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Chapter 6: Problem 19

Let \(X\) be a random variable taking on values \(a_{1}, a_{2}, \ldots, a_{r}\) with probabilities \(p_{1}, p_{2}, \ldots, p_{r}\) and with \(E(X)=\mu .\) Define the spread of \(X\) as follows: $$\bar{\sigma}=\sum_{i=1}^{r}\left|a_{i}-\mu\right| p_{i}$$ This, like the standard deviation, is a way to quantify the amount that a random variable is spread out around its mean. Recall that the variance of a sum of mutually independent random variables is the sum of the individual variances. The square of the spread corresponds to the variance in a manner similar to the correspondence between the spread and the standard deviation. Show by an example that it is not necessarily true that the square of the spread of the sum of two independent random variables is the sum of the squares of the individual spreads.

Short Answer

Expert verified
The squares of spreads do not add up: \((\bar{\sigma}_W)^2 \neq (\bar{\sigma}_Y)^2 + (\bar{\sigma}_Z)^2\).

Step by step solution

01

Define two independent random variables

Consider two independent random variables, \(Y\) and \(Z\). Let \(Y\) take values 1 and 2 with probabilities \(\frac{1}{2}\) for each and \(Z\) take values 3 and 4 with probabilities \(\frac{1}{2}\) for each.
02

Calculate expected values

The expected value of \(Y\) is \(E(Y) = 1\cdot\frac{1}{2} + 2\cdot\frac{1}{2} = 1.5\). Similarly, the expected value of \(Z\) is \(E(Z) = 3\cdot\frac{1}{2} + 4\cdot\frac{1}{2} = 3.5\).
03

Calculate spreads for Y and Z

The spread of \(Y\) is \(\bar{\sigma}_Y = \sum_{i=1}^{2} \left| a_i - \mu_Y \right| p_i = |1-1.5| \cdot \frac{1}{2} + |2-1.5| \cdot \frac{1}{2} = 0.25 + 0.25 = 0.5\). Similarly, the spread of \(Z\) is \(\bar{\sigma}_Z = |3-3.5| \cdot \frac{1}{2} + |4-3.5| \cdot \frac{1}{2} = 0.25 + 0.25 = 0.5\).
04

Consider the sum of the random variables

Calculate the new random variable \(W = Y + Z\). \(W\) can take values 4 (\(1+3\)), 5 (\(1+4\) or \(2+3\)), and 6 (\(2+4\)). The probabilities are \(P(W=4) = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}\), \(P(W=5) = \frac{1}{2}\cdot\frac{1}{2} + \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{2}\), \(P(W=6) = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}\).
05

Calculate expected value and spread of W

The expected value of \(W\) is \(E(W) = 4\cdot\frac{1}{4} + 5\cdot\frac{1}{2} + 6\cdot\frac{1}{4} = 5\). The spread of \(W\) is \(\bar{\sigma}_W = |4-5| \cdot \frac{1}{4} + |5-5| \cdot \frac{1}{2} + |6-5| \cdot \frac{1}{4} = \frac{1}{4} + 0 + \frac{1}{4} = 0.5\).
06

Compare squares of spreads

The square of the spread of \(Y\) plus \(Z\) sums to \(0.5^2 + 0.5^2 = 0.5\). However, the square of the spread of \(W\) is \(0.5^2 = 0.25\). Thus, \((\bar{\sigma}_W)^2 eq (\bar{\sigma}_Y)^2 + (\bar{\sigma}_Z)^2\).

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Key Concepts

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

Expected Value
The expected value of a random variable is a fundamental concept in probability. It represents the average or mean value that the random variable is expected to take. In essence, if you could repeat an experiment with the random variable many times, the expected value is the long-run average outcome.

For a discrete random variable like the ones in our exercise, the expected value is calculated by multiplying each possible value of the random variable by its probability, and summing the results:

\[ E(X) = a_1 \cdot p_1 + a_2 \cdot p_2 + \ldots + a_r \cdot p_r \]
In the exercise, we calculated the expected values for random variables \(Y\) and \(Z.\) For instance, \(E(Y) = 1 \cdot \frac{1}{2} + 2 \cdot \frac{1}{2} = 1.5.\)

This calculation shows us the central tendency of each random variable, which is important for comparing differences when looking at spread and variance.
Independent Random Variables
Two random variables are independent if the outcome of one does not affect the outcome of the other. This property is crucial when calculating probabilities and expected values for sums of random variables.

In our example, \(Y\) and \(Z\) are independent because the value of \(Y\) does not influence the value of \(Z.\) Knowing that these random variables are independent allows us to calculate the probability distribution of their sum \(W\) by simply multiplying the probabilities of the events happening together.

More formally, two random variables \(Y\) and \(Z\) are independent if for all \(a\) and \(b\),

\[ P(Y = a, Z = b) = P(Y = a) \cdot P(Z = b) \]
In our step-by-step solution, we used this principle to find the probability distribution of \(W = Y+Z\). This is essential in confirming the independence of the two variables and in performing calculations for the expected values and other statistical measures.
Variance
Variance is a measure of how much the values of a random variable deviate from their mean, and it's a key component for understanding the spread of data around the expected value.

Unlike the spread, variance is not derived using absolute deviation but rather squared deviations to ensure all values contribute positively to the measure:

\[ \sigma^2 = \sum_{i=1}^{r}(a_i - \mu)^2 \cdot p_i \]
This formula quantifies how much variation or "spread" exists from the average value. Larger variances indicate that values are more spread out.

One important property of variance is that for independent random variables, the variance of their sum is equal to the sum of their variances. However, as demonstrated in this exercise, the square of the spread does not have the same property. While the spread is similar to standard deviation, used for quantifying spread, it does not translate to variance in the same way when dealing with multiple random variables.

Understanding these differences can help in comparing different measures of variability and deciding which to use in different scenarios.

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Most popular questions from this chapter

Assume that the lifetime of a diesel engine part is a random variable \(X\) with density \(f_{X}\). When the part wears out, it is replaced by another with the same density. Let \(N(t)\) be the number of parts that are used in time \(t\). We want to study the random variable \(N(t) / t\). Since parts are replaced on the average every \(E(X)\) time units, we expect about \(t / E(X)\) parts to be used in time \(t\). That is, we expect that $$\lim _{t \rightarrow \infty} E\left(\frac{N(t)}{t}\right)=\frac{1}{E(X)}$$ This result is correct but quite difficult to prove. Write a program that will allow you to specify the density \(f_{X},\) and the time \(t,\) and simulate this experiment to find \(N(t) / t\). Have your program repeat the experiment 500 times and plot a bar graph for the random outcomes of \(N(t) / t\). From this data, estimate \(E(N(t) / t)\) and compare this with \(1 / E(X) .\) In particular, do this for \(t=100\) with the following two densities: (a) \(f_{X}=e^{-t}\) (b) \(f_{X}=t e^{-t}\)

The lifetime, measure in hours, of the ACME super light bulb is a random variable \(T\) with density function \(f_{T}(t)=\lambda^{2} t e^{-\lambda t},\) where \(\lambda=.05 .\) What is the expected lifetime of this light bulb? What is its variance?

For a sequence of Bernoulli trials, let \(X_{1}\) be the number of trials until the first success. For \(j \geq 2,\) let \(X_{j}\) be the number of trials after the \((j-1)\) st success until the \(j\) th success. It can be shown that \(X_{1}, X_{2}, \ldots\) is an independent trials process. (a) What is the common distribution, expected value, and variance for \(X_{j} ?\) (b) Let \(T_{n}=X_{1}+X_{2}+\cdots+X_{n} .\) Then \(T_{n}\) is the time until the \(n\) th success. Find \(E\left(T_{n}\right)\) and \(V\left(T_{n}\right)\) (c) Use the results of (b) to find the expected value and variance for the number of tosses of a coin until the \(n\) th occurrence of a head.

In a second version of roulette in Las Vegas, a player bets on red or black. Half of the numbers from 1 to 36 are red, and half are black. If a player bets a dollar on black, and if the ball stops on a black number, he gets his dollar back and another dollar. If the ball stops on a red number or on 0 or 00 he loses his dollar. Find the expected winnings for this bet.

Gerolamo Cardano in his book, The Gambling Scholar, written in the early 1500 s, considers the following carnival game. There are six dice. Each of the dice has five blank sides. The sixth side has a number between 1 and 6 -a different number on each die. The six dice are rolled and the player wins a prize depending on the total of the numbers which turn up. (a) Find, as Cardano did, the expected total without finding its distribution. (b) Large prizes were given for large totals with a modest fee to play the game. Explain why this could be done.
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