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Consider the following two experiments: the first has outcome \(X\) taking on the values \(0,1,\) and 2 with equal probabilities; the second results in an (independent) outcome \(Y\) taking on the value 3 with probability \(1 / 4\) and 4 with probability \(3 / 4\). Find the distribution of (a) \(Y+X\) (b) \(Y-X\).

Short Answer

Expert verified
Distribution of \(Y+X\): \(3: \frac{1}{12}, 4: \frac{1}{3}, 5: \frac{1}{3}, 6: \frac{1}{4}\). Distribution of \(Y-X\): \(3: \frac{1}{3}, 2: \frac{1}{4}, 1: \frac{1}{12}, 4: \frac{1}{4}\).

Step by step solution

01

Understand the Random Variables and Probabilities

We are given two random variables, \(X\) and \(Y\). \(X\) takes values \(0, 1,\) and \(2\) with equal probabilities \(\frac{1}{3}\). \(Y\) takes values \(3\) and \(4\) with probabilities \(\frac{1}{4}\) and \(\frac{3}{4}\) respectively.
02

Determine the Values of Y+X

We need to determine all possible values of \(Y+X\). These values result from the sum of each possible value of \(X\) with each possible value of \(Y\). The possible sums are: \(3+0, 3+1, 3+2, 4+0, 4+1, 4+2\), which results in the set \(\{3, 4, 5, 6\}\).
03

Calculate Probabilities for Y+X

Calculate the probability of each sum:1. \(P(Y+X=3) = P(Y=3)P(X=0) = \frac{1}{4} \cdot \frac{1}{3} = \frac{1}{12}\)2. \(P(Y+X=4) = P(Y=3)P(X=1) + P(Y=4)P(X=0) = \frac{1}{4} \cdot \frac{1}{3} + \frac{3}{4} \cdot \frac{1}{3} = \frac{1}{12} + \frac{3}{12} = \frac{1}{3}\)3. \(P(Y+X=5) = P(Y=3)P(X=2) + P(Y=4)P(X=1) = \frac{1}{4} \cdot \frac{1}{3} + \frac{3}{4} \cdot \frac{1}{3} = \frac{1}{12} + \frac{3}{12} = \frac{1}{3}\)4. \(P(Y+X=6) = P(Y=4)P(X=2) = \frac{3}{4} \cdot \frac{1}{3} = \frac{1}{4}\)Thus, the distribution of \(Y+X\) is: \(P(Y+X=3)=\frac{1}{12}, \) \(P(Y+X=4)=\frac{1}{3}, \) \(P(Y+X=5)=\frac{1}{3}, \) \(P(Y+X=6)=\frac{1}{4}\).
04

Determine the Values of Y-X

Now determine all possible values of \(Y-X\). These values result from subtracting each possible value of \(X\) from each possible value of \(Y\). The possible differences are \(3-0, 3-1, 3-2, 4-0, 4-1, 4-2\), which results in the set \(\{3, 2, 1, 4\}\).
05

Calculate Probabilities for Y-X

Calculate the probability of each difference:1. \(P(Y-X=3) = P(Y=3)P(X=0) + P(Y=4)P(X=1) = \frac{1}{4} \cdot \frac{1}{3} + \frac{3}{4} \cdot \frac{1}{3} = \frac{1}{12} + \frac{1}{4} = \frac{1}{3}\)2. \(P(Y-X=2) = P(Y=3)P(X=1) + P(Y=4)P(X=2) = \frac{1}{4} \cdot \frac{1}{3} + \frac{1}{4}\cdot \frac{1}{3} = \frac{1}{12} + \frac{3}{12} = \frac{1}{4}\)3. \(P(Y-X=1) = P(Y=3)P(X=2)= \frac{1}{4} \cdot \frac{1}{3} = \frac{1}{12}\)4. \(P(Y-X=4) = P(Y=4)P(X=0)= \frac{3}{4} \cdot \frac{1}{3} = \frac{1}{4}\)Thus, the distribution of \(Y-X\) is: \(P(Y-X=3)=\frac{1}{3}, \) \(P(Y-X=2)=\frac{1}{4}, \) \(P(Y-X=1)=\frac{1}{12}, \) \(P(Y-X=4)=\frac{1}{4}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Random Variables
Random variables are fundamental elements within the field of probability and statistics. They are used to encapsulate the outcome of statistical experiments. In essence, a random variable is a numerical description of the outcome of a random phenomenon.

Think of it as a way to assign numbers to the outcomes of experiments. For example, when you roll a die, the outcome is a number between 1 and 6, making it a discrete random variable. In our specific exercise, there are two random variables:
  • X: Represents an experiment with outcomes 0, 1, and 2, each having equal likelihoods.
  • Y: Corresponds to another experiment with outcomes 3 and 4 but with different probabilities of occurring.
The beauty of random variables is in how they allow us to quantitatively analyze the possibilities and probabilities of different outcomes occurring. They are integral to creating probability distributions, like in our exercise, where we analyzed combinations of outcomes from two random variables.
Joint Distribution
The concept of joint distribution revolves around understanding the probability of two or more random variables taking on a specific combination of values. It is crucial when dealing with situations involving more than one random variable, as in our exercise.

In the context of the exercise, joint distribution helps us figure out the probability of different outcomes occurring when the two variables, X and Y, are considered together.
  • To calculate joint distribution, determine all possible combinations of outcomes between the random variables.
  • Then, compute the probability of each combination happening.
For instance, with the values given for X and Y, a joint distribution calculation would involve figuring out how often different pairs of X and Y values occur. This is essential for computing other probabilities, such as those related to the sums and differences, as we did in the exercise. Joint distributions allow us to see the big picture of all possible outcomes when considering multiple random variables together.
Probability Calculation
Probability calculation is the backbone of decision-making in uncertainty and forms the basis of statistical inference. It involves determining the likelihood of different outcomes, based on known probabilities of underlying variables.

In our exercise, we calculated the probability of the sum and the difference of the random variables Y and X. The steps were:
  • Identify all possible outcomes for the operation (sum or difference in this case).
  • Compute probabilities for each individual step by applying probability rules and using the known individual probabilities of X and Y.
For example, to find the probability of a sum of 5 in Y+X, you'd calculate each scenario forming a sum of 5, add their probabilities together, and hence derive the overall probability. This calculation was repeated for all possible outcomes. The probability calculations let us quantify the uncertainty and give us an evidence-based view of the likelihood of different outcomes.
Discrete Probability Distribution
Discrete probability distributions are specific types of probability distributions related to discrete random variables. In our exercise, X and Y are both discrete random variables since their possible values are countable, such as 0, 1, 2, 3, and 4.

Each of these discrete outcomes has an associated probability, and these together form a probability distribution.
  • The sum of probabilities for all potential outcomes in a discrete distribution must equal 1, reflecting total certainty that one of the possible outcomes will occur.
  • For the exercises with Y+X and Y-X, we used these distributions to find the probability of each possible resulting sum or difference, forming a new discrete probability distribution which details the probability of each potential result.
This principle is vital in statistics and data analysis because it enables precise prediction and analysis. Discrete probability distributions simplify complex datasets and allow for straightforward computational analysis of potential outcomes.

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Most popular questions from this chapter

Assume that \(X_{1}\) and \(X_{2}\) are independent random variables, each having an exponential density with parameter \(\lambda\). Show that \(Z=X_{1}-X_{2}\) has density $$ f_{Z}(z)=(1 / 2) \lambda e^{-\lambda|z|} $$.

In one play of a certain game you win an amount \(X\) with distribution $$ p_{X}=\left(\begin{array}{ccc} 1 & 2 & 3 \\ 1 / 4 & 1 / 4 & 1 / 2 \end{array}\right) $$ Using the program NFoldConvolution find the distribution for your total winnings after ten (independent) plays. Plot this distribution.

Assume that the service time for a customer at a bank is exponentially distributed with mean service time 2 minutes. Let \(X\) be the total service time for 10 customers. Estimate the probability that \(X>22\) minutes.

Suppose that \(X\) and \(Y\) are independent and \(Z=X+Y\). Find \(f_{Z}\) if (a) $$ \begin{array}{l} f_{X}(x)=\left\\{\begin{array}{ll} \lambda e^{-\lambda x}, & \text { if } x>0 \\ 0, & \text { otherwise } \end{array}\right. \\ f_{Y}(x)=\left\\{\begin{array}{ll} \mu e^{-\mu x}, & \text { if } x>0 \\ 0, & \text { otherwise. } \end{array}\right. \end{array} $$ (b) $$ f_{X}(x)=\left\\{\begin{array}{ll} \lambda e^{-\lambda x}, & \text { if } x>0 \\ 0, & \text { otherwise } \end{array}\right. $$ $$ f_{Y}(x)=\left\\{\begin{array}{ll} 1, & \text { if } 0

Suppose we want to test a coin for fairness. We flip the coin \(n\) times and record the number of times \(X_{0}\) that the coin turns up tails and the number of times \(X_{1}=n-X_{0}\) that the coin turns up heads. Now we set $$ Z=\sum_{i=0}^{1} \frac{\left(X_{i}-n / 2\right)^{2}}{n / 2} $$ Then for a fair \(\operatorname{coin} Z\) has approximately a chi-squared distribution with \(2-1=1\) degree of freedom. Verify this by computer simulation first for a fair \(\operatorname{coin}(p=1 / 2)\) and then for a biased \(\operatorname{coin}(p=1 / 3)\).

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