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Let \(X\) have the \(\operatorname{pmf} p(x)=1 / 3, x=-1,0,1\). Find the pmf of \(Y=X^{2}\).

Short Answer

Expert verified
The pmf of \(Y=X^{2}\) is \(P(Y = y) = \frac{2}{3}\) for \(y = 1\), \(P(Y = y) = \frac{1}{3}\) for \(y = 0\), and \(P(Y = y) = 0\) for any other value of \(y\).

Step by step solution

01

Identify the Possible Outcomes

The first step is to identify the possible outcomes of \(Y\). Since \(Y = X^2\), and \(X\) can take the values -1, 0, and 1, the possible values of \(Y\) are 1, 0, and 1, respectively.
02

Compute the pmf of \(Y\)

The second step is to compute the pmf of \(Y\). The probability that \(Y = y\) is the sum of the probabilities that \(X = x\), for every \(x\) such that \(x^2 = y\). If \(y = 0\), then \(P(Y = 0) = P(X = 0) = \frac{1}{3}\). If \(y = 1\), then \(P(Y = 1) = P(X = -1) + P(X = 1) = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}\).
03

Write the pmf of \(Y\)

The final step is to write down the pmf of \(Y\). The pmf of \(Y\) is then \(P(Y = y) = \frac{2}{3}\), for \(y = 1\), and \(P(Y = y) = \frac{1}{3}\), for \(y = 0\). For any other value of \(y\), \(P(Y = y) = 0\).

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Key Concepts

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

Discrete Random Variables
In probability theory, a discrete random variable is a type of variable that can take on a countable number of distinct values. Think of it as having a handful of marbles, each with a number on it. These variables are the cornerstone of many probability calculations and are especially important when we talk about events with specific outcomes, such as rolling a die or flipping a coin.

Understanding discrete random variables is crucial to grasping the broader concepts of probability because they lay the groundwork for defining probability distributions, which tell us how likely different outcomes are. In the context of our exercise, the random variable X is discrete since it can only take on the values -1, 0, and 1. Notice how there are a finite number of possibilities? That's a telltale sign that we're dealing with a discrete random variable.
PMF Calculation
The probability mass function (pmf) is a function that gives us the probability that a discrete random variable is exactly equal to some value. Think of it as a recipe that tells us how to mix different probabilities. The pmf is essential to efficiently communicating the probabilities of the outcomes of a discrete random variable.

To calculate the pmf of a new variable like Y in our exercise, you have to map the probabilities from the original variable X. Since Y = X^2, some outcomes of X correspond to the same outcome of Y, which is why they must be combined (as seen with Y = 1 from X = -1 and X = 1). This step often involves some addition, as probabilities from different outcomes contributing to the same final result are summed. This calculation step is an excellent exercise in seeing how different variable transformations affect the probabilities of outcomes.
Probability Theory
At its core, probability theory is the branch of mathematics concerned with analyzing random phenomena and uncertainty. It allows us to make sense of the world's intrinsic randomness, providing a mathematical framework to quantify the likelihood of events. From coin flips to stock market fluctuations, probability theory gives us the tools to understand and predict outcomes.

Crucial to this is the concept of a probability distribution — a description of how probabilities are spread over the possible values of a random variable. The probability mass function we calculated for variable Y exemplifies such a distribution specific to discrete random variables. Probability theory not only encompasses the calculations of these probabilities but also underpins important principles such as the law of large numbers and the central limit theorem, which have profound implications in statistics and the real world.

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Most popular questions from this chapter

At the beginning of a study of individuals, \(15 \%\) were classified as heavy smokers, \(30 \%\) were classified as light smokers, and \(55 \%\) were classified as nonsmokers. In the five-year study, it was determined that the death rates of the heavy and light smokers were five and three times that of the nonsmokers, respectively. A randomly selected participant died over the five- year period: calculate the probability that the participant was a nonsmoker.

Suppose we are playing draw poker. We are dealt (from a well-shuffled deck) five cards, which contain four spades and another card of a different suit. We decide to discard the card of a different suit and draw one card from the remaining cards to complete a flush in spades (all five cards spades). Determine the probability of completing the flush.

Players \(A\) and \(B\) play a sequence of independent games. Player \(A\) throws a die first and wins on a "six." If he fails, \(B\) throws and wins on a "five" or "six." If he fails, \(A\) throws and wins on a "four," "five," or "six." And so on. Find the probability of each player winning the sequence.

After a hard-fought football game, it was reported that, of the 11 starting players, 8 hurt a hip, 6 hurt an arm, 5 hurt a knee, 3 hurt both a hip and an arm, 2 hurt both a hip and a knee, 1 hurt both an arm and a knee, and no one hurt all three. Comment on the accuracy of the report.

For each of the following, find the constant \(c\) so that \(p(x)\) satisfies the condition of being a pmf of one random variable \(X\). (a) \(p(x)=c\left(\frac{2}{3}\right)^{x}, x=1,2,3, \ldots\), zero elsewhere. (b) \(p(x)=c x, x=1,2,3,4,5,6\), zero elsewhere.

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