Problem 262 - An Introduction to Mathematical Statistics and Its Applications

Question

. Let \(Y\) have pdf $$ f_{Y}(y)= \begin{cases}y, & 0 \leq y \leq 1 \\ 2-y, & 1 \leq y \leq 2 \\ 0, & \text { elsewhere }\end{cases} $$ Find \(M_{Y}(t)\).

Step-by-Step Solution

Answer

The moment generating function \(M_{Y}(t)\) of random variable \(Y\) is given by \[M_{Y}(t) = \frac{3e^t - e^{2t} - 1 - 3t}{t^2}\] if \(t \neq 0\) and 1 if \(t = 0\).

Step 1: Understanding the pdf of Y

The pdf of the random variable \( Y \) is piecewise-defined as: \( f_{Y}(y) = y \) for \( y \) in range \([0,1]\), \( f_{Y}(y) = 2 - y \) when \( y \) is in range \([1,2]\), and \( f_{Y}(y) = 0 \) elsewhere.

Step 2: Understanding Moment Generating Function \(M_{Y}(t)\)

The moment generating function is a powerful tool that can be used to help calculate probabilities and characteristics of a random variable, given its probability density function. The \(M_{Y}(t)\) of a random variable Y is defined as \(M_{Y}(t) = E[e^{tY}]\), which is derived by integrating the expression \( e^{ty}f_{Y}(y) \) over the domain where \( f_{Y}(y) \) is non-zero. In this case, it means integrating over the intervals [0,1] and [1,2].

Step 3: Computing \(M_{Y}(t)\) over interval [0,1]

To calculate \(M_{Y}(t)\), we start by finding the integral over the interval [0,1]. This integral becomes:\[\int_{0}^{1} e^{ty}y dy\]This is a straightforward integral that, by using integration by parts, gives the result: \[\frac{e^t - 1 - t}{t^2}\] if \(t \neq 0\) and 0.5 if \(t = 0\).

Step 4: Computing \(M_{Y}(t)\) over interval [1,2]

Next, compute the integral over the second interval [1,2]. \[\int_{1}^{2} e^{ty}(2-y) dy\]This is a more complex integral that involves both integration by parts and substitution, which gives \[\frac{2e^t - e^{2t} - 2t}{t^2}\] if \(t \neq 0\) and 0.5 if \(t = 0\).

Step 5: Adding the results

Find \(M_{Y}(t)\) by adding the results of the two integrals. For \(t \neq 0\), this gives: \[M_{Y}(t) = \frac{e^t - 1 - t}{t^2} + \frac{2e^t - e^{2t} - 2t}{t^2} = \frac{3e^t - e^{2t} - 1 - 3t}{t^2}\]And for \(t = 0\), you have \(M_{Y}(0) = 0.5 + 0.5 = 1\) which is the expected result, as the moment generating function evaluated at 0 is always 1 for a random variable.

Key Concepts

Piecewise-Defined Function

A piecewise-defined function is a function that is defined by different expressions over different intervals of its domain. This allows us to describe complex behaviors with simple, manageable parts. For the probability density function (pdf) of the random variable \( Y \), it is defined piecewise as:

• \( f_{Y}(y) = y \) for \( y \in [0, 1] \)
• \( f_{Y}(y) = 2 - y \) for \( y \in [1, 2] \)
• \( f_{Y}(y) = 0 \) for \( y \) outside [0, 2]

This means that the function changes its definition depending on the range of \( y \). It's essential to carefully follow these intervals to accurately compute functions like moment generating functions.

Probability Density Function

A probability density function (pdf) is a function that describes the likelihood of a random variable to take on a particular value. For a continuous random variable, the pdf must satisfy a few conditions:

• The pdf must be non-negative for all values.
• The integral of the pdf over its entire range must be equal to 1.

In our case, the pdf \( f_{Y}(y) \) combines two linear functions that handle the intervals separately. Understanding the shape and properties of the pdf helps in further calculations that involve probabilities and expectations related to \( Y \).

Integration by Parts

Integration by parts is a technique used to integrate products of functions. It stems from the product rule for differentiation. The formula for integration by parts is given by: \[ \int u \, dv = uv - \int v \, du \] where \( u \) and \( dv \) are chosen from the original integrand. This technique is particularly useful when calculating the moment generating function, specifically when the integrands are products of exponential functions and polynomials, as seen in this exercise. When integrating over the interval \([0, 1]\) and \([1, 2]\), integration by parts simplifies the process of finding \( M_{Y}(t) \), ensuring accurate calculations.

Random Variable Characteristics

Understanding random variable characteristics is key in statistics and probability. Characteristics such as the pdf, mean (expected value), and variances provide insights into the behavior of the random variable \( Y \). The moment generating function \( M_{Y}(t) \) is particularly crucial as it encodes all the moments of the random variable. Evaluating this at different \( t \) values can yield means and variances, which describe the distribution's spread and central tendency. For \( t=0 \), the moment generating function always equals 1, reassuring that our computations are consistent with fundamental probability rules.