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Chapter 3: Problem 12

Let \(Y\) be a random variable with \(p(y)\) given in the accompanying table. Find \(E(Y), E(1 / Y)\), \(E\left(Y^{2}-1\right),\) and \(V(Y)\). $$\begin{array}{c|cccc}y & 1 & 2 & 3 & 4 \\\\\hline \rho(y) & .4 & .3 & .2 & .1\end{array}$$

Short Answer

Expert verified
E(Y) = 2.0, E(1/Y) = 0.642, E(Y虏-1) = 4.0, V(Y) = 1.0

Step by step solution

01

Calculate E(Y)

To calculate the expected value, or mean, of the random variable \(Y\), use the formula \(E(Y) = \sum(y \times p(y))\). For each possible value of \(y\), multiply \(y\) by its probability, then sum the results. \(E(Y) = 1 \times 0.4 + 2 \times 0.3 + 3 \times 0.2 + 4 \times 0.1 = 0.4 + 0.6 + 0.6 + 0.4 = 2.0\).
02

Calculate E(1/Y)

Use the formula \(E\left(\frac{1}{Y}\right) = \sum\left(\frac{1}{y} \times p(y)\right)\). To do this, compute \(\frac{1}{y} \times p(y)\) for each respective value: \(\frac{1}{1} \times 0.4 = 0.4\), \(\frac{1}{2} \times 0.3 = 0.15\), \(\frac{1}{3} \times 0.2 \approx 0.067\), \(\frac{1}{4} \times 0.1 = 0.025\). Summing these values gives \(0.642\).
03

Calculate E(Y虏-1)

For this, use the linearity of expectation: \(E(Y^2 - 1) = E(Y^2) - E(1)\). First calculate \(E(Y^2)\): find \(\sum(y^2 \times p(y))\). \((1^2 \times 0.4) + (2^2 \times 0.3) + (3^2 \times 0.2) + (4^2 \times 0.1) = 0.4 + 1.2 + 1.8 + 1.6 = 5.0\). Then, \(E(1)\) is simply 1. Therefore, \(E(Y^2 - 1) = 5.0 - 1 = 4.0\).
04

Calculate V(Y)

The variance of \(Y\) is given by \(V(Y) = E(Y^2) - [E(Y)]^2\). From previous steps, \(E(Y^2) = 5.0\) and \(E(Y) = 2.0\). So, \(V(Y) = 5.0 - (2.0)^2 = 5.0 - 4.0 = 1.0\).

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Key Concepts

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

Random Variables
A random variable, often denoted by a letter such as \( Y \), is a variable that can take on different values, each with a certain probability. These variables are called "random" because they are inherently unpredictable until an outcome is realized. In probability theory, random variables are crucial as they allow us to model real-world scenarios and calculate various outcomes and their probabilities.
  • They can be discrete, having a countable number of possible values, or continuous, with an infinite number of possible values within a range.
  • In our exercise, \( Y \) is a discrete random variable, meaning it can only take specific values: 1, 2, 3, and 4.
  • Each of these values has a corresponding probability \( p(y) \), which together form the probability distribution of the random variable.
Probability Distribution
The probability distribution of a random variable is a mathematical function that provides the probabilities of occurrence of different possible outcomes. It essentially describes how the total probability is distributed over the possible values of the random variable.
  • For discrete random variables, the probability distribution is often represented as a probability mass function (PMF). This PMF is a table or formula that provides the likelihood for each discrete value.
  • In our case, we have a PMF with \( p(y) \) values assigned to \( y = 1, 2, 3, 4 \) as 0.4, 0.3, 0.2, and 0.1, respectively.
  • The sum of all probabilities in a probability distribution must equal 1, ensuring that one of the possible outcomes will occur.
Understanding probability distributions allows us to calculate more complex metrics such as expected values and variances.
Variance Calculation
Variance is a measure of how much the values of a random variable differ from the expected value or mean. It provides insight into the variability or spread of the data around the mean, telling us whether the data points are tightly clustered or widely scattered.
  • Variance is calculated using the formula: \( V(Y) = E(Y^2) - [E(Y)]^2 \).
  • The expectation \( E(Y) \) represents the average value of \( Y \), while \( E(Y^2) \) represents the average of the squares of \( Y \).
  • In our example, we calculated \( E(Y^2) \) as 5.0 and \( E(Y) \) as 2.0. Thus, the variance \( V(Y) \) is \( 5.0 - 4.0 = 1.0 \).
A small variance indicates that the random variable is usually close to its mean, while a large variance signifies that values are spread out widely.

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