/𝜆²)] - (𝛼/𝜆)²
• 𝑉𝑎𝑟(𝑋) = 𝛼²/𝜆²+ 𝛼/𝜆² - (𝛼/𝜆)²
• 𝑉𝑎𝑟(𝑋) = (𝛼/𝜆)²+ 𝛼/𝜆² - (𝛼/𝜆)²
• 𝑉𝑎𝑟(𝑋) = 𝛼/𝜆²

First, note that ₀∫

Cumulative Distribution Function (CDF)

Moment-Generating Function

When the PDF of gamma variable 𝑋 is:

• 𝑓(𝑥) = [1/(𝛤(𝛼)𝜆^𝛼)]·𝑥^𝛼-1·𝑒^-𝑥/𝜆# for 𝑥 > 0

Then the moment-generating function of 𝑋 is:

• $M_X(t)=(1-𝜆t{)}^{-𝛼}$

See: Moment-Generating Function - Gamma Distribution

Example 1

Users visit a certain internet site at the average rate of 12 hits per minute. Every sixth visitor receives some promotion that comes in a form of a flashing banner. Then the time between consecutive promotions has Gamma distribution with parameters 𝛼 = 6 and 𝜆 = 12

Example 2

Compilation of a computer program consists of 3 blocks that are processed sequentially, one after another. Each block takes Exponential time with the mean of 5 minutes, independently of other blocks

• 𝛼 = 3
• 𝜆 = 1/5

Compute the expectation and variance of the total compilation time

solution

• 𝐄[𝑋] = 𝛼/𝜆

• 𝐄[𝑋] = 3/(1/5)

• 𝐄[𝑋] = 15

• 𝑉𝑎𝑟(𝑋) = 𝛼/𝜆²

• 𝑉𝑎𝑟(𝑋) = 3/(1/5)²

• 𝑉𝑎𝑟(𝑋) = 75

Compute the probability for the entire program to be compiled in less than 12 minutes

solution

• 𝑷 {𝑋 < 12} = 𝐶𝐷𝐹(𝑥=12)
• 𝑷 {𝑋 < 12} = 𝐶𝐷𝐹(𝑥=12)
• 𝑷 {𝑋 < 12} = ₀∫¹²((1/5)³/𝛤(3)) (𝑥^3-1𝑒^-(1/5)𝑥)𝑑𝑥
• 𝑷 {𝑋 < 12} = (1/5)³/𝛤(3) ₀∫¹²𝑥²𝑒^-𝑥/5𝑑𝑥
• 𝑷 {𝑋 < 12} = (1/5)³/𝛤(3) … requires 2 rounds of integration by parts

Subpages

• Gamma Distribution vs Poisson Distribution