Problem 39 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The function is a valid joint pdf. Both \( \mu_X \) and \( \mu_Y \) are \( \frac{7}{12} \).

Step 1: Check Non-Negativity

For a joint probability density function (pdf), the function must be non-negative for all values. Here, \( f(x, y) = x + y \). Since both \( x \) and \( y \) are between 0 and 1, \( f(x, y) \) ranges from 0 to 2. Outside the given region, \( f(x, y) = 0 \). Therefore, by definition, the non-negativity condition is satisfied.

Step 2: Integrate Over the Entire Space

We need to integrate \( f(x, y) \) over the entire space to ensure it equals 1. This integral is given by the double integral: \[\int_{0}^{1} \int_{0}^{1} (x+y) \, dx \, dy\].

Step 3: Integrate with Respect to x

First, integrate with respect to \( x \):\[ \int_{0}^{1} (x + y) \, dx = \left[ \frac{x^2}{2} + yx \right]_{0}^{1} = \frac{1}{2} + y \].

Step 4: Integrate with Respect to y

Now, integrate the result from Step 3 with respect to \( y \): \[\int_{0}^{1} \left(\frac{1}{2} + y\right) \, dy = \left[ \frac{y}{2} + \frac{y^2}{2} \right]_{0}^{1} = \frac{1}{2} + \frac{1}{2} = 1\]. This confirms that \( f(x, y) \) integrates to 1 over the entire space, verifying it is a joint pdf.

Step 5: Find Expected Value \(\mu_X\)

The expected value \( \mu_X \) is given by the integral:\[\mu_X = \int_{0}^{1} \int_{0}^{1} x (x+y) \, dx \, dy\].First, integrate with respect to \( x \):\[ \int_{0}^{1} x (x+y) \, dx = \left[ \frac{x^3}{3} + \frac{yx^2}{2} \right]_{0}^{1} = \frac{1}{3} + \frac{y}{2}\]Integrate this with respect to \( y \) next:\[\int_{0}^{1} \left( \frac{1}{3} + \frac{y}{2} \right) \, dy = \left[ \frac{y}{3} + \frac{y^2}{4} \right]_{0}^{1} = \frac{1}{3} + \frac{1}{4} = \frac{7}{12}\].

Step 6: Find Expected Value \(\mu_Y\)

Similarly, the expected value \( \mu_Y \) is calculated as:\[\mu_Y = \int_{0}^{1} \int_{0}^{1} y (x+y) \, dx \, dy\].First, integrate with respect to \( x \):\[ \int_{0}^{1} y(x+y) \, dx = \left[ \frac{yx^2}{2} + y^2x \right]_{0}^{1} = \frac{y}{2} + y^2\]Next, integrate with respect to \( y \):\[\int_{0}^{1} \left( \frac{y}{2} + y^2 \right) \, dy = \left[ \frac{y^2}{4} + \frac{y^3}{3} \right]_{0}^{1} = \frac{1}{4} + \frac{1}{3} = \frac{7}{12}\].

Key Concepts

Expected Value

The expected value, denoted as \(\mu\), is a fundamental concept in probability that represents the average or mean value of a random variable. For two random variables, \(X\) and \(Y\), the expected value is determined using their joint probability density function \(f(x, y)\).

The expected value of \(X\), \(\mu_X\), involves integrating the product of \(x\) and the joint pdf over the entire space. Mathematically, it is expressed as: \[\mu_X = \int_{0}^{1} \int_{0}^{1} x f(x, y) \, dx \, dy\]
Following a similar process, the expected value of \(Y\), \( \mu_Y\), is: \[\mu_Y = \int_{0}^{1} \int_{0}^{1} y f(x, y) \, dx \, dy\]
For our function \(f(x, y) = x + y\), both expected values resulted in \(\frac{7}{12}\), indicating the mean positions of the variables within their limits.

Joint PDF Verification

In probability theory, verifying that a function is a joint probability density function (joint pdf) is crucial for analysis. A joint pdf must satisfy two conditions: the non-negativity condition, and its integral over the entire space must equal 1.

The function \(f(x, y) = x + y\) for \(x, y\) in the range [0, 1] is a candidate for a joint pdf. Firstly, the function must be non-negative which it is, since both \(x\) and \(y\) in [0, 1] make \(x + y\) range from 0 to 2. Outside this range, \(f(x, y) = 0\), ensuring non-negativity outside the region too.

The second check involves calculating the double integral \(\int_{0}^{1}\int_{0}^{1}(x+y)\,dx\,dy\). Through integration, we verified this results in 1, demonstrating that our function is indeed a joint pdf.

Double Integration

Double integration involves calculating the integral of a function of two variables over a specific region. It's a powerful tool for finding probabilities and expected values when dealing with two-dimensional problems.

The process of double integrating generally involves:

• First, performing an integration with respect to one variable, holding the other constant,
• Then, integrating the result with respect to the second variable.

In the case of our function \(f(x, y) = x + y\), the double integration was carried out by first integrating over \(x\), followed by \(y\): \[\int_{0}^{1} \int_{0}^{1} (x+y) \, dx \, dy\]
The computation reliably identified that the entire integral evaluates to 1, confirming \(f(x, y)\) as a valid joint pdf.

Non-Negativity Condition

The non-negativity condition is one of the foundational requirements for any probability density function, stating it must always be non-negative across its domain. This implies that no probability can be negative because probabilities inherently range from 0 to 1.

For a joint pdf, every evaluation within its valid range should result in a non-negative value. With \(f(x, y) = x + y\), the function is non-negative in the interval \(0 \leq x, y \leq 1\), since the smallest possible sum of \(x\) and \(y\) is 0 and the largest is 2. Outside this range, the function is explicitly 0, fitting the non-negativity stipulation universally.

Ensuring this condition holds is the first step in affirming a joint pdf is mathematically and conceptually sound.