Problem 40 - Thomas Calculus

Question

Verify that \(f\) gives a joint probability density function. Then find the expected values \(\mu_{X}\) and \(\mu_{Y}\) . $$ f(x, y)=\left\\{\begin{array}{ll}{4 x y,} & {\text { if } 0 \leq x \leq 1 \text { and } 0 \leq y \leq 1} \\ {0,} & {\text { otherwise. }}\end{array}\right. $$

Step-by-Step Solution

Answer

Yes, \( f(x, y) \) is a joint probability density function; \( \mu_X = \mu_Y = \frac{2}{3} \).

Step 1: Verify Joint Probability Density

To verify that the function is a joint probability density function, we need to check that the integral over the entire space equals 1.Calculate the double integral of \(f(x, y) = 4xy\) over the region \([0, 1] \times [0, 1]\):\[\int_0^1 \int_0^1 4xy \, dx \, dy = \int_0^1 \left( \int_0^1 4xy \, dx \right) \, dy\]First, compute the inner integral:\[\int_0^1 4xy \, dx = 4y \int_0^1 x \, dx = 4y \left[ \frac{x^2}{2} \right]_0^1 = 4y \times \frac{1}{2} = 2y\]Now, compute the outer integral:\[\int_0^1 2y \, dy = \left[ \frac{2y^2}{2} \right]_0^1 = y^2 \bigg|_0^1 = 1 - 0 = 1\]Since the integral equals 1, \(f(x, y)\) is a valid joint probability density function.

Step 2: Find Expected Value \( \mu_X \)

The expected value \( \mu_X \) is given by:\[\mu_X = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} x f(x, y) \, dx \, dy = \int_0^1 \int_0^1 x \cdot 4xy \, dx \, dy\]Let's compute the inner integral first:\[\int_0^1 4x^2 y \, dx = 4y \int_0^1 x^2 \, dx = 4y \left[ \frac{x^3}{3} \right]_0^1 = 4y \cdot \frac{1}{3} = \frac{4y}{3}\]Now compute the outer integral:\[\int_0^1 \frac{4y}{3} \, dy = \frac{4}{3} \left[ \frac{y^2}{2} \right]_0^1 = \frac{4}{3} \cdot \frac{1}{2} = \frac{2}{3}\]Thus, \( \mu_X = \frac{2}{3} \).

Step 3: Find Expected Value \( \mu_Y \)

The expected value \( \mu_Y \) is calculated in a similar manner:\[\mu_Y = \int_0^1 \int_0^1 y \cdot 4xy \, dx \, dy\]First, solve the inner integral:\[\int_0^1 4xy^2 \, dx = 4y^2 \int_0^1 x \, dx = 4y^2 \left[ \frac{x^2}{2} \right]_0^1 = 4y^2 \cdot \frac{1}{2} = 2y^2\]Now, compute the outer integral:\[\int_0^1 2y^2 \, dy = 2 \left[ \frac{y^3}{3} \right]_0^1 = 2 \cdot \frac{1}{3} = \frac{2}{3}\]Thus, \( \mu_Y = \frac{2}{3} \).

Key Concepts

Expected Value

The expected value, often denoted as \( E[X] \) or \( \mu \), is a fundamental concept in probability theory representing the average or mean value of a random variable if the experiment were repeated an infinite number of times under the same conditions. For a continuous random variable with a given probability density function (PDF), the expected value can be calculated using integration. If you have a joint probability density function like \( f(x, y) \), the expected value of \( X \), denoted by \( \mu_X \), is calculated over the domain of \( x \) and \( y \) by integrating:

• \( \mu_X = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} x f(x, y) \, dx \, dy \)

This involves finding the inner integral first with respect to \( x \), followed by the outer integral with respect to \( y \). Similarly, the expected value of \( Y \), \( \mu_Y \), can be calculated as:

• \( \mu_Y = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} y f(x, y) \, dx \, dy \)

Expected values help us grasp the center of the distribution, providing insight into where values of \( X \) and \( Y \) are likely to cluster.

Double Integral

Double integrals enable us to calculate the integral of a function over a two-dimensional area. When dealing with joint probability density functions, the double integral helps verify that the total probability over the entire probability space equals 1. This is a crucial step in affirming that a function is indeed a valid probability density function.For our example, the task involved calculating the double integral of \( f(x, y) = 4xy \) over the square region \([0, 1] \times [0, 1]\). This necessitates setting up the integral as follows:

• Compute the inner integral, usually with respect to \( x \): \( \int_0^1 4xy \, dx \).
• Integrate the result with respect to \( y \) over its range: \( \int_0^1 2y \, dy \).

This process ensures that the entire probability mass adds up to 1, verifying the integrity of the probability distribution. Double integrals are immensely helpful not just in probability, but in fields like physics and engineering, where multiple dimensions of data must be processed and analyzed.

Probability Theory

Probability theory forms the backbone of understanding how likely events are to occur. A joint probability density function (pdf) like \( f(x, y) \) represents the likelihood of different combinations of \( X \) and \( Y \) occurring together. This is part of a broader family of distribution functions used to model and analyze random variables' behavior.In our exercise, we start by establishing that \( f(x, y) \) is a proper joint pdf by ensuring it integrates to 1 over the entire sample space. This reflects one of the critical properties of probability distributions - that the sum (or in this case, integral) of all possible events within the valid range must equal 1.Not only do these functions describe random events, but they are also instrumental in other applications, such as statistical inference, where they provide insights based on empirical data. By understanding and manipulating these probabilistic models, we can make informed predictions and decisions in uncertain environments.