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Let \(X_{1}\) and \(X_{2}\) be independent random variables with common distribution $$ p_{X}=\left(\begin{array}{ccc} 0 & 1 & 2 \\ 1 / 8 & 3 / 8 & 1 / 2 \end{array}\right) $$ Find the distribution of the sum \(X_{1}+X_{2}\).

Short Answer

Expert verified
The distribution of the sum \(X_1 + X_2\) is \(\begin{array}{cccc|c} 0 & 1 & 2 & 3 & 4 \\ \frac{1}{64} & \frac{6}{64} & \frac{15}{64} & \frac{20}{64} & \frac{22}{64} \end{array}\).

Step by step solution

01

Understand the Problem

We are given two independent random variables, \(X_1\) and \(X_2\), both with the same probability distribution. We need to find the probability distribution of their sum, \(X_1 + X_2\). We will need to calculate the probability of each possible value that the sum \(X_1 + X_2\) can take.
02

List Possible Outcomes for the Sum

First, we identify all possible values that \(X_1\) and \(X_2\) can take from the given distribution which are \(0, 1, 2\). Then, we list the possible values for \(X_1 + X_2\), i.e., \(0+0, 0+1, 0+2, 1+0, 1+1, 1+2, 2+0, 2+1, 2+2\) which are \(0, 1, 2, 3, 4\). Therefore, our target is to find the probability for each sum.
03

Calculate Probabilities for Each Sum

Utilize the independence of \(X_1\) and \(X_2\) to compute the joint probabilities. For instance, \(P(X_1 + X_2 = 0) = P(X_1 = 0) \times P(X_2 = 0) = \frac{1}{8} \times \frac{1}{8} = \frac{1}{64}\). Calculate similarly for all possible sums: \(1, 2, 3, 4\).
04

Compute Final Distribution

Sum the probabilities from the previous step for each outcome forming the total probability of that outcome. For \(X_1 + X_2 = 0\), we already calculated it as \(\frac{1}{64}\). For \(X_1 + X_2 = 1\), \(P(0 + 1) + P(1 + 0)= \frac{1}{8} \times \frac{3}{8} + \frac{3}{8} \times \frac{1}{8} = \frac{6}{64}\). Similarly, compute for other sums. Ensure all probabilities sum up to 1.
05

Verify and Conclude

After computing the probabilities for each sum correctly, write the final distribution of \(X_1 + X_2\) as: \[p_{X_1+X_2} = \begin{array}{cccc|c} 0 & 1 & 2 & 3 & 4 \ \frac{1}{64} & \frac{6}{64} & \frac{15}{64} & \frac{20}{64} & \frac{22}{64} \end{array}\] Ensure the probabilities add up to 1 to verify the calculation.

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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 crucial in probability and statistics. They can be thought of as a function that assigns a real number to each outcome of a random experiment. In the case of discrete random variables such as in the given problem with variables \(X_1\) and \(X_2\), they take on a countable number of distinct values.
Habitually, random variables are designated by capital letters like \(X\) and can symbolize scenarios such as rolling a die or flipping a coin. When solving problems involving random variables, it's important to recognize the range of potential values and the probabilities associated with each. These probabilities form the probability distribution.
For \(X_1\) and \(X_2\), their common distribution was:
  • \(0\) with probability \(\frac{1}{8}\),
  • \(1\) with probability \(\frac{3}{8}\),
  • \(2\) with probability \(\frac{1}{2}\).
Understanding this distribution helps us compute probabilities of other related random variables, such as their sums in independent scenarios.
Independent Events
Independence in probability is a fascinating and fundamental concept. Two events are independent if the occurrence of one does not affect the occurrence of the other. In terms of random variables like \(X_1\) and \(X_2\), they are independent if knowing the value of \(X_1\) gives no information about \(X_2\), and vice versa.
When working with independent events, the joint probability of their intersection can be computed by multiplying the probabilities of each event individually. For example, \(P(X_1 = 0)\) with \(P(X_2 = 0)\) gives the joint probability \(P(X_1 = 0, X_2 = 0) = \frac{1}{8} \times \frac{1}{8} = \frac{1}{64}\).
Independence makes calculating joint distributions straightforward but assumes that there is no relationship between variables. Such assumptions are crucial when modeling real-world situations.
Joint Probability
Joint probability refers to the probability of two (or more) events happening at the same time. In statistical terms, it is essential when dealing with the interactions of multiple random variables as seen with \(X_1\) and \(X_2\).
In the exercise, we evaluated every combination of \(X_1\) and \(X_2\) to find their joint probabilities. This process involves computing the probability of each pair of outcomes from \(X_1\) and \(X_2\). The independence of the random variables simplifies finding the joint probability: it's merely the product of individual probabilities. For instance, \(P(X_1 = 1, X_2 = 2) = \frac{3}{8} \times \frac{1}{2} = \frac{3}{16}\).
In the context of random variables, this concept elucidates how multiple variables interact, providing insights into complex scenarios where combined outcomes are needed.
Discrete Probability
Discrete probability involves scenarios where the set of possible outcomes is countable. This could be finite or infinite, and in our problem involving \(X_1\) and \(X_2\), we deal with a finite set: \([0, 1, 2]\).
Understanding discrete probability distributions is pivotal to solving problems such as the distribution of the sum \(X_1 + X_2\). The probabilities of the sum result from aggregating the probabilities of all combinations that lead to each possible outcome.
For example, to find the probability of the sum being \(2\), we computed:
  • \(P(0 + 2)\),
  • \(P(1 + 1)\),
  • \(P(2 + 0)\).
Adding these probabilities gives the distribution for that particular sum.
Working with discrete probabilities allows precise calculations, where each outcome is clearly defined, offering clarity that can be particularly advantageous in statistical modeling.

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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 $$ f_{Z}(z)=(1 / 2) \lambda e^{-\lambda|z|} $$.

The price of a stock on a given trading day changes according to the distribution $$ p_{X}=\left(\begin{array}{cccc} -1 & 0 & 1 & 2 \\ 1 / 4 & 1 / 2 & 1 / 8 & 1 / 8 \end{array}\right) $$ Find the distribution for the change in stock price after two (independent) trading days.

Assume that you are playing craps with dice that are loaded in the following way: faces two, three, four, and five all come up with the same probability \((1 / 6)+r\). Faces one and six come up with probability \((1 / 6)-2 r,\) with \(0<\) \(r<.02\). Write a computer program to find the probability of winning at craps with these dice, and using your program find which values of \(r\) make craps a favorable game for the player with these dice.

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 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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