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Chapter 5: Problem 8

A random variable \(X\) has the gamma distribution $$ f(x)=\frac{\lambda}{\Gamma(r)}(\lambda x)^{r-1} e^{-\lambda x}, \quad x>0 $$ a. Show that the moment-generating function for \(t<\lambda\) of \(X\) is $$ M_{X}(t)=\left(1-\frac{t}{\lambda}\right)^{-r} $$ b. Find the mean and variance of \(X\).

Short Answer

Expert verified
The MGF is \( M_X(t) = \left(1 - \frac{t}{\lambda}\right)^{-r} \). The mean is \( \frac{r}{\lambda} \) and the variance is \( \frac{r}{\lambda^2} \).

Step by step solution

01

Understanding the Moment-Generating Function Definition

The moment-generating function (MGF) of a random variable is defined as \( M_X(t) = E[e^{tX}] \). For a gamma-distributed random variable with probability density function \( f(x) \), the MGF is calculated as \[ M_X(t) = \int_{0}^{\infty} e^{tx} f(x) \, dx. \]
02

Writing the Integral for MGF

Substitute the given probability density function of the gamma distribution into the integral for the MGF: \[ M_X(t) = \int_{0}^{\infty} e^{tx} \frac{\lambda}{\Gamma(r)}(\lambda x)^{r-1} e^{-\lambda x} \, dx = \frac{\lambda^r}{\Gamma(r)} \int_{0}^{\infty} x^{r-1} e^{-(\lambda - t)x} \, dx. \]
03

Simplifying the Integral to Match Gamma Integral Form

To simplify the integral, observe that it is in the form needed for a gamma integral with parameterization. Recognize that \[ \int_{0}^{\infty} x^{r-1} e^{-\alpha x} \, dx = \frac{\Gamma(r)}{\alpha^r}, \] where \( \alpha = \lambda - t \).
04

Applying the Gamma Integral Result

Apply the gamma integral result to the equation: \[ \frac{\lambda^r}{\Gamma(r)} \int_{0}^{\infty} x^{r-1} e^{-(\lambda - t)x} \, dx = \frac{\lambda^r}{\Gamma(r)} \times \frac{\Gamma(r)}{(\lambda-t)^r}. \] The \( \Gamma(r) \) terms cancel out, leading to \[ M_X(t) = \left( \frac{\lambda}{\lambda-t} \right)^r = \left( 1 - \frac{t}{\lambda} \right)^{-r}. \]
05

Using the MGF to Find Mean and Variance

The mean of a gamma distribution \(X\) with shape parameter \(r\) and rate \(\lambda\) is \( E[X] = \frac{r}{\lambda} \). This can be derived using the MGF derivative at \(t=0\), which is \( M_X'(0) \). The variance is found similarly, as \( \text{Var}(X) = \frac{r}{\lambda^2} \).
06

Confirming Mean and Variance Formulas

To confirm: Using the MGF's first derivative, \( M_X'(t) \), calculate at \( t = 0 \) to find mean. Similarly, use second derivative \( M_X''(0) \) to derive variance, confirming \[ E[X^2] = \frac{r(r+1)}{\lambda^2}. \] Then, \( \text{Var}(X) = E[X^2] - (E[X])^2 = \frac{r}{\lambda^2}. \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Moment-Generating Function
In probability theory, the moment-generating function (MGF) is a powerful tool that helps us understand the properties of a random variable. It is defined as the expectation of the exponential function of the random variable, namely \( M_X(t) = E[e^{tX}] \). For a gamma-distributed random variable, the MGF allows us to decipher key characteristics like mean and variance.
The process to derive the MGF involves substituting the gamma probability density function (pdf) into the formula for the MGF and then solving the resulting integral.
  • First, express the MGF for a gamma distribution as \( M_X(t) = \int_{0}^{\infty} e^{tx} f(x) \, dx \), where \( f(x) = \frac{\lambda}{\Gamma(r)}(\lambda x)^{r-1} e^{-\lambda x} \).
  • Next, simplify the integral to fit the form of a standard gamma integral, \( \int_{0}^{\infty} x^{r-1} e^{-\alpha x} \, dx = \frac{\Gamma(r)}{\alpha^r} \) with \( \alpha = \lambda - t \).
  • Finally, applying the properties of gamma integrals gives \( M_X(t) = \left( 1 - \frac{t}{\lambda} \right)^{-r} \), demonstrating how the MGF captures essential characteristics of the distribution.
Understanding the MGF simplifies the process of finding expected values and variances specialized for the gamma distribution.
Mean and Variance
The gamma distribution is defined by its shape parameter \( r \) and its rate parameter \( \lambda \). Using the properties of the MGF helps derive the mean and variance straightforwardly.
Here is how you can calculate these properties:
  • The mean or expected value of a gamma-distributed variable \( X \) is given by \( E[X] = \frac{r}{\lambda} \). This is derived directly from evaluating the first derivative of the MGF, \( M_X'(t) \), at \( t = 0 \).
  • The variance, which gives insight into the spread of the distribution, is \( \text{Var}(X) = \frac{r}{\lambda^2} \). This is determined by taking the second derivative of the MGF, \( M_X''(t) \), and calculating at \( t = 0 \), then using the formula \( \text{Var}(X) = E[X^2] - (E[X])^2 \).
The mean reflects the center of the distribution, while the variance shows how much the data might vary around that mean. Knowing these helps in understanding the distribution's behavior in diverse scenarios, such as reliability analysis and queuing models.
Probability Density Function
The probability density function (pdf) of the gamma distribution is the backbone of understanding the behavior of the distribution. The formula for the gamma pdf is expressed as:
\[ f(x) = \frac{\lambda}{\Gamma(r)}(\lambda x)^{r-1} e^{-\lambda x}, \quad x>0 \]
This form emphasizes several important factors:
  • The term \( \Gamma(r) \) is the gamma function evaluated at \( r \), which is a generalization of the factorial function to the continuous case.
  • The expression \( (\lambda x)^{r-1} \) governs the shape, influenced by \( r \), the shape parameter.
  • The exponential term \( e^{-\lambda x} \) ensures the function properly describes the likelihood over continuous positive variables.
The gamma pdf is crucial for modeling situations where you have to deal with continuous data that follows a skewed pattern, often where waiting times and life durations are involved. It acts as the foundation for probability calculations, allowing us to derive expectation and variance, making it a versatile tool in statistical analysis.

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Most popular questions from this chapter

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