Problem 131 - An Introduction to Mathematical Statistics and Its Applications

Question

If \(p_{X, Y}(x, y)=c x y\) at the points \((1,1),(2,1),(2,2)\), and \((3,1)\), and equals 0 elsewhere, find \(c\).

Step-by-Step Solution

Answer

The constant \(c\) is \(1 / 10\).

Step 1: Understanding the information given

From the scenario given, we can tell that the joint probability mass function is \(p_{X, Y}(x, y) = c x y\) at the points \((1,1), (2,1), (2,2), (3,1)\) and zero elsewhere. This information will be used to setup an expression for \(c\).

Step 2: Set up the sum of the probabilities

The sum of all probabilities in a distribution should equate to 1. Therefore, we can form the equation by substituting the given coordinates into \(p_{X, Y}(x, y) = c x y\) and then summing them. The equation will therefore be, \(c(1 * 1) + c(2 * 1) + c(2 * 2) + c(3 * 1) = 1\).

Step 3: Solve for \(c\)

Simplify and solve for \(c\). The simplified equation becomes: \(c + 2c + 4c + 3c = 1\), which adds up to \(10c = 1\). Solving for \(c\), we get \(c = 1 / 10\).

Key Concepts

Probability Distribution

Understanding a probability distribution is crucial when dealing with random variables in probability theory. A probability distribution provides a comprehensive description of how probabilities are assigned to different values that a random variable can take on. There are different types of probability distributions based on whether the random variable is discrete or continuous. For discrete random variables, the distribution is referred to as a probability mass function (PMF). PMFs assign probabilities to each potential outcome in a way that the sum of all probabilities is 1.
This means that the probability of all possible outcomes must total 100%. For example, in the given exercise, we are working with a joint probability mass function, where probabilities are assigned to pairs of values \((x, y)\).
The key takeaway is that for any probability distribution, you must ensure that the total probability sums to 1. This forms the basis of determining the constant \(c\) in the exercise.

Discrete Random Variables

Discrete random variables represent quantities that can take on only a finite or countably infinite set of values. Unlike continuous variables that can take on any value within a range, discrete variables are usually associated with counts or distinct events.Some common examples of discrete random variables include:

• The roll of a die, which can result in any integer value from 1 to 6.
• The number of heads in a series of coin flips.
• The number of students in a classroom.

In the joint probability mass function given in the exercise, \((x, y)\) represents a pair of discrete random variables. Here, each pair is a specific scenario with its assigned probability determined by the formula \(p_{X, Y}(x, y) = c \cdot x \cdot y\). This formula indicates how likely each pair of outcomes is, based on the values of \(x\) and \(y\).
This discrete nature means that the probabilities are concentrated on specific combinations, not spread across an interval.

Probability Theory

Probability theory is the branch of mathematics that deals with the analysis of random phenomena. It provides the tools and frameworks to understand how likely events are to occur. This theory is foundational to a variety of fields, including statistics, finance, and science.One of the core principles of probability theory is ensuring that all probabilities in a given scenario must sum up to 1. This is a reflection of certainty, meaning that the occurrence of some outcome within the considered set is guaranteed.
In the exercise, solving for the constant \(c\) relies on this principle, where each potential outcome's probability is part of an overall distribution that must sum to 1. Another important aspect is working with joint probabilities, which extend the ideas of individual probabilities to situations involving two or more random variables. These enable us to analyze and predict correlations or dependencies between different events.