/*! 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 15 Let \(X\) be a random variable d... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(X\) be a random variable distributed uniformly over \([0,20] .\) Define a new random variable \(Y\) by \(Y=\lfloor X\rfloor\) (the greatest integer in \(X) .\) Find the expected value of \(Y\). Do the same for \(Z=\lfloor X+.5\rfloor\). Compute \(E(|X-Y|)\) and \(E(|X-Z|) . \quad\) (Note that \(Y\) is the value of \(X\) rounded off to the nearest smallest integer, while \(Z\) is the value of \(X\) rounded off to the nearest integer. Which method of rounding off is better? Why?)

Short Answer

Expert verified
The expected value for \(Y\) is 9.5, for \(Z\) is 10. \(E(|X-Y|) = 0.5\), \(E(|X-Z|) = 0.25\). Rounding using \(Z\) is better.

Step by step solution

01

Understand the Uniform Distribution

The random variable \(X\) is uniformly distributed over the interval \([0, 20]\). This means each value between 0 and 20 is equally likely to be chosen. The probability density function (PDF) for a uniform distribution on \([a, b]\) is given by \( f(x) = \frac{1}{b-a} \). Thus, for our case, \(f(x) = \frac{1}{20} \) for \( 0 \leq x \leq 20 \).
02

Define and Calculate E(Y)

The random variable \(Y = \lfloor X \rfloor \) means \(Y\) takes integer values from 0, 1, ..., 19 based on the interval \([0, 20)\). The probability that \(Y = k\) for \(k = 0, 1, ..., 19\) is \(\frac{1}{20}\), because \(X\) being in \([k, k+1)\) gives \(\lfloor X \rfloor = k\). Therefore, the expected value of \(Y\) is \(E(Y) = \sum_{k=0}^{19} k \frac{1}{20} = \frac{1}{20} \sum_{k=0}^{19} k \). Use the formula for the sum of an arithmetic series to find \( \sum_{k=0}^{19} k = \frac{19 \cdot 20}{2} = 190\). Thus, \(E(Y) = \frac{190}{20} = 9.5\).
03

Define and Calculate E(Z)

The variable \(Z = \lfloor X + 0.5 \rfloor\) rounds \(X\) to the nearest integer. If \(X\) is in \([k - 0.5, k + 0.5)\), then \(Z = k\) for \(k = 1, 2, ..., 19\). For \(k = 0\), \(Z = 0\) when \(X \in [0, 0.5)\), and \(Z = 20\) when \(X \in [19.5, 20]\) (but these intervals are half-width). For \(Y = 1, 2, ..., 19\), the probability is \(\frac{1}{20}\), while it's \(\frac{1}{40}\) for \(Z= 0\) and \(Z= 20\). So, \(E(Z)\) is calculated by: \(E(Z) = \sum_{k=1}^{19} k \cdot \frac{1}{20} + 0 \cdot \frac{1}{40} + 20 \cdot \frac{1}{40}\). The sum is \(\frac{1}{20} \sum_{k=1}^{19} k = 9.5\). Then add \(20 \times \frac{1}{40} = 0.5\), giving \(E(Z) = 10\).
04

Calculate E(|X - Y|)

The random variable \( |X - Y| \) measures how far \(X\) is from \(Y\). Since \(Y = \lfloor X \rfloor\), \(|X - Y|\) is uniformly distributed between \([0, 1\), with an average deviation of \(0.5\). Therefore, \(E(|X - Y|) = 0.5\).
05

Calculate E(|X - Z|)

The random variable \( |X - Z| \) measures the deviation when rounding to the nearest integer. The worst case for deviation is 0.5, given the symmetry about 0.5 within each interval \([k-0.5, k+0.5)\). On average, since \(X\) deviates equally within each interval, this also turns out to be \(0.25\). Thus, \(E(|X - Z|) = 0.25\).
06

Compare and Conclude

Comparing the deviations, \(E(|X-Y|) = 0.5\) and \(E(|X-Z|) = 0.25\). Since \(E(|X-Z|)\) is less than \(E(|X-Y|)\), rounding \(X\) using \(Z\) (to the nearest integer) is "better" as it minimizes the average deviation.

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Key Concepts

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

Uniform Distribution
Uniform distribution is a fundamental concept in probability and statistics. It describes a situation where each outcome in a range is equally likely. In the exercise, the random variable \(X\) follows a uniform distribution over the interval \([0, 20]\). This means that any value within this interval, whether it is 0, 5, 11.23, or 19.99, is equally probable.

The probability density function (PDF) of a uniform distribution on an interval \([a, b]\) is given by the expression \( f(x) = \frac{1}{b-a} \). For \(X\), the PDF becomes \( f(x) = \frac{1}{20} \) since the length of the interval \([0, 20]\) is 20. This constant probability across the interval is what defines uniform distribution.

Understanding uniform distribution helps us calculate expectations and deviations, as each point is considered with equal weight. This forms the basis for solving problems that rely on finding averages and deviations.
Rounding Methods
Rounding is a technique to simplify numbers, and it can be done in various ways. In this exercise, two different rounding methods are used: flooring and traditional rounding.

For the random variable \(Y\), which is defined by \(Y = \lfloor X \rfloor\), the rounding method used is called 'flooring'. This method rounds down \(X\) to the greatest integer less than or equal to \(X\). For example, if \(X = 7.89\), \(Y\) will be 7.

On the other hand, \(Z = \lfloor X + 0.5 \rfloor\) reflects rounding \(X\) to the nearest integer. This method involves adding 0.5 to \(X\) before applying the floor function. So, if \(X = 7.2\), adding 0.5 makes it 7.7, and flooring it gives \(Z = 7\), but if \(X = 7.6\), it becomes 8.1 after adding 0.5, thus \(Z = 8\).

These methods affect how we perceive deviations from true values and are pivotal in determining the "best" method in minimizing average deviations.
Random Variable
A random variable is a variable whose possible values result from a random phenomenon. In the context of this exercise, \(X\) is a random variable uniformly distributed over \([0, 20]\). This means \(X\) can take any real value in the interval, each equally likely.

Random variables are categorized if they have measurable values within an occurrence of a random experiment. \(Y\) and \(Z\) are derived random variables contingent upon \(X\). \(Y\) is determined by flooring \(X\), while \(Z\) is determined by rounding \(X + 0.5\).

In probability, knowing if or how a variable is random helps in predicting and calculating expected results, average deviations, and understanding the behavior of outcomes under specified conditions.
Probability Density Function
A Probability Density Function (PDF) provides the likelihood of a random variable taking a particular value. Specifically, it is used for continuous random variables like \(X\) in this exercise.

For the uniform distribution of \(X\), the PDF is \( f(x) = \frac{1}{20} \) for the interval \(0 \leq x \leq 20\). This function indicates that each small range of values within \( \ [0, 20)\ \) is equally probable. The area under the PDF curve within any interval gives the probability of \(X\) falling within that range.

PDFs are crucial for defining and understanding continuous distributions. They allow us to calculate expectations, variances, and probabilities that guide us in probability assessments and help to solve problems like finding the expected value of rounded variables \(Y\) and \(Z\).

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Most popular questions from this chapter

Write a computer program to calculate the mean and variance of a distribution which you specify as data. Use the program to compare the variances for the following densities, both having expected value 0 : $$\begin{array}{l}p_{X}=\left(\begin{array}{ccccc} -2 & -1 & 0 & 1 & 2 \\\3 / 11 & 2 / 11 & 1 / 11 & 2 / 11 & 3 / 11\end{array}\right) \\\p_{Y}=\left(\begin{array}{ccccc} -2 & -1 & 0 & 1 & 2 \\\1 / 11 & 2 / 11 & 5 / 11 & 2 / 11 & 1 / 11\end{array}\right)\end{array}$$

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.

Show that, if \(X\) and \(Y\) are random variables taking on only two values each, and if \(E(X Y)=E(X) E(Y),\) then \(X\) and \(Y\) are independent.

A deck of ESP cards consists of 20 cards each of two types: say ten stars, ten circles (normally there are five types). The deck is shuffled and the cards turned up one at a time. You, the alleged percipient, are to name the symbol on each card before it is turned up. Suppose that you are really just guessing at the cards. If you do not get to see each card after you have made your guess, then it is easy to calculate the expected number of correct guesses, namely ten. If, on the other hand, you are guessing with information, that is, if you see each card after your guess, then, of course, you might expect to get a higher score. This is indeed the case, but calculating the correct expectation is no longer easy. But it is easy to do a computer simulation of this guessing with information, so we can get a good idea of the expectation by simulation. (This is similar to the way that skilled blackjack players make blackjack into a favorable game by observing the cards that have already been played. See Exercise 29.) (a) First, do a simulation of guessing without information, repeating the experiment at least 1000 times. Estimate the expected number of correct answers and compare your result with the theoretical expectation. (b) What is the best strategy for guessing with information? (c) Do a simulation of guessing with information, using the strategy in (b). Repeat the experiment at least 1000 times, and estimate the expectation in this case. (d) Let \(S\) be the number of stars and \(C\) the number of circles in the deck. Let \(h(S, C)\) be the expected winnings using the optimal guessing strategy in (b). Show that \(h(S, C)\) satisfies the recursion relation $$h(S, C)=\frac{S}{S+C} h(S-1, C)+\frac{C}{S+C} h(S, C-1)+\frac{\max (S, C)}{S+C}$$ and \(h(0,0)=h(-1,0)=h(0,-1)=0 .\) Using this relation, write a program to compute \(h(S, C)\) and find \(h(10,10)\). Compare the computed value of \(h(10,10)\) with the result of your simulation in (c). For more about this exercise and Exercise 26 see Diaconis and Graham. \(^{11}\)

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.

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