Problem 42 - Thomas Calculus

Question

Verify that \(f\) gives a joint probability density function. Then find the expected values \(\mu_{X}\) and \(\mu_{Y}\) . $$ f(x, y)=\left\\{\begin{array}{ll}{\frac{3}{2}\left(x^{2}+y^{2}\right),} & {\text { if } 0 \leq x \leq 1 \text { and } 0 \leq y \leq 1} \\ {0,} & {\text { otherwise. }}\end{array}\right. $$

Step-by-Step Solution

Answer

The function is a joint probability density function. The expected values are \(\mu_X = \frac{5}{8}\) and \(\mu_Y = \frac{5}{8}\).

Step 1: Verify the Joint Probability Density Function

To confirm that \(f(x, y)\) is a joint probability density function, we need to verify two conditions. First, the function should be non-negative for all \(x, y\). In our case, \(f(x, y) = \frac{3}{2}(x^2 + y^2)\) is non-negative when \(0 \leq x \leq 1\) and \(0 \leq y \leq 1\) because both \(x^2\) and \(y^2\) are non-negative and have non-negative coefficients. Second, the integral of \(f(x, y)\) over the entire space should equal 1: \[ \int_{0}^{1} \int_{0}^{1} \frac{3}{2}(x^2 + y^2) \ dx \ dy \].

Step 2: Set Up the Double Integral

To find the integral over the defined region, calculate: \[ \int_{0}^{1} \int_{0}^{1} \frac{3}{2}(x^2 + y^2) \ dx \ dy \]. This involves integrating with respect to \(x\) first, over the interval from 0 to 1, and then integrating with respect to \(y\).

Step 3: Evaluate the Inner Integral with respect to x

Solve \( \int_{0}^{1} \frac{3}{2}(x^2 + y^2) \ dx \). This separates into two integrals: \[ \int_{0}^{1} \frac{3}{2}x^2 \ dx + \int_{0}^{1} \frac{3}{2}y^2 \ dx \]. We integrate \( \frac{3}{2}x^2 \) to get \( \frac{3}{2} \times \frac{x^3}{3} \) evaluated from 0 to 1, which results in \( \frac{1}{2} \times 1 = \frac{1}{2} \). For \( \frac{3}{2}y^2 \), the integral is simply \( \frac{3}{2}y^2 \) evaluated over 0 to 1, resulting in \( \frac{3}{2}y^2 \).

Step 4: Evaluate the Outer Integral with respect to y

Now calculate \( \int_{0}^{1} \left( \frac{1}{2} + \frac{3}{2}y^2 \right) \ dy \). This separates into \( \int_{0}^{1} \frac{1}{2} \ dy + \int_{0}^{1} \frac{3}{2}y^2 \ dy \). The first integral results in \( \frac{1}{2}y \) evaluated from 0 to 1, giving \( \frac{1}{2} \). The second integral results in \( \frac{3}{2} \times \frac{y^3}{3} \) evaluated from 0 to 1, giving \( \frac{1}{2} \). Sum the results to verify the total probability is \(1\).

Step 5: Calculate the Expected Value \(\mu_X\)

The expected value \(\mu_X\) is calculated as: \[ \mu_X = \int_{0}^{1} \int_{0}^{1} x \cdot f(x,y) \ dx \ dy \]. Substitute \(f(x,y)\) into the equation, and evaluate the integral as before. Simplified, it becomes: \( \int_{0}^{1} \int_{0}^{1} \frac{3}{2}x(x^2 + y^2) \ dx \ dy \).

Step 6: Evaluate \(\mu_X\)

Compute the integrals: \( \int_{0}^{1} x^3 \ dx = \frac{x^4}{4}\) from 0 to 1, which is \( \frac{1}{4} \), and for the \(x^2y^2\) term stored over \(x\), \( \int_{0}^{1} x^2 \ dx = \frac{1}{3} \). This makes \( \mu_X = \frac{1}{4} + \frac{y^2}{2} \) evaluated from 0 to 1 resulting \(\frac{5}{8}\) after solving \(y_term\)

Step 7: Calculate the Expected Value \(\mu_Y\)

Following a similar approach for \(\mu_Y\), calculate: \[ \mu_Y = \int_{0}^{1} \int_{0}^{1} y \cdot f(x,y) \ dx \ dy \] resulting to \( \frac{5}{8}\).

Key Concepts

Expected Value

The expected value, often symbolized as \( \mu \), is a fundamental concept in statistics and probability that provides a measure for the center of a probability distribution. It is essentially the long-run average value of random variables, representing a 'mean' of the expected outcomes. When dealing with the joint probability density function, calculating the expected value involves looking at the probabilities across multiple variables—in this case, \( X \) and \( Y \).

For our function \( f(x, y) = \frac{3}{2}(x^2 + y^2) \), the expected values \( \mu_X \) and \( \mu_Y \) are found by integrating over all possible values, multiplying the variable in question by the probability density function. This can be expressed as:

• \( \mu_X = \int_{0}^{1} \int_{0}^{1} x \cdot f(x,y) \ dx \ dy \)
• \( \mu_Y = \int_{0}^{1} \int_{0}^{1} y \cdot f(x,y) \ dx \ dy \)

This process involves setting up and evaluating the double integral appropriate for each expected value component.

Double Integral

A double integral is an extension of a single integral that allows integration over a two-dimensional area. It is used to accumulate values over a region defined by two variables, say \( x \) and \( y \). In probability, double integrals help calculate the total probability over a certain space for continuous random variables.

The function \( f(x, y) = \frac{3}{2}(x^2 + y^2) \) requires us to integrate over the range \( 0 \leq x \leq 1 \) and \( 0 \leq y \leq 1 \). This is set up as:
\[ \int_{0}^{1} \int_{0}^{1} \frac{3}{2}(x^2 + y^2) \ dx \ dy \]

To evaluate it, we first perform the integration with respect to \( x \) (the inner integral), then with respect to \( y \) (the outer integral). This process is crucial for verifying the joint probability density function and for finding expected values.

Probability Density

Probability density refers to a function that describes the relative likelihood of a random variable to take on a particular value. Unlike probabilities in a discrete case, where specific values have a cumulative probability, in the continuous case, the probability density function (PDF) defines the density over an interval.

For function \( f(x, y) = \frac{3}{2}(x^2 + y^2) \), the joint PDF shows how probability is distributed over the plane defined by \( x \) and \( y \).

• The joint PDF must satisfy that it is always non-negative.
• The entire space under the function must integrate to 1, confirming it's a valid probability model.

This is why setting up and evaluating the integral over the specified domain to prove it sums to one is critical in such exercises.

Integration

Integration is the mathematical process of finding the integral of a function, which represents the accumulation of quantities over a specific interval. In the context of probability, integration is used to find areas under the curve of a probability density function.

Integration comes in various forms, but here for \( f(x, y) = \frac{3}{2}(x^2 + y^2) \), we focus on definite integrals over specified bounds, \( 0 \) to \( 1 \) for both variables, \( x \) and \( y \). The steps involved include:

• Setting up the integral bounds and expression based on the function's domain.
• Solving the inner integral first and using this solution to tackle the outer integral.

These steps ensure the integration yields the correct probabilities or expected values, confirming that our function behaves as a valid joint probability density function.