# Intro

Within this post, I’ll explore the properties of the Gamma distribution. The results presented here are interesting as they ripple throughout mathematical statistics. Each result has a proof associated with it in hopes of better how the result came to be. Below is a preview of the posts contents.

• Gamma in real life
• Parameterizations of Gamma
• Definition 1: $X \sim Gamma\left({\alpha,\beta}\right), \quad f\left( x \right) = \frac{1}{ {\Gamma \left( \alpha \right){\theta ^\alpha } } }{x^{\alpha - 1} }\exp \left( { - \frac{x}{\theta } } \right), \, x > 0$
• Definition 2: $X \sim Gamma\left({\alpha,\frac{1}{\lambda} }\right), \quad f\left( x \right) = \frac{ { {\lambda ^\alpha } } }{ {\Gamma \left( \alpha \right)} }{x^{\alpha - 1} }\exp \left( { - \lambda x} \right), \, x > 0$
• Derivation of Gamma Distribution
• Examples with Gamma
• Integrating the Gamma PDF
• Exploiting the Gamma-Poisson Relationship
• Gamma as a means to an end for integration by parts.
• Gamma Function
• Proposition 1: $\Gamma \left( 1 \right) = 1$
• Proposition 2: For any $\alpha >0$, we have: $\Gamma \left( {\alpha + 1} \right) = \alpha \Gamma \left( \alpha \right)$.
• Proposition 3: For any integer $\alpha$, we have: $\Gamma \left( {\alpha + 1} \right) = \alpha !$
• (Bonus) Proposition 4: $\Gamma \left( {\frac{1}{2} } \right) = \sqrt \pi$.
• Lemma 1: Guassian Integral $I = \int\limits_0^\infty {\exp \left( { - {x^2} } \right)dx} = \frac{ {\sqrt \pi } }{2}$
• Verifying the Gamma PDF
• Theorem 1: If $X \sim Gamma \left({\alpha,\theta}\right)$ with $\alpha >0$ and $\theta >0$, then $\int\limits_0^\infty {\frac{1}{ {\Gamma \left( \alpha \right){\theta ^\alpha } } }{x^{\alpha - 1} }\exp \left( { - \frac{x}{\theta } } \right)dx} = 1$
• Gamma Cumulative Distribution Function (CDF): $P\left( {X \le x} \right) = \frac{ {\gamma \left( {\alpha ,\frac{x}{\theta } } \right)} }{ {\Gamma \left( \alpha \right)} }$
• Exploring the Moments
• Theorem 2: The $k$-th moment of a Gamma distribution is $\left( {\alpha + k - 1} \right)\left( {\alpha + k - 2} \right) \cdots \alpha {\theta ^k}$
• Theorem 3: The Moment Generating Function of a Gamma distribution is $\frac{1}{ { { {\left( {1 - \theta t} \right)}^\alpha } } }$ if $t < \frac{1}{\theta}$.
• Theorem 4: The Characteristic Function of a Gamma distribution is $\frac{1}{ { { {\left( {1 - it\theta } \right)}^\alpha } } }$.
• Gamma Distribution Properties
• Scaling Property: $cX \sim Gamma\left({\alpha, c\theta}\right)$
• Mean: $\alpha \theta$
• Variance: $\alpha {\theta ^2}$
• Skewness: $\frac{2}{ {\sqrt \alpha } }$
• Kurtosis: $3 + \frac{6}{\alpha }$
• Excess Kurtosis: $\frac{6}{\alpha }$
• Estimators
• Maximum Likelihood Estimation: ${ {\hat \theta }_{MLE} } = \frac{ {\bar x} }{\alpha }$, no closed form for $\alpha$.
• Method of Moments Estimation: $\tilde \alpha = \frac{ {n{ {\bar X}^2} } }{ {\sum\limits_{i = 1}^n { { {\left( { {X_i} - \bar X} \right)}^2} } } },\tilde \theta = \frac{ {\sum\limits_{i = 1}^n { { {\left( { {X_i} - \bar X} \right)}^2} } } }{ {n\bar X} }$
• Lemma 2: $\sum\limits_{i = 1}^n {X_i^2} - n{ {\bar X}^2} = \sum\limits_{i = 1}^n { { {\left( { {X_i} - \bar X} \right)}^2} }$
• Related distributions
• If $X \sim Gamma\left({1, \frac{1}{\lambda } }\right)$, then $X$ has an exponential distribution with rate parameter $\lambda$.
• If $X \sim Gamma\left({\frac{\nu}{2}, 2}\right)$, then $X$ is a chi-squared distribution with $\nu$ degrees of freedom, $\chi ^2\left({\nu}\right)$.
• If $X_i \sim Gamma\left({\alpha_i, \theta }\right)$, then $\sum\limits_{i = 1}^n { {X_i} } \mathop \sim \limits^{iid} Gamma\left({ \sum\limits_{i = 1}^n { {\alpha_i} } , \theta }\right)$.
• If $X_i \sim Exponential\left({ {\lambda } }\right)$, then $\sum\limits_{i = 1}^n { {X_i} } \mathop \sim \limits^{iid} Gamma\left({n, \frac{1}{\lambda } }\right)$.
• Gamma in R
• Misc
• Misc 1: $0 = \sum\limits_{i = 1}^n {\left( { {X_i} - \bar X} \right)}$

## Gamma in real life

Examples of the Gamma random variable are:

1. The time it takes for an event to occur, e.g. waiting to check out or for assistance.
2. Accumulation of a quantity in a given time period, e.g. the amount of goals scored or accidents in a work place.

## Parameterizations of Gamma

We say that a random variable $X$ has a Gamma distribution with parameters $\alpha > 0$ and $\theta > 0$ if its probability density function has the form:

### Definition 1: $X \sim Gamma\left({\alpha,\beta}\right), \quad f\left( x \right) = \frac{1}{ {\Gamma \left( \alpha \right){\theta ^\alpha } } }{x^{\alpha - 1} }\exp \left( { - \frac{x}{\theta } } \right), \, x > 0$

Alternatively, we can parameterize Gamma in a different way by using $\theta = \frac{1}{\lambda}$ giving the probability density function the form:

### Definition 2: $X \sim Gamma\left({\alpha,\frac{1}{\lambda} }\right), \quad f\left( x \right) = \frac{ { {\lambda ^\alpha } } }{ {\Gamma \left( \alpha \right)} }{x^{\alpha - 1} }\exp \left( { - \lambda x} \right), \, x > 0$

Either of these parameterizations is okay!

We often refer to $\alpha$ as the shape parameter and $\theta$ as the scale parameter. Note $\alpha$ and $\theta$ are not necessarily integers.

Note: $\Gamma \left( \alpha \right)$ is the Gamma function!

## Derivation of Gamma Distribution

Suppose $X \sim Poisson\left(\lambda\right)$. That is, events happen within a poisson process at rate $\lambda$ for a unit of time. So, $f\left(x\right) = {\frac{ { { {\left( {\lambda} \right)}^x}\exp \left( { - \lambda} \right)} }{ {x!} } }$

Now, let $T$ represent the length of time until $\alpha$ event. Since $T$ is representing time, we are working with a continuous random variable. So, the cumulative distribution function (CDF) that models the probability that there are at least $\alpha$ events in the time $t$ is given by: $F\left( t \right) = P\left( {T \le t} \right) = 1 - P\left( {T > t} \right)$

Note $P\left( {T > t} \right)$ is the probability of fewer than $\alpha$ changes in the interval $\left[0, t\right]$. To find $T$ we need to use the Poisson process exhibited by $X$ with an extension to time.

When speaking about an extension to time, remember that $X$ was only acting on a unit of time. However, to support $T$, $X$ must act on the interval $\left[0, t\right]$. Using the fact that the events are Poisson and independent of each other, the rate is updated to be: $\lambda + \lambda + \cdots + \lambda = \lambda t$. So, the poisson process is now $X\sim Poisson\left({\lambda t}\right)$. Thus, $f\left(x\right) = {\frac{ { { {\left( { \lambda t} \right)}^x}\exp \left( { - \lambda t} \right)} }{ {x!} } }$

With this knowledge, the connection of the CDF to the poisson process is possible:

In a nutshell, the above statement represents equality of probabilities between observing the $\alpha$-th event after time $t$ and observing less than $\alpha$ events from now until time $t$.

Taking the derivative with respect to $t$ yields the probability density function of $T$:

Simplifying the series gives:

Returning:

Note: $\Gamma \left( \alpha \right) = \left( {\alpha - 1} \right)!$ is the Gamma function for integers.

Why are we using integers? Well, $\alpha$ is defined to be the number of events. Since poisson is a discrete random variable, the events must be discrete (i.e. integers). See the Gamma function section for more information!

Now, if we set $\lambda = \frac{1}{\theta }$, then we receive the first parameterization of the Gamma Distribution:

## Examples with Gamma

### Integrating the Gamma PDF

Students ask questions in class according to Poisson process with the average rate of one question per 20 minutes. Find the probability that the third question is asked during the last 10 minutes of a 50-minute class.

Using only the Gamma Distribution’s integral definition we have:

### Exploiting the Gamma-Poisson Relationship

Within the derivation section of Gamma, the starting process used to create the gamma distribution was that of a poisson process. From the poisson process, we are able to obtain the following shortcuts that enable us to avoid calculating out the gamma distribution via integration by parts.

Therefore:

### Gamma as a means to an end for integration by parts.

When doing integration by parts, if the integral is set up as follows: $\int\limits_0^\infty { {x^{\alpha - 1} }\exp \left( { - \frac{x}{\theta } } \right)dx}$ We are able to coerce the integral to be one using the definition of the gamma distribution, while in turn solving it!

General method for solving the integral:

To see the verification of the Gamma distribution integrating to 1, see Theorem 1. Here’s an example of using the integral against itself:

## Gamma Function

Before we begin, note that there is a function called the Gamma function, which is defined as:

Definition 3: $\Gamma \left( \alpha \right) = \int\limits_0^\infty { {x^{\alpha - 1} }\exp \left( { - x} \right)dx} , \quad \alpha > 0$

From here we have:

Proof:

Proof:

### Proposition 3: For any integer $\alpha$, we have: $\Gamma \left( {\alpha + 1} \right) = \alpha !$

Using the results from 1. and 2. in addition to the assumption that $\alpha$ is an integer we have: $\Gamma \left( {\alpha + 1} \right) = \alpha ! = \alpha \cdot \left( {\alpha - 1} \right) \cdot \left( {\alpha - 2} \right) \cdots 3 \cdot 2 \cdot 1$

, which is the factorial function.

Proof:

### (Bonus) Proposition 4: $\Gamma \left( {\frac{1}{2} } \right) = \sqrt \pi$.

For dealing with half-integers, the Gamma function is able to have the same recusive process applied. In this case, the base value for the Gamma function is given by: $\Gamma \left( {\frac{1}{2} } \right) = \sqrt \pi$.

Before we delve into the proof, we will need to prove a lemma regarding the Guassian Integral.

### Lemma 1: Guassian Integral $I = \int\limits_0^\infty {\exp \left( { - {x^2} } \right)dx} = \frac{ {\sqrt \pi } }{2}$

Using polar coordinates we have:

Proof for (Bonus) Proposition 4:

## Verifying the Gamma PDF

### Theorem 1: If $X \sim Gamma \left({\alpha,\theta}\right)$ with $\alpha >0$ and $\theta >0$, then $\int\limits_0^\infty {\frac{1}{ {\Gamma \left( \alpha \right){\theta ^\alpha } } }{x^{\alpha - 1} }\exp \left( { - \frac{x}{\theta } } \right)dx} = 1$

Proof:

Theorem 1 is very powerful for using Gamma integrals against themselves as you will see…

## Exploring the Moments

### Theorem 3: The Moment Generating Function of a Gamma distribution is $\frac{1}{ { { {\left( {1 - \theta t} \right)}^\alpha } } }$ if $t < \frac{1}{\theta}$.

The restriction on $t$ is avoids negative values and zero, which are not supported by the Gamma function.

## Gamma Distribution Properties

### Scaling Property: $cX \sim Gamma\left({\alpha, c\theta}\right)$

Consider the standard Gamma Distribution, $X\sim Gamma\left( {\alpha ,\theta } \right)$,

We are looking to perform a transformation such that: $Y = g\left({X}\right) = cX$. To apply the transformation formula:

We need to find $X = g^{-1}\left({Y}\right)$ and the Jacobian.

Therefore,

### Mean: $\alpha \theta$

Using $k$-th moment given by Theorem 2 we have the mean as: $E\left[ { {X^1} } \right] = \alpha \theta$

Alternatively,

Using the moment generating function given by Theorem 3 we have the mean as:

### Variance: $\alpha {\theta ^2}$

Using $k$-th moment given by Theorem 2 we have the second moment as as: $E\left[ { {X^2} } \right] = \left( {\alpha + 1} \right)\alpha {\theta ^2} = {\alpha ^2}{\theta ^2} + \alpha {\theta ^2}$ and using the result from the calculation of the mean from the $k$-th moment we have:

Alternatively,

Using the moment generating function given by Theorem 3 we have:

and using the result from the calculation of the mean from the moment generating function we have:

### Skewness: $\frac{2}{ {\sqrt \alpha } }$

To obtain a distribution’s skewness, we need to find the third standardized moment: $E\left[ { { {\left( {\frac{ {X - \mu } }{\sigma } } \right)}^3} } \right]$

The moment can be rewritten as: $E\left[ { { {\left( {\frac{ {X - \mu } }{\sigma } } \right)}^3} } \right] = \frac{1}{ { {\sigma ^3} } }E\left[ { { {\left( {X - \mu } \right)}^3} } \right]$

Focusing on the centeralized third moment given by: $E\left[ { { {\left( {X - \mu } \right)}^3} } \right] = E\left[ { {X^3} - 3\mu {X^2} + 3{\mu ^2}X - {\mu ^3} } \right] = E\left[ { {X^3} } \right] - 3\mu E\left[ { {X^2} } \right] + 2{\mu ^3}$

Using Theorem 2, we obtain the $k$-th moments for 1, 2, and 3: %

Therefore,

Note, the variance of a gamma random variable is given by $Var\left( X \right) = \alpha {\theta ^2}$. Therefore, the standard deviation for gamma is: $SD\left( X \right) = \theta \sqrt{\alpha}$

Returning:

### Kurtosis: $3 + \frac{6}{\alpha }$

To obtain a distribution’s Kurtosis: $\beta_2 = \frac{ {E\left[ { { {\left( {X - \mu } \right)}^4} } \right]} }{ { { {\left( {E\left[ { { {\left( {X - \mu } \right)}^2} } \right]} \right)}^2} } } = \frac{ {E\left[ { { {\left( {X - \mu } \right)}^4} } \right]} }{ { { {\left( {Var\left( X \right)} \right)}^2} } }$

First, we obtain focus on the centeralized fourth moment: %

Using Theorem 2, we obtain the $k$-th moments for 1, 2, 3, and 4: %

Therefore,

Note, the variance of a gamma random variable is given by $Var\left( X \right) = \alpha {\theta ^2}$.

Returning,

### Excess Kurtosis: $\frac{6}{\alpha }$

Excess Kurtosis is a slight modification of Kurtosis. ${\beta _2} = \frac{ {E\left[ { { {\left( {X - \mu } \right)}^4} } \right]} }{ { { {\left( {Var\left( X \right)} \right)}^2} } } - 3$

The notable difference between Kurtosis and Excess Kurtosis is the later has “minus 3” appended at the end of its formula. The correction is often attributed to the fact that the kurtosis of a Normal Distribution is 3. Thus, to make the Kurtosis of the Normal equal to zero, 3 is subtracted.

Hence, the Gamma Distribution’s Excess Kurtosis is: %

## Estimators

### Maximum Likelihood Estimation: ${ {\hat \theta }_{MLE} } = \frac{ {\bar x} }{\alpha }$, no closed form for $\alpha$.

1. Obtain the likelihood function: %

2. Obtain the log likelihood function: %

3. Find the maximum value for $\theta$: %

4. Substitute in maximum value for $\theta$ to likelihood function: %

Note: $\frac{\partial }{ {\partial \alpha } }\log \left( {\Gamma \left( \alpha \right)} \right) = \frac{ {\frac{\partial }{ {\partial \alpha } }\left( {\Gamma \left( \alpha \right)} \right)} }{ {\Gamma \left( \alpha \right)} } = \frac{ {\Gamma \left( \alpha \right){\psi ^{\left( 0 \right)} }\left( \alpha \right)} }{ {\Gamma \left( \alpha \right)} } = {\psi ^{\left( 0 \right)} }\left( \alpha \right)$

1. Find the maximum value for $\alpha$.

There is no closed-form solution for maximizing $\alpha$. =(

### Method of Moments Estimation: $\tilde \alpha = \frac{ {n{ {\bar X}^2} } }{ {\sum\limits_{i = 1}^n { { {\left( { {X_i} - \bar X} \right)}^2} } } },\tilde \theta = \frac{ {\sum\limits_{i = 1}^n { { {\left( { {X_i} - \bar X} \right)}^2} } } }{ {n\bar X} }$

First, obtain the 1st and 2nd theoretical moment using Theorem 2 to obtain:

Secondly, obtain the 1st and 2nd sample moments:

Thirdly, equate the theoretical moments with their respective sample moment:

Fourth, solve for the most straight forward parameter. In this case, solve for $\theta$.

Fiveth, substitute the parameter into the second equation and solve:

Note the other form of: $\tilde \theta = \frac{ {\frac{1}{n}\sum\limits_{i = 1}^n {X_i^2} - { {\bar X}^2} } }{ {\bar X} }$ is okay. However, the preferred form is that of the difference since it slows the amount of floating point errors (i.e. numerical unstability).

Therefore, the method of moments estimators for Gamma are:

Proof:

### If $X \sim Gamma\left({1, \frac{1}{\lambda } }\right)$, then $X$ has an exponential distribution with rate parameter $\lambda$.

Proof

First, note that the alternative parameterization of Gamma gives:

Then by setting $\alpha = 1$, we have:

### If $X \sim Gamma\left({\frac{\nu}{2}, 2}\right)$, then $X$ is a chi-squared distribution with $\nu$ degrees of freedom, $\chi ^2\left({\nu}\right)$.

Proof

Setting $\alpha = \frac{v}{2}$ gives: %

### If $X_i \sim Gamma\left({\alpha_i, \theta }\right)$, then $\sum\limits_{i = 1}^n { {X_i} } \mathop \sim \limits^{iid} Gamma\left({ \sum\limits_{i = 1}^n { {\alpha_i} } , \theta }\right)$.

Proof

Proof by Induction using convolution theorem

Basis: $X _1$ is a single gamma random variable. Therefore, $X _1$ has a gamma distribution of $Gamma\left({ \alpha_1, \theta }\right)$.

Induction: Suppose that $S_n = \sum\limits_{i = 1}^n { {X_i} }$ is of independent and identically distributed gamma random variables so that it has the distribution:

Let ${X_{n + 1} }$ a single gamma random variable independent from $S _n$ and identically distributed. Then, we expect the combined distribution of $S _n + X _{n+1}$ to be:

Proof

Proof by Induction using MGFs

Corollary

## Gamma in R

Command Result
dgamma(x, shape=alpha, scale=beta) $f\left( x \right)$
pgamma(q, shape=alpha, scale=beta) $P\left( {X \le x} \right)$
qgamma(p, shape=alpha, scale=beta) ${\phi _p} \mathrel\backepsilon P\left( {X \le {\phi _p} } \right) = p$
rgamma(x, shape=alpha, scale=beta) ${X_1}, \ldots ,{X_n}\sim Gamma\left( {\alpha ,\beta } \right)$

Proof: %

Proof: %