Problem 139 - An Introduction to Mathematical Statistics and Its Applications

Question

Suppose that two fair dice are tossed one time. Let \(X\) denote the number of 2 's that appear, and \(Y\) the number of 3 's. Write the matrix giving the joint probability density function for \(X\) and \(Y\). Suppose a third random variable, \(Z\), is defined, where \(Z=X+Y\). Use \(p_{X, Y}(x, y)\) to find \(p_{Z}(z)\).

Step-by-Step Solution

Answer

After calculating, the joint pdf \(p_{X, Y}(x, y)\) will be a 3x3 matrix with rows representing values of \(X\) and columns representing values of \(Y\). The pdf for \(Z=X+Y\) will contain probabilities for \(Z=0, 1, 2, 3, 4\), finding each by summing the probabilities of pairs (x, y) in the joint pdf for which \(x+y=z\). The exact values depend on the calculations made in Steps 2 and 4.

Step 1: Create the joint PDF

Firstly, create a matrix to represent the joint probability density function (pdf), \(p_{X, Y}(x, y)\). For this, calculate the probabilities of all combinations of \(x\) (number of 2s) and \(y\) (number of 3s) which can be 0, 1 or 2. Remember, the probability of rolling a '2' or '3' with a die is \(\frac{1}{6}\) and the rolls are independent.

Step 2: Calculate joint probabilities

For each possible pair (x, y), calculate the joint probability. For instance, probability of getting no '2' and one '3' (x=0, y=1) is equal to probability of not getting '2' in both dice times probability of getting '3' in one of the dice, resulting is \((\frac{5}{6})^2 * 2*\frac{1}{6}\). Repeat this process for all cells.

Step 3: Define the new random variable \(Z=X+Y\)

The third random variable \(Z\) is a sum of \(X\) and \(Y\), it can take values from 0 to 4.

Step 4: Identify probabilities for Z

For each possible value of \(z\), identify the pairs (x, y) for which \(x+y = z\), and add their corresponding probabilities from \(p_{X,Y}(x, y)\) to find \(p_Z(z)\).

Step 5: Make the final check

As a final check, ensure that the sum of all probabilities in \(p_Z(z)\) equals 1. This is because \(p_Z(z)\) is a probability mass function, and the sum of probabilities for all outcomes in a pdf must equal 1.

Key Concepts

Random Variables

In probability theory, a random variable is a numerical outcome of a random phenomenon. For example, when tossing a fair six-sided die, the potential outcomes are the numbers 1 through 6. However, a random variable allows us to assign numeric outcomes not directly linked to the value shown on the dice. In our exercise, we have two random variables:

• \(X\): represents the count of '2's rolled when two dice are tossed.
• \(Y\): represents the count of '3's rolled.

These random variables can take any value in the set \(\{0, 1, 2\}\) because, with two dice, the maximum number displaying either a '2' or a '3' is 2. Additionally, we define a third variable, \(Z\), as the sum of \(X\) and \(Y\). This is an example of how random variables can be combined to form new variables by associating their outcomes.

Probability Density Function

The probability density function (PDF) is a function that describes the likelihood of a random variable to take on a certain value.
In the context of discrete random variables, such as our dice problem, we often refer to the joint probability density function. This joint PDF, \(p_{X, Y}(x, y)\), represents the probability of \(X\) and \(Y\) taking on specific values simultaneously.

To create this function, we calculate probabilities for each combination of outcomes for \(X\) and \(Y\).
For example, if you want to know the probability of rolling exactly one '2' and one '3', you would determine this as the product of the probabilities of not rolling '2' or '3' on the other die rolls necessary for this event.

• A critical aspect is ensuring our calculated probabilities across all potential outcomes sum up to 1, a requirement for any probability function.

Dice Probability

Dice probability refers to the probabilistic outcomes related to dice-rolling scenarios. Understanding these probabilities is essential for our problem, where the dice are fair. This means each face of the dice, whether it's a '1', '2', '3', etc., has an equal chance of appearing, specifically a probability of \(\frac{1}{6}\).
For two dice rolls, each face combination clearly supports a joint probability for occurrences, which requires the independence of each roll.

• The outcome of one die does not affect the other. Hence, if the probability of rolling a '2' is \(\frac{1}{6}\) on each die, the independence allows using multiplication for joint probabilities.

For instance, the probability of not rolling a '2' on both dice is the product of each individual non-'2' probability, or \(\left( \frac{5}{6} \right)^2\). Applying such logic enables stacking probabilities in multi-roll scenarios.

Probability Mass Function

A probability mass function (PMF) concerns discrete random variables and tells us the probability of the variable taking certain distinct values. In the case of a joint distribution, such as \(p_{X, Y}(x, y)\), each cell in our probability matrix represents this function's value, calculated using product probabilities.
For any other random variable like \(Z\), which is a function of \(X\) and \(Y\), we also have a PMF, \(p_Z(z)\). This PMF is derived from summing the joint probabilities where the sum of \(X\) and \(Y\) equals \(z\).

• Using this sum, each \(p_{Z}(z)\) value is found by identifying pairs \((x, y)\) such that \(x+y=z\).
• Finally, the overall PMF must also sum up to 1, confirming that it appropriately represents all possible outcomes for \(Z\).

Hence, creating these PMFs involves aggregating probabilities, ensuring cohesiveness with probability laws and the behavior of dice rolls.