Problem 95 - Applied Statistics and Probability for Engineers

Question

A continuous random variable \(X\) has the following probability distribution: $$ f(x)=4 x e^{-2 x}, \quad x>0 $$ (a) Find the moment-generating function for \(X\). (b) Find the mean and variance of \(X\).

Step-by-Step Solution

Answer

(a) MGF is \( \frac{4}{(2-t)^2} \) for \( t < 2 \); (b) Mean is 1, variance is 0.5.

Step 1: Identify the Moment-Generating Function (MGF) Formula

The moment-generating function (MGF) for a continuous random variable is defined as \( M_X(t) = E[e^{tX}] = \int_{-fty}^{fty} e^{tx}f(x)dx \). For our given density function \( f(x) = 4x e^{-2x} \), we integrate over the range \( x>0 \). Thus, the expression becomes \( M_X(t) = \int_0^{fty} e^{tx} \cdot 4x e^{-2x} dx \).

Step 2: Simplify and Solve the Integral for MGF

The integral becomes \( \int_0^{fty} 4x e^{(t-2)x} dx \). This is a standard form \( \int x e^{ax} dx \), which evaluates to: \( \left[ \frac{x e^{ax}}{a} - \frac{e^{ax}}{a^2} \right]_0^{fty} \). Calculating with limits, we finally get the simplified form of the MGF, which evaluates to \( M_X(t) = \frac{4}{(2-t)^2} \) for \( t < 2 \).

Step 3: Find the Mean Using the MGF

The mean \( \mu \) of the distribution can be found using the MGF by taking the first derivative at \( t = 0 \). This is given by \( \mu = M_X'(0) \). Therefore, we have \( M_X'(t) = \frac{8}{(2-t)^3} \). Evaluating this at \( t = 0 \), \( \mu = \frac{8}{8} = 1 \).

Step 4: Find the Variance Using the MGF

To find the variance \( \sigma^2 \), we first take the second derivative of the MGF: \( M_X''(t) = \frac{24}{(2-t)^4} \). Then, evaluate \( M_X''(0) \), which gives \( \frac{24}{16} = 1.5 \). We've \( \text{Var}(X) = M_X''(0) - [M_X'(0)]^2 = 1.5 - 1^2 = 0.5 \).

Key Concepts

Moment-Generating Function

The Moment-Generating Function (MGF) is a powerful tool in probability and statistics used to derive moments of a probability distribution. For a continuous random variable like our example, the MGF is defined as \( M_X(t) = E[e^{tX}] \). This translates to evaluating the integral of \( e^{tx} \) multiplied by the probability density function \( f(x) \).
In our case, we start with the given density function \( f(x) = 4x e^{-2x} \) for \( x > 0 \). The MGF is computed by solving the integral

• \( M_X(t) = \int_0^{\infty} e^{tx} \, 4x e^{-2x} \, dx \)
• This integral, after applying integration techniques like substitution or by recognizing the pattern, simplifies to \( M_X(t) = \frac{4}{(2-t)^2} \)

It's crucial to note that this result holds only for \( t < 2 \), ensuring the convergence of the integral.
Thus, MGFs are extremely useful in finding other properties like the mean and variance efficiently.

Mean and Variance

Once we have the MGF, determining the mean and variance becomes straightforward. These are fundamental characteristics of the distribution of a random variable, highlighting its central tendency and spread, respectively.
To find the mean \( \mu \), we take the first derivative of the MGF with respect to \( t \) and evaluate it at \( t = 0 \). For the given MGF \( M_X(t) = \frac{4}{(2-t)^2} \), the first derivative is

• \( M_X'(t) = \frac{8}{(2-t)^3} \)
• When \( t = 0 \), \( M_X'(0) = \frac{8}{8} = 1 \), indicating a mean of 1

Variance \( \sigma^2 \) is derived from the second derivative. It indicates how much values vary from the mean.
We first calculate

• \( M_X''(t) = \frac{24}{(2-t)^4} \)
• Evaluate it at zero, \( M_X''(0) = \frac{24}{16} = 1.5 \)

Variance is then given by the formula: \( \text{Var}(X) = M_X''(0) - [M_X'(0)^2] = 1.5 - 1^2 = 0.5 \)
Mean and variance, obtained from derivatives of the MGF, provide a clear, concise measure of the distribution characteristics.

Continuous Random Variables

Continuous random variables are variables that can take an infinite number of values within a given range. These are contrasted with discrete random variables, which can only take particular values. Understanding continuous random variables is essential in probability theory, as they describe phenomena that are not restricted to distinct outcomes.
For continuous random variables, a probability density function (PDF), as opposed to a probability mass function in discrete cases, describes the likelihood of different outcomes. Such a function must satisfy

• It integrates to 1 over all possible values
• Function values are always non-negative\( f(x) \geq 0 \)

The given function \( f(x) = 4x e^{-2x} \) outlines a specific example, where the range is \( x > 0 \). You integrate this function across its range to calculate probabilities, such as in deriving the MGF.
Continuous random variables play a vital role across numerous applications, from natural sciences to engineering, giving a more precise model for non-discrete phenomena.