Problem 21 - A First Course in Probability

Question

Let $$f(x, y)=24 x y \quad 0 \leq x \leq 1,0 \leq y \leq 1,0 \leq x+y \leq 1$$ and let it equal 0 otherwise. (a) Show that \(f(x, y)\) is a joint probability density function. (b) Find \(E[X]\). (c) Find \(E[Y]\).

Step-by-Step Solution

Answer

(a) \(f(x, y)\) is a joint PDF because it satisfies the two conditions: \(f(x, y) \geq 0\) for all \(x, y\), and the integral of \(f(x, y)\) over the entire region is equal to 1. (b) \(E[X] = \frac{2}{3}\) (c) \(E[Y] = \frac{2}{3}\)

Step 1: Check if \(f(x, y) \geq 0\)

The given function \(f(x, y)\) is defined as \(24xy\) for \(0 \leq x \leq 1, 0 \leq y \leq 1, 0 \leq x + y \leq 1\) and equals 0 otherwise. Since \(x\) and \(y\) are non-negative within the specified range, the product \(24xy\) will also be non-negative for all values of \(x\) and \(y\). Hence, the first condition is satisfied.

Step 2: Check if the integral of \(f(x, y)\) is equal to 1

To show that the integral of \(f(x, y)\) over the entire region is equal to 1, we have to compute: \[\iint_{-\infty}^{\infty} f(x, y) dx dy = \iint_{R} 24xy dx dy\] where \(R\) is the region defined by \(0 \leq x \leq 1, 0 \leq y \leq 1, 0 \leq x+y \leq 1\). The limits for the integral must be adjusted to the region \(R\): \[\int_0^1 \int_0^{1-x} 24xy dy dx\] Now, we compute the integral: \[\iint_{R} 24xy dx dy = \int_0^1 \int_0^{1-x} 24xy dy dx = 1\] Since the integral of \(f(x, y)\) is equal to 1, the second condition is satisfied. Thus, \(f(x, y)\) is a joint PDF. (b) Find \(E[X]\).

Step 1: Compute the integral for \(E[X]\)

We want to compute the following integral to find the expected value of \(X\): \[E[X] = \iint_{-\infty}^{\infty} x f(x, y) dx dy\] Considering the region \(R\) and using the joint PDF \(f(x, y)\), we get: \[E[X] = \int_0^1 \int_0^{1-x} 24x^2y dy dx\]

Step 2: Compute \(E[X]\)

Now, we compute the integral: \[E[X] = \int_0^1 \int_0^{1-x} 24x^2y dy dx = \frac{2}{3}\] Therefore, \(E[X] = \frac{2}{3}\). (c) Find \(E[Y]\).

Step 1: Compute the integral for \(E[Y]\)

We want to compute the following integral to find the expected value of \(Y\): \[E[Y] = \iint_{-\infty}^{\infty}) y f(x, y) dx dy\] Considering the region \(R\) and using the joint PDF \(f(x, y)\), we get: \[E[Y] = \int_0^1 \int_0^{1-x} 24xy^2 dy dx\]

Step 2: Compute \(E[Y]\)

Now, we compute the integral: \[E[Y] = \int_0^1 \int_0^{1-x} 24xy^2 dy dx = \frac{2}{3}\] Therefore, \(E[Y] = \frac{2}{3}\).

Key Concepts

Expected Value

The expected value, often referred to as the mean, is a fundamental concept in probability and statistics. It helps us find the average or the center of a random process. Imagine having a random variable, say variables like height, weight, or even dice rolls. The expected value is the long-term average we would expect if we repeated the process an infinite number of times.
For a continuous random variable like the ones we see here, the expected value can be calculated using integration. Specifically, when dealing with joint probability density functions (PDFs), we use double integration to handle both variables involved. To find \(E[X]\), the expected value of \(X\), we calculate:

• First, find the expression \(x \cdot f(x, y)\).
• Next, set up the double integral over the region \(R\).
• Lead to solve \(\int_0^1 \int_0^{1-x} 24x^2y \, dy \, dx\) which gives \(\frac{2}{3}\).

The same approach applies to \(E[Y]\) by adjusting the integration to prioritize \(y\). Understanding this concept empowers you to predict outcomes more accurately.

Double Integration

Double integration is an extension of the concept of single integrals to functions of two variables. It's a tool that allows us to calculate the volume under a surface over a given region. In the context of probability, this calculation is essential when working with joint PDFs.In this exercise, double integration helps in proving whether a given function is a joint PDF. There are a couple of key steps:

• First, verify the region over which the function is defined, ensuring the function is non-negative.
• Second, establish if the integral over this region equals 1, confirming it is indeed a legitimate PDF.
  In this case, computing \(\int_0^1 \int_0^{1-x} 24xy \, dy \, dx\) validates the function.

Thus, double integration is not only a mathematical technique, but it becomes the language through which we articulate and solve such probability-based problems.

Probability Theory

Probability theory is the mathematical framework used for quantifying uncertain outcomes. It assists in modeling various random phenomena, be it the flipping of a coin or a more complex scenario like joint distributions for two continuous random variables.A joint probability density function (PDF) gives us the likelihood that two variables, say \(X\) and \(Y\), will occur simultaneously. Validating a function like \(f(x, y) = 24xy\) as a joint PDF involves checking two conditions:

• The function must be non-negative over its defined region.
• The total probability, calculated through integration over its domain, must equal one, which ensures it adheres to the definition of a PDF.

These principles demonstrate how probability theory provides the underlying structure for understanding randomness and supports tasks such as predicting outcomes and creating statistical models. This foundation is vital in fields ranging from engineering to artificial intelligence, where data-driven decision-making is key.