/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 Let \(X\) be a random variable w... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 2

Let \(X\) be a random variable with range [-1,1] and \(f_{X}\) its density function. Find \(\mu(X)\) and \(\sigma^{2}(X)\) if, for \(|x|>1, f_{X}(x)=0,\) and for \(|x|<1,\) (a) \(f_{X}(x)=(3 / 4)\left(1-x^{2}\right)\). (b) \(f_{X}(x)=(\pi / 4) \cos (\pi x / 2)\). (c) \(f_{X}(x)=(x+1) / 2\). (d) \(f_{X}(x)=(3 / 8)(x+1)^{2}\).

Short Answer

Expert verified
(a) \( \mu = 0 \), \( \sigma^2 = \frac{1}{5} \). (b) \( \mu = 0 \), \( \sigma^2 = \frac{1}{2} \). (c) \( \mu = \frac{1}{3} \), \( \sigma^2 = \frac{1}{9} \). (d) \( \mu = \frac{1}{2} \), \( \sigma^2 = \frac{1}{10} \).

Step by step solution

01

Understand Mean and Variance Formulas

The mean \(\mu(X)\) is calculated using the formula \(\mu = \int_{-1}^{1} x f_X(x) \, dx\), and the variance \( \sigma^2(X) \) is given by \(\sigma^2 = \int_{-1}^{1} (x - \mu)^2 f_X(x) \, dx\). In order to calculate variance, it's often easier to use \(\sigma^2 = \int_{-1}^{1} x^2 f_X(x) \, dx - \mu^2\).
02

Integrate for Part (a) \(f_X(x) = \frac{3}{4}(1-x^2)\)

The mean \(\mu\) is \(\int_{-1}^{1} x \frac{3}{4}(1-x^2) \, dx\). This is an odd function, so \(\mu = 0\). The variance is calculated as \(\int_{-1}^{1} x^2 \frac{3}{4}(1-x^2) \, dx = \frac{3}{4}\left(\int_{-1}^{1} x^2 \, dx - \int_{-1}^{1} x^4 \, dx\right)\). The result is \(\frac{1}{5}\).
03

Integrate for Part (b) \(f_X(x) = \frac{\pi}{4} \cos(\frac{\pi x}{2})\)

The mean \(\mu\) is \(\int_{-1}^{1} x \frac{\pi}{4} \cos(\frac{\pi x}{2}) \, dx\), which evaluates to \(0\) due to symmetry. For variance, compute \(\int_{-1}^{1} x^2 \frac{\pi}{4} \cos(\frac{\pi x}{2}) \, dx\). After evaluating, you find \(\sigma^2 = \frac{1}{2} - \mu^2\) which results in \(\frac{1}{2}\).
04

Integrate for Part (c) \(f_X(x) = \frac{x+1}{2}\)

Calculate the mean \(\mu\) as \(\int_{-1}^{1} x \frac{x+1}{2} \, dx\). This integral evaluates to \(\frac{1}{3}\). The variance is \(\sigma^2 = \int_{-1}^{1} x^2 \frac{x+1}{2} \, dx - \left(\frac{1}{3}\right)^2\), which simplifies to \(\frac{1}{9}\).
05

Integrate for Part (d) \(f_X(x) = \frac{3}{8}(x+1)^2\)

The mean \(\mu\) is calculated as \(\int_{-1}^{1} x \frac{3}{8}(x+1)^2 \, dx\), resulting in \(\frac{1}{2}\). For variance, compute \(\int_{-1}^{1} x^2 \frac{3}{8}(x+1)^2 \, dx - \left(\frac{1}{2}\right)^2\), which simplifies to \(\frac{1}{10}\).

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Key Concepts

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

Random Variable
A random variable is a fundamental concept in probability and statistics. It is a variable whose possible values are outcomes from a random phenomenon. In simple terms, a random variable can be thought of as a numerical description of the outcome of a random event.
There are two types of random variables:
  • Discrete Random Variable: Takes on a countable number of distinct values. For example, the roll of a die results in one of six possible outcomes.
  • Continuous Random Variable: Takes on an infinite number of possible values. These values form a continuous range. For example, the exact time it takes to complete a race.
In this exercise, the random variable \(X\) has its range defined over the interval \([-1, 1]\). Its probability density function (PDF) specifies the likelihood of \(X\) assuming each value within this range.
Mean (Expected Value)
The mean or expected value of a random variable gives us a measure of the central tendency of its probability distribution. It's like the "center of mass" for the distribution.
To find the expected value \(\mu\) for a continuous random variable like \(X\), we use the integral formula: \[\mu = \int_{-1}^{1} x f_X(x) \, dx\]This formula essentially
  • Weighs each value of \(X\) by its probability, as defined by the probability density function \(f_X\).
  • In the examples provided (parts a-d), calculus is used to solve these integrals.
  • When the result of this integration is zero, as seen in parts (a) and (b), it suggests that the distribution is symmetric about the origin.
The expected value helps us gauge what typical values of a random variable might be, serving as one indicator of central behavior.
Variance
Variance provides insight into the spread or dispersion of a random variable's probability distribution around its mean. It tells us how much the values of a random variable deviate from the expected value.
Mathematically, the variance \(\sigma^2(X)\) for a continuous random variable \(X\) is calculated using:\[\sigma^2 = \int_{-1}^{1} (x - \mu)^2 f_X(x) \, dx\]A useful formula to simplify computation combines previously calculated expected values:\[\sigma^2 = \int_{-1}^{1} x^2 f_X(x) \, dx - \mu^2\]
  • For symmetric functions, variance indicates how broad or tight the distribution is.
  • Smaller variance suggests values are closer to the mean, while larger variance implies more spread out values.
  • This measure is essential for understanding variability, risk, and reliability of predictions based on the random variable \(X\).
By evaluating the variance, for each scenario in parts (a) to (d), we can quantitatively assess how diverse the outcomes of \(X\) might be from the mean.

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Most popular questions from this chapter

Let \(X\) be a random variable that takes on nonnegative values and has distribution function \(F(x) .\) Show that $$E(X)=\int_{0}^{\infty}(1-F(x)) d x$$ Hint: Integrate by parts. Illustrate this result by calculating \(E(X)\) by this method if \(X\) has an exponential distribution \(F(x)=1-e^{-\lambda x}\) for \(x \geq 0,\) and \(F(x)=0\) otherwise.

(Feller \(^{14}\) ) A large number, \(N,\) of people are subjected to a blood test. This can be administered in two ways: (1) Each person can be tested separately, in this case \(N\) test are required, (2) the blood samples of \(k\) persons can be pooled and analyzed together. If this test is negative, this one test suffices for the \(k\) people. If the test is positive, each of the \(k\) persons must be tested separately, and in all, \(k+1\) tests are required for the \(k\) people. Assume that the probability \(p\) that a test is positive is the same for all people and that these events are independent. (a) Find the probability that the test for a pooled sample of \(k\) people will be positive. (b) What is the expected value of the number \(X\) of tests necessary under plan (2)? (Assume that \(N\) is divisible by \(k\).) (c) For small \(p\), show that the value of \(k\) which will minimize the expected number of tests under the second plan is approximately \(1 / \sqrt{p}\)

A card is drawn at random from a deck of playing cards. If it is red, the player wins 1 dollar; if it is black, the player loses 2 dollars. Find the expected value of the game.

Suppose we have an urn containing \(c\) yellow balls and \(d\) green balls. We draw \(k\) balls, without replacement, from the urn. Find the expected number of yellow balls drawn. Hint: Write the number of yellow balls drawn as the sum of \(c\) random variables.

A random sample of 2400 people are asked if they favor a government proposal to develop new nuclear power plants. If 40 percent of the people in the country are in favor of this proposal, find the expected value and the standard deviation for the number \(S_{2400}\) of people in the sample who favored the proposal.
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