Gamma Distribution

Recap

• Exponential distribution models waiting time until the next event in a Poisson process.
• The number of events in time \(T\) follows a Poisson(λT) distribution.

• Events occur at times \(T₁, T₂, T₃, ...\)

  The waiting times between them are exponential.

Beyond the First Event

• What if we want the waiting time until the r-th event, not just the next one?
• This leads to the Gamma distribution.

Definition: Gamma Distribution (Generalizing Exponential Waiting Times)

Let:

• \(Tᵣ = W₁ + W₂ + ... + Wᵣ\) be the waiting time for the r-th event.
• Each \(Wⱼ\) is independent and follows Exponential(λ).

Then:

\[ Tᵣ \sim \text{Gamma}(r, \lambda) \]

• \(r\): number of events we are waiting for
• \(\lambda\): event rate

Gamma has two parameters: \(r\) and \(\lambda\)

Exponential is a special case when \(r = 1\).

Derivation Sketch (without messy math)

To compute:

Continuous waiting time for \(r^{th}\) arrival

\[ P(Tᵣ > t) \]

This is the same as:

\[ P(\text{fewer than r events occur in time } t) \]

Which is:

of Poisson events that occur in time \(t\)

\[ P(N(t) \leq r - 1) \]

Where \(N(t)\) is the number of Poisson events in time \(t\).

\[ P(Tᵣ > t) = \sum_{k=0}^{r-1} \frac{(λt)^k e^{-λt}}{k!} \]

This is the CDF (complement) of the Gamma distribution.

Take the derivative to get the PDF:

\[ f(t) = \frac{λ^r t^{r-1} e^{-λt}}{(r-1)!} \]

This is the Gamma probability density function.

Descriptive Statistics

Type	Formula
Support	\(t \in [0, \infty)\)
Shape Parameter	\(r > 0\) (number of events)
Rate Parameter	\(\lambda > 0\)
Mean	\(\mathbb{E}[T] = \frac{r}{\lambda}\)
Variance	\(\text{Var}(T) = \frac{r}{\lambda^2}\)
Standard Deviation	\(\sigma = \sqrt{\frac{r}{\lambda^2}}\)
Probability Density Function (PDF)	\(f(t) = \frac{\lambda^r t^{r-1} e^{-\lambda t}}{(r-1)!} \quad \text{for } t \geq 0\)
Cumulative Distribution Function (CDF)	\(F(t) = \mathbb{P}(T \leq t) = \int_0^t f(s)\, ds \quad or \quad \frac{\gamma(r, \lambda t)}{\Gamma(r)}\)

Use Case Example

DMV Analogy (Series Case)

• You must speak to three agents, one after another.
• (Mean of Exp. Distribution) Each has a 15-minute exponential wait ⇒ \(λ = \frac{1}{15}\)

Total wait time is the sum of 3 exponential variables:

\[ T \sim \text{Gamma}(3, 1/15) \]

Clarification: Series vs Parallel

Series (Gamma): You wait for 3 stages one after another

e.g. DMV → agent A → agent B → agent C

→ Waiting time follows Gamma

\[ T \sim \text{Gamma}(3, 1/15) \]

Mean and Variance

For \(T \sim \text{Gamma}(r, \lambda)\), where:

• \(r=3\)
• \(\lambda = \frac{1}{15}\)

Then:

• Mean: \(\mathbb{E}[T] = \frac{r}{\lambda} = \frac{3}{1/15} = 45 \text{ minutes}\)
• Variance: \(\text{Var}(T) = \frac{r}{\lambda^2} = \frac{3}{(1/15)^2} = 3 \cdot 225 = 675\)

Probability Density Function (PDF)

The PDF of the Gamma distribution is:

\[ f(t) = \frac{\lambda^r t^{r-1} e^{-\lambda t}}{(r-1)!} \]

Substitute values:

\[ f(t) = \frac{(1/15)^3 \cdot t^2 \cdot e^{-t/15}}{2!} = \frac{1}{2} \cdot \left(\frac{1}{3375}\right) \cdot t^2 \cdot e^{-t/15} \]

So:

\[ f(t) = \frac{t^2}{6750} \cdot e^{-t/15} \]

This gives the probability density at time \(t\), in minutes.

Parallel (Not Gamma, Exponential):

3 open lanes, choose the shortest

→ Minimum of exponential variables

→ Not Gamma, uses Exponential Distribution

For the parallel case, you are taking the minimum of 3 independent exponential wait times, each with: \(\lambda = \frac{1}{15}\)

This gives:

\[ T_{\min} \sim \text{Exponential}(3 \cdot \lambda) = \text{Exponential}\left(\frac{1}{5}\right) \]

Mean and Variance

For \(T \sim \text{Exponential}(\lambda)\), we have:

• Mean: \(\mathbb{E}[T] = \frac{1}{\lambda} = 5 \text{ minutes}\)
• Variance: \(\text{Var}(T) = \frac{1}{\lambda^2} = 25\)

Probability Density Function (PDF)

The PDF is: \(f(t) = \lambda e^{-\lambda t}\)

Substitute \(\lambda = \frac{1}{5}\): \(f(t) = \frac{1}{5} e^{-t/5}\)

This gives the probability density for being helped at time \(t\), when choosing the fastest of 3 lanes.