Problem 10 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

Let \(X\) and \(Y\) be two continuous random variables with joint probability density function $$ f(x, y)=\frac{12}{5} x y(1+y) \quad \text { for } 0 \leq x \leq 1 \text { and } 0 \leq y \leq 1 $$ and \(f(x, y)=0\) otherwise. a. Find the probability \(\mathrm{P}\left(\frac{1}{4} \leq X \leq \frac{1}{2}, \frac{1}{3} \leq Y \leq \frac{2}{3}\right)\). b. Determine the joint distribution function of \(X\) and \(Y\) for \(a\) and \(b\) between 0 and \(1 .\) c. Use your answer from \(\mathbf{b}\) to find \(F_{X}(a)\) for \(a\) between 0 and \(1 .\) d. Apply the rule on page 122 to find the probability density function of \(X\) from the joint probability density function \(f(x, y)\). Use the result to verify your answer from \(\mathbf{c}\). e. Find out whether \(X\) and \(Y\) are independent.

Step-by-Step Solution

Answer

Calculate the integrals for parts a, b, and c. Verify the results with the joint and marginal density functions for independence.

Step 1: Understanding the Problem

The problem is composed of several parts: (a) To find the probability for given ranges of \(X\) and \(Y\), (b) To determine the joint distribution function,(c) To find the marginal distribution function of \(X\), (d) To find the probability density function of \(X\) and verify part (c), and(e) To check if \(X\) and \(Y\) are independent.

Step 2: Probability Calculation (Part a)

The probability \( \mathrm{P}(\frac{1}{4} \leq X \leq \frac{1}{2}, \frac{1}{3} \leq Y \leq \frac{2}{3}) \) is found by integrating the joint density function \( f(x,y) = \frac{12}{5}xy(1+y) \) over the specified bounds:\[ \int_{1/4}^{1/2} \int_{1/3}^{2/3} \frac{12}{5} x y (1+y) \, dy \, dx \]. First integrate with respect to \( y \), then with respect to \( x \). Calculating these, you get the probability after evaluating the definite integrals.

Step 3: Compute Joint Distribution Function (Part b)

The joint distribution function \( F(a, b) \) is found by integrating the joint density function \( f(x,y) \) as follows:\[ F(a,b) = \int_0^a \int_0^b \frac{12}{5} xy(1+y) \, dy \, dx \].Evaluate this double integral to find \( F(a, b) \).

Step 4: Marginal Distribution Function (Part c)

To find the marginal distribution function \( F_X(a) \), integrate the joint distribution function \( F(a, y) \) with respect to \( y \):\[ F_X(a) = \int_0^1 f(a, y) \, dy \].Substitute \( f(a, y) = \frac{12}{5} a y (1+y) \) and integrate over \( y \) from 0 to 1 to find \( F_X(a) \).

Step 5: Probability Density Function of X (Part d)

The marginal probability density function of \( X \), \( f_X(x) \), is found by integrating the joint density over all possible values of \( y \):\[ f_X(x) = \int_0^1 \frac{12}{5} x y (1+y) \, dy \].Perform the integration to obtain \( f_X(x) \). Confirm that this result is the derivative of \( F_X(a) \) with respect to \( a \).

Step 6: Independence Check (Part e)

Check if random variables \( X \) and \( Y \) are independent. They are independent if and only if:\[ f(x, y) = f_X(x) \cdot f_Y(y) \].Calculate \( f_Y(y) \) by integrating \( f(x, y) \) over \( x \). Compare \( f(x, y) \) with \( f_X(x) \cdot f_Y(y) \) to check for equality.

Key Concepts

Continuous Random Variables

Continuous random variables are variables that can take any value within a certain range. Unlike discrete random variables, which are limited to specific values, continuous random variables have values that form a continuous spectrum.

• They are used to model quantities that cannot be counted but measured.
• Examples include height, weight, and in our exercise, the joint values of variables \(X\) and \(Y\).

For continuous random variables, probabilities of exact values are zero because the possibilities are infinite. Instead, we speak of probabilities over intervals, like the range from 0 to 1 in our given problem.
To calculate probabilities for continuous variables, we use integrals of their probability density functions across the desired range.

Marginal Distribution

The marginal distribution refers to the probability distribution of a subset of a collection of random variables, essentially reducing a multivariable distribution to focus on one variable. This is crucial in our exercise as we need to understand the behavior of \(X\) and \(Y\) individually.
To find the marginal distribution of \(X\), we integrate the joint density function over all values of \(Y\). This provides the density function of \(X\) without considering \(Y\).

• Similarly, the marginal distribution of \(Y\) is obtained by integrating over \(X\).
• Marginal distributions help in understanding how each variable behaves on its own.

By focusing on one random variable, the joint complexities are reduced, providing important insights for further analyses.

Independent Random Variables

Random variables are said to be independent if the occurrence of one doesn’t influence the other. In probability terms, two variables \(X\) and \(Y\) are independent if their joint probability density function \(f(x, y)\) equals the product of their marginal probability density functions \(f_X(x)\) and \(f_Y(y)\).
Our exercise leads us to explore this independence:

• If \(f(x, y) = f_X(x) \cdot f_Y(y)\), \(X\) and \(Y\) are independent.
• If not, they have some dependence, as changes in one affect the distribution of the other.

Understanding independence allows us to break down complex problems involving multiple variables into simpler, single-variable problems.

Probability Density Function

The probability density function (pdf) is a key concept in dealing with continuous random variables. It describes the likelihood of a random variable to take on a specific value.
Unlike in discrete cases, where a probability mass function is used, the pdf integrates over intervals to calculate probabilities:

• The area under the pdf curve between two points gives the probability that the variable falls within that interval.
• The pdf itself can have any shape, but the area under the entire curve must be equal to 1, representing a total probability of 100%.

In this exercise, the joint probability density function of \(X\) and \(Y\) \(f(x, y) = \frac{12}{5}xy(1+y)\) is essential for understanding how likely each pair of \(X\) and \(Y\) values is, within the given range. Understanding and using the pdf correctly is crucial for solving many probability problems involving continuous random variables.