Question

The following density function defines the gamma distribution: $$ \mathrm{f}(\mathrm{x})=\left[\lambda^{\alpha} /(\Gamma(\alpha))\right] \mathrm{X}^{\alpha-1} \mathrm{e}^{-\lambda \mathrm{x}}, \text { for } \mathrm{X}>0 $$ where \(\alpha\) and \(\lambda\) are positive parameters. \(\Gamma(\mathrm{t})\), the gamma function, is defined by $$ \Gamma(t)={ }^{\infty} \int_{0} x^{t-1} e^{-x} d x \text { for } t>0 $$ Integration by parts yields the following recursion relations: \(\Gamma(\mathrm{t}+1)=\mathrm{t} \Gamma(\mathrm{t})\), and if \(\mathrm{n}\) is an integer \(\Gamma(\mathrm{n}+1)=\mathrm{n} !\). Find the moment generating function, mean, and variance of the gamma distribution.

Short answer

The moment generating function (MGF) for the gamma distribution is \(M(t) = \left(\frac{\lambda}{t-\lambda}\right)^{\alpha}\), the mean is \(\mu = \frac{\alpha}{\lambda}\), and the variance is \(\sigma^2 = \frac{\alpha}{\lambda^2}\).

Step-by-step solution

Solve for the Moment Generating Function (MGF)

We can find the MGF by computing the expected value of \(e^{tx}\) with respect to the gamma distribution's density function \(f(x)\): $$ M(t) = \int_{0}^{\infty} e^{tx}f(x) dx = \int_{0}^{\infty} e^{tx}\left[\frac{\lambda^{\alpha}}{\Gamma(\alpha)}x^{\alpha-1}e^{-\lambda x}\right] dx = \frac{\lambda^{\alpha}}{\Gamma(\alpha)}\int_{0}^{\infty}x^{\alpha-1}e^{(t-\lambda)x} dx $$ To continue with the integration, we consider the integral: $$ \int_{0}^{\infty}x^{\alpha-1}e^{(t-\lambda)x} dx $$ This integral is a gamma function in disguise with parameters \(\alpha\) and \(t-\lambda\). We can rewrite it as: $$ \frac{\Gamma(\alpha)}{(t-\lambda)^{\alpha}} $$ So, the moment generating function is: $$ M(t) = \frac{\lambda^{\alpha}}{\Gamma(\alpha)} \cdot \frac{\Gamma(\alpha)}{(t-\lambda)^{\alpha}} = \left(\frac{\lambda}{t-\lambda}\right)^{\alpha} $$

Find the Mean

To find the mean, we need to take the first derivative of the MGF: $$ \frac{dM(t)}{dt} = \frac{d}{dt}\left(\frac{\lambda}{t-\lambda}\right)^{\alpha} $$ Using the chain rule, we get: $$ \alpha\left(\frac{\lambda}{t-\lambda}\right)^{\alpha-1} \cdot \frac{-\lambda}{(t-\lambda)^2} $$ Now, we evaluate it at \(t=0\): $$ \mu = \frac{dM(t)}{dt}\Big|_{t=0} = \alpha\left(\frac{\lambda}{-\lambda}\right)^{\alpha-1} \cdot \frac{-\lambda}{(-\lambda)^2} = \frac{\alpha}{\lambda} $$

Find the Second Moment

We can find the second moment by taking the second derivative of the MGF: $$ \frac{d^2M(t)}{dt^2} = \frac{d^2}{dt^2}\left(\frac{\lambda}{t-\lambda}\right)^{\alpha} $$ Calculating the second derivative using the chain rule twice, we get: $$ \alpha(\alpha-1)\left(\frac{\lambda}{t-\lambda}\right)^{\alpha-2} \cdot \frac{\lambda^2}{(t-\lambda)^4} $$ Now, we evaluate it at \(t=0\): $$ \mu_2 = \frac{d^2M(t)}{dt^2}\Big|_{t=0} = \alpha(\alpha-1)\left(\frac{\lambda}{-\lambda}\right)^{\alpha-2} \cdot \frac{\lambda^2}{(-\lambda)^4} = \frac{\alpha(\alpha-1)}{\lambda^2} $$

Find the Variance

To find the variance, we use the second moment and the mean: $$ \sigma^2 = \mu_2 - \mu^2 = \frac{\alpha(\alpha-1)}{\lambda^2} - \left(\frac{\alpha}{\lambda}\right)^2 $$ Simplifying, we get: $$ \sigma^2 = \frac{\alpha}{\lambda^2} $$ In conclusion, the moment generating function for the gamma distribution is \(M(t) = \left(\frac{\lambda}{t-\lambda}\right)^{\alpha}\), the mean is \(\mu = \frac{\alpha}{\lambda}\), and the variance is \(\sigma^2 = \frac{\alpha}{\lambda^2}\).

Key concepts

Moment Generating Function

The Moment Generating Function (MGF) is crucial for understanding the distribution of random variables. For the gamma distribution, we start by finding this function. The MGF, denoted as \( M(t) \), is the expected value of \( e^{tx} \) with respect to the distribution's density function \( f(x) \).

To obtain the MGF for the gamma distribution, we need to solve the integral:

\[M(t) = \int_{0}^{\infty} e^{tx}f(x) dx = \int_{0}^{\infty} e^{tx}\left[\frac{\lambda^{\alpha}}{\Gamma(\alpha)}x^{\alpha-1}e^{-\lambda x}\right] dx\]

Inside the integral, notice the simplified form:

\[\int_{0}^{\infty}x^{\alpha-1}e^{(t-\lambda)x} dx = \frac{\Gamma(\alpha)}{(t-\lambda)^{\alpha}}\]

This actually represents a gamma function with parameters \( \alpha \) and \( t-\lambda \). Plugging this back into our formula gives the MGF:

\[M(t) = \left(\frac{\lambda}{t-\lambda}\right)^{\alpha}\]

The MGF provides a pathway to computing the mean and variance of the distribution. Understanding the form of MGF can help you identify the distribution and its properties quickly.

• The MGF shows how the gamma distribution scales with \( \lambda \) and \( \alpha \).
• The gamma function inside displays its connection to factorials, especially when dealing with integer shape parameters.

Mean and Variance Calculations

Mean and variance offer statistical insight into the distribution of random variables. Calculating these from the MGF of the gamma distribution involves differentiation.

**Mean Calculation**
To find the mean of a gamma distribution, we derive the MGF once and evaluate it at \( t = 0 \). Differentiating the MGF gives:

\[\frac{dM(t)}{dt} = \alpha \left(\frac{\lambda}{t-\lambda}\right)^{\alpha-1} \cdot \frac{-\lambda}{(t-\lambda)^2}\]

Evaluating at \( t=0 \) gives the mean \( \mu \):

\[\mu = \frac{\alpha}{\lambda}\]

This shows the balance between the shape parameter \( \alpha \) and the rate parameter \( \lambda \).

**Variance Calculation**
The variance provides insight into the spread of the distribution. We calculate the second moment by differentiating the MGF twice:

\[\frac{d^2M(t)}{dt^2} = \alpha(\alpha-1)\left(\frac{\lambda}{t-\lambda}\right)^{\alpha-2} \cdot \frac{\lambda^2}{(t-\lambda)^4}\]

Evaluating this at \( t=0 \) gives the second moment \( \mu_2 \). Then, variance \( \sigma^2 \) is:

\[\sigma^2 = \mu_2 - \mu^2 = \frac{\alpha}{\lambda^2}\]

These calculations show how the gamma distribution's shape and scale parameters determine its mean and variance.

• Mean is directly proportional to the shape parameter \( \alpha \).
• Variance increases with \( \alpha \) showing increased spread for larger \( \alpha \).

Gamma Function Recursion Relations

The gamma function \( \Gamma(t) \) is a continuous extension of the factorial function \((n!)\), essential in defining the gamma distribution. Its recursion relation plays a key role in simplifying and computing various aspects of the gamma distribution.

The gamma recursion relation is given by:

\[\Gamma(t+1) = t \Gamma(t)\]

This relation allows for calculating the gamma function for various values of \( t \) by reducing it step by step until it reaches a known value, like \( \Gamma(1) = 1 \). For integer values \( n \), the gamma function simplifies to:

\[\Gamma(n+1) = n!\]

These relations make computing MGFs and other properties of distributions involving the gamma function much simpler. In practice:

• When \( t \) is a whole number, \( \Gamma(t) \) gives \( (t-1)! \), mirroring factorial calculations.
• For non-integers, the recursion helps in computing values without directly integrating the gamma function definition.
  
  The role of the gamma function is major in probability and statistics, especially for handling continuous and Poisson processes.