Problem 159 - An Introduction to Mathematical Statistics and Its Applications

Question

Find and graph \(f_{X, Y}(x, y)\) if the joint cdf for random variables \(X\) and \(Y\) is $$ F_{X, Y}(x, y)=x y, \quad 0 \leq x \leq 1, \quad 0 \leq y \leq 1 $$

Step-by-Step Solution

Answer

The joint probability density function \(f_{X,Y}(x,y)\) is 1 for \(0 \leq x \leq 1, 0 \leq y \leq 1\) and 0 otherwise. The function can be represented graphically as a square prism of height 1 centered on the range [0, 1] x [0, 1].

Step 1: Calculating the Partial Derivatives

We need to find the first order partial derivatives of the function \(F_{X,Y}(x,y) = xy\) with respect to both \(x\) and \(y\). To obtain the joint pdf, we differentiate \(F_{X, Y}(x, y)\) partially with respect to both \(x\) and \(y\). Using the definition of a derivative, we have: \[f_{X,Y}(x,y) = \frac{\partial^2 F_{X,Y}(x,y)}{\partial x \partial y}.\] Calculating these derivatives, we get \[f_{X,Y}(x,y) = 1\] for \(0 \leq x \leq 1, 0 \leq y \leq 1\). Outside this region, the function is 0 as the probabilities are 0.

Step 2: Graphing the Function

To graph this function, we note that it is a constant function equal to 1 inside the unit square [0, 1] x [0, 1] and zero outside. Thus we obtain a square of side length 1 with height 1 within this range, and a height of 0 outside this range. The graph is a square prism centered on the range [0, 1] x [0, 1] with height 1.

Step 3: Conclusion

We have successfully found that the joint probability density function \(f_{X,Y}(x,y)=1\) inside the unit square and 0 outside. We represented this graphically with a square prism of height 1.

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in calculus used to study functions of several variables. They represent the rate at which the function changes as we vary one of its variables while keeping the others constant. Here, we have a function of two variables, specifically the cumulative distribution function (cdf) \(F_{X,Y}(x,y) = xy\). To find the joint probability density function (pdf), \(f_{X,Y}(x,y)\), we need to take partial derivatives of the cdf with respect to both \(x\) and \(y\). This involves computing the second order partial derivative: \[f_{X,Y}(x,y) = \frac{\partial^2 F_{X,Y}(x,y)}{\partial x \partial y}.\]For \(F_{X,Y}(x,y) = xy\):

• The partial derivative with respect to \(x\) is \(\frac{\partial}{\partial x}(xy) = y\).
• The partial derivative of this result with respect to \(y\) gives \(\frac{\partial}{\partial y}(y) = 1\).

Thus, the joint pdf is \(f_{X,Y}(x,y) = 1\) within the specified domain \(0 \leq x \leq 1,\ 0 \leq y \leq 1\), and 0 otherwise. These derivatives tell us that the joint distribution is flat across the desired regions.

Graphing Functions

Graphing functions helps visualize the relationship defined by the mathematical expression. In this exercise, once we found that the joint pdf \(f_{X,Y}(x,y) = 1\) within the square \([0,1] \times [0,1]\) and 0 otherwise, we could graphically illustrate it.The graph consists of a square prism or a cuboid:

• The base of the prism is the unit square defined by the range \(0 \leq x \leq 1\) and \(0 \leq y \leq 1\).
• The height of the prism within this base is 1, representing the constant value of the joint pdf in this region.
• Outside of this base, the height drops to 0, indicating that the probability is zero.

Drawing this function involves sketching a rectangle of height 1 from the base \([0, 1] \times [0, 1]\) and nothing else beyond this region. This is a visual confirmation of how the joint pdf behaves in the given range and is a crucial step for understanding distributions and their representations.

Continuous Random Variables

Continuous random variables are distinct as they can take any real-valued number within a specified range. In this case, random variables \(X\) and \(Y\) are both continuous, constrained within \(0 \leq x \leq 1\) and \(0 \leq y \leq 1\).For continuous cases:

• The joint cdf \(F(x,y)\) provides the probability that both \(X\) and \(Y\) are less than or equal to certain values. Here, it was given as \(xy\).
• The joint pdf \(f(x,y)\), found by differentiating the cdf, gives the probability density at each point, which helps evaluate probabilities over infinitesimally small intervals within the specified range.
• Inside the domain \([0,1] \times [0,1]\), the pdf remains constant at 1, indicating uniform distribution.

This interpretation shows how continuous random variables distribute values across their range, and their joint pdf helps in understanding that distribution by identifying the density over specific intervals.