Problem 43 - Thomas Calculus

Question

Suppose that \(f\) is a uniform joint probability density function on \(0 \leq x < 2,0 \leq y < 3 .\) What is the formula for \(f ?\) What is the probability that \(X < Y ?\)

Step-by-Step Solution

Answer

The formula for \( f(x, y) \) is \( \frac{1}{6} \), and \( P(X < Y) = \frac{2}{3} \).

Step 1: Determine the Area of the Region

The function is defined over a rectangular region where the range for variable \( x \) is from 0 to 2, and the range for \( y \) is from 0 to 3. Therefore, the area of this region is calculated as follows:\[\text{Area} = (2 - 0) \times (3 - 0) = 6.\]

Step 2: Determine the Formula for the Uniform Density Function

Since \(f\) is a joint uniform probability density function over the region, and the total probability must be 1, the formula for \(f\) is given by the inverse of the area:\[f(x, y) = \frac{1}{6} \quad \text{for} \quad 0 \leq x < 2, 0 \leq y < 3.\]

Step 3: Set Up the Probability Expression for \(X < Y\)

The condition \(X < Y\) means that we are looking for the probability over the triangular region in our rectangle where \(x < y\). This region is bounded by the line \(y = x\) and the rectangle's limits. Graphically, this is the triangle below \(y = x\) and above \(x = 0\) and \(y = 0\) but within the rectangle.

Step 4: Determine the Limits for Integration

To find the bounds for \(X < Y\), we note that for each vertical slice \(x\) from 0 to 2, \(y\) starts from \(x\) up to 3. Therefore, the limits are:- \(x\) goes from 0 to 2- \(y\) goes from \(x\) to 3.

Step 5: Integrate Using the Density Function

Calculate the probability by integrating the joint density function \( f(x, y) = \frac{1}{6} \) over the specified region:\[P(X < Y) = \int_0^2 \int_x^3 \frac{1}{6} \, dy \, dx.\]

Step 6: Perform the Inner Integral

First, integrate with respect to \(y\):\[\int_x^3 \frac{1}{6} \, dy = \left[ \frac{y}{6} \right]_x^3 = \frac{3}{6} - \frac{x}{6} = \frac{3-x}{6}.\]

Step 7: Perform the Outer Integral

Now, integrate with respect to \(x\):\[\int_0^2 \frac{3-x}{6} \, dx = \frac{1}{6} \left( \left[ 3x - \frac{x^2}{2} \right]_0^2 \right) = \frac{1}{6} \left((6 - 2) \right) = \frac{4}{6} = \frac{2}{3}.\]

Step 8: Conclusion

The probability that \( X < Y \) given the uniform distribution over the specified area is \( \frac{2}{3} \).

Key Concepts

Probability Calculation

When dealing with a uniform joint probability density function, calculating probability involves understanding how the distribution behaves over a defined area. The goal is to determine how likely certain conditions, such as "\(X < Y\)," are satisfied.

With uniform distributions, probabilities are calculated by integrating the density function over the region of interest. For the exercise, the uniform density function is constant, simplifying the computation.

We ultimately integrate over the triangular region where \(X < Y\) within the rectangular bounds of the density function, which provides a clear method for finding probabilities in such scenarios.

Integration

Integration is the cornerstone of calculating probabilities for joint distributions. It helps to find the likelihood of events occurring under certain conditions. In the context of this exercise, integration allows us to compute the probability that \(X < Y\).

Two integrals are performed sequentially here: the inner integral is calculated with respect to \(y\) and the outer integral is performed with respect to \(x\). Integrating a constant function over a defined region involves determining the limits of integration for \(x\) and \(y\).

The integration spans from the line indicating \(y = x\) up to the edge of the rectangle, simplifying to \ \[ P(X < Y) = \int_0^2 \int_x^3 \frac{1}{6} \, dy \, dx.\ \] This setup efficiently computes the probability contained in the desired region.

Area of Region

Understanding the area of a region is a fundamental step when dealing with joint probability density functions as it directly affects the computation of the uniform density function. Here, the area over which the joint density function is uniformly distributed is determined first.

The area is simply the product of the lengths of the sides of the rectangle: from \(x=0\) to \(x=2\) and \(y=0\) to \(y=3\).

This produces an area of 6, which is critical since the uniform nature of the density function dictates that the probability density must sum to 1 over the entire region. Hence, \( f(x, y) = \frac{1}{6} \) for all \( (x, y) \) in that area.

Joint Density Function

A joint density function describes how two random variables, such as \(X\) and \(Y\), are distributed in relation to one another over a certain region. For a uniform joint density function, like in this exercise, this relationship is defined over a rectangular area where each outcome is equally likely.

The key characteristic of a uniform joint density function is its constant value over the specified range. This means the probability is spread evenly across all possible outcomes within the bounds.

This function, \( f(x, y) = \frac{1}{6}\), emanates from the area of the region being 6, allowing us to effectively analyze and calculate probabilities for specific conditions such as \(X < Y\). This thorough understanding of joint density functions facilitates easier calculation and interpretation of probabilities across complex scenarios.