Problem 8
Question
Let \(X\) and \(Y\) be independent random variables with probability distributions given by $$ \mathrm{P}(X=0)=\mathrm{P}(X=1)=\frac{1}{2} \quad \text { and } \quad \mathrm{P}(Y=0)=\mathrm{P}(Y=2)=\frac{1}{2} . $$ a. Compute the distribution of \(Z=X+Y\). b. Let \(\tilde{Y}\) and \(\tilde{Z}\) be independent random variables, where \(\tilde{Y}\) has the same distribution as \(Y\), and \(\tilde{Z}\) the same distribution as \(Z\). Compute the distribution of \(\tilde{X}=\tilde{Z}-\tilde{Y}\).
Step-by-Step Solution
Verified Answer
Z has values 0, 1, 2, 3 with probabilities 1/4, 1/4, 1/2, 1/4. \(\tilde{X}\) has values 0, 1, 2, 3 with probabilities 1/2, 3/8, 1/4, 1/8.
1Step 1: Understand the problem
We have two independent random variables, \(X\) and \(Y\), with given probability distributions. Our task is twofold: first, to find the distribution of the sum \(Z = X + Y\), and second, to find the distribution of \(\tilde{X} = \tilde{Z} - \tilde{Y}\), where \(\tilde{Y}\) and \(\tilde{Z}\) mimic the distributions of \(Y\) and \(Z\) respectively.
2Step 1: Calculate distribution of Z = X + Y
Since \(X\) and \(Y\) are independent, we calculate \(Z\) by combining their possible sums. The possible values for \(Z\) are \(0 + 0 = 0\), \(0 + 2 = 2\), \(1 + 0 = 1\), and \(1 + 2 = 3\). We then calculate each probability: \(\mathrm{P}(Z=0) = \mathrm{P}(X=0)\mathrm{P}(Y=0) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\), \(\mathrm{P}(Z=1) = \mathrm{P}(X=1)\mathrm{P}(Y=0) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\), \(\mathrm{P}(Z=2) = \mathrm{P}(X=0)\mathrm{P}(Y=2) + \mathrm{P}(X=1)\mathrm{P}(Y=0) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\), and \(\mathrm{P}(Z=3) = \mathrm{P}(X=1)\mathrm{P}(Y=2) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\).
3Step 2: Define distributions of \(\tilde{Y}\) and \(\tilde{Z}\)
\(\tilde{Y}\) has the same distribution as \(Y\): \(\mathrm{P}(\tilde{Y}=0) = \mathrm{P}(\tilde{Y}=2) = \frac{1}{2}\). \(\tilde{Z}\) has the same distribution as \(Z\): \(\mathrm{P}(\tilde{Z}=0)=\frac{1}{4}, \mathrm{P}(\tilde{Z}=1)=\frac{1}{4}, \mathrm{P}(\tilde{Z}=2)=\frac{1}{2}, \text{ and } \mathrm{P}(\tilde{Z}=3)=\frac{1}{4}\).
4Step 3: Calculate distribution of \(\tilde{X} = \tilde{Z} - \tilde{Y}\)
Calculate possible values of \(\tilde{X}\) given \(\tilde{Z} = \tilde{Y} + \tilde{X}\): When \(\tilde{Z} = 0\) and \(\tilde{Y} = 0\), \(\tilde{X} = 0\); when \(\tilde{Z} = 2\), \(\tilde{X} = 0\) if \(\tilde{Y} = 2\) or \(\tilde{X} = 2\) if \(\tilde{Y} = 0\). Similarly, when \(\tilde{Z} = 1\), \(\tilde{X} = 1\) only if \(\tilde{Y} = 0\). When \(\tilde{Z} = 3\), \(\tilde{X} = 1\) if \(\tilde{Y} = 2\) or \(\tilde{X} = 3\) if \(\tilde{Y} = 0\). Collect these results and calculate probabilities: \(\mathrm{P}(\tilde{X}=0)=\frac{1}{2}, \mathrm{P}(\tilde{X}=1)=\frac{1}{4} + \frac{1}{8}=\frac{3}{8}\), \(\mathrm{P}(\tilde{X}=2)=\frac{1}{4}\), and \(\mathrm{P}(\tilde{X}=3)=\frac{1}{8}\).
Key Concepts
Random VariablesIndependent EventsProbability TheorySum of Random Variables
Random Variables
In probability theory, a random variable is a fundamental concept. It represents a variable whose possible values are numerical outcomes of a random phenomenon. For example, if we toss a coin, we can assign a value of 0 to heads and 1 to tails, making the outcome of the coin toss a random variable. Random variables can be discrete, where outcomes are countable, like the number of cars passing through a toll booth in a day, or continuous, where outcomes are uncountably infinite, such as the exact time a train arrives at a station.
When dealing with random variables like in this exercise, they are often described by their probability distributions. A probability distribution assigns probabilities to each possible value of the random variable, guiding our understanding of how likely different outcomes are. For example, the random variable X in our exercise can take two discrete values, 0 and 1, each with a probability of \(\frac{1}{2}\). This simple setup is foundational in making predictions and understanding chance events.
When dealing with random variables like in this exercise, they are often described by their probability distributions. A probability distribution assigns probabilities to each possible value of the random variable, guiding our understanding of how likely different outcomes are. For example, the random variable X in our exercise can take two discrete values, 0 and 1, each with a probability of \(\frac{1}{2}\). This simple setup is foundational in making predictions and understanding chance events.
Independent Events
Independent events in probability theory refer to events whose occurrence doesn't affect the probability of occurrence of the other. This is a key consideration when calculating joint probabilities. If two events, A and B, are independent, then the probability of both events A and B occurring is the product of their individual probabilities: \(P(A \cap B) = P(A) \times P(B)\).
This concept is essential in our exercise. The random variables X and Y are independent, meaning the probability of any combination of outcomes (e.g., X = 0 and Y = 0) can be calculated by multiplying their individual probabilities. For example, the probability that both X equals 0 and Y equals 0 is calculated as \(P(X=0 \text{ and } Y=0) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\). Understanding this independence allows us to confidently blend the distributions of X and Y to find combined outcomes, such as in the calculation of Z = X + Y.
This concept is essential in our exercise. The random variables X and Y are independent, meaning the probability of any combination of outcomes (e.g., X = 0 and Y = 0) can be calculated by multiplying their individual probabilities. For example, the probability that both X equals 0 and Y equals 0 is calculated as \(P(X=0 \text{ and } Y=0) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\). Understanding this independence allows us to confidently blend the distributions of X and Y to find combined outcomes, such as in the calculation of Z = X + Y.
Probability Theory
Probability theory is the branch of mathematics that deals with quantifying uncertainty. It provides the framework for understanding random variables and the occurrence of complex events. Through probability distributions, it helps us predict how likely specific outcomes are. A fundamental principle is that probabilities are always between 0 and 1, where 0 indicates impossibility and 1 certainty.
In practice, like in this exercise, probability theory allows us to analyze and interpret random processes by defining the likelihood of various results. Calculating the distribution of Z = X + Y in the problem involves applying the basic rules of probability to add, multiply and combine individual probabilities of different outcomes of X and Y. By using the principles of probability theory, we can derive the complete set of outcomes for Z, along with their respective probabilities. This deep grasping of how probabilities interact is key for probabilistic modeling and sound decision-making in uncertain environments.
In practice, like in this exercise, probability theory allows us to analyze and interpret random processes by defining the likelihood of various results. Calculating the distribution of Z = X + Y in the problem involves applying the basic rules of probability to add, multiply and combine individual probabilities of different outcomes of X and Y. By using the principles of probability theory, we can derive the complete set of outcomes for Z, along with their respective probabilities. This deep grasping of how probabilities interact is key for probabilistic modeling and sound decision-making in uncertain environments.
Sum of Random Variables
The sum of random variables is an important concept in probability theory. It involves determining the probability distribution of a random variable that is the sum of two or more other random variables. This sum can reveal patterns and outcomes that are not apparent when considering each variable separately.
In our exercise, we consider the sum of two independent random variables, X and Y, denoted as Z = X + Y. To find the probability distribution of Z, we list all possible sums and calculate the probability of each sum occurring. Since X and Y are independent, calculating the probability for each sum involves multiplying the probabilities of the contributing values of X and Y. For example, we find \(P(Z=2)\) by adding probabilities of the independent combinations leading to the sum of 2: either X = 0, Y = 2 or X = 1, Y = 0, giving us \(\frac{1}{2}\).
This calculation provides insight into how outcomes combine to form new probabilities, broadening our understanding of the random variables at play and the broader picture of uncertainty in modeling.
In our exercise, we consider the sum of two independent random variables, X and Y, denoted as Z = X + Y. To find the probability distribution of Z, we list all possible sums and calculate the probability of each sum occurring. Since X and Y are independent, calculating the probability for each sum involves multiplying the probabilities of the contributing values of X and Y. For example, we find \(P(Z=2)\) by adding probabilities of the independent combinations leading to the sum of 2: either X = 0, Y = 2 or X = 1, Y = 0, giving us \(\frac{1}{2}\).
This calculation provides insight into how outcomes combine to form new probabilities, broadening our understanding of the random variables at play and the broader picture of uncertainty in modeling.
Other exercises in this chapter
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