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Chapter 5: Q34E (page 220)

a. Recalling the definition of \({{\rm{\sigma }}^{\rm{2}}}\) for a single \({\rm{rv X}}\), write a formula that would be appropriate for computing the variance of a function \({\rm{h(X,Y)}}\) of two random variables. (Hint: Remember that variance is just a special expected value.)

b. Use this formula to compute the variance of the recorded score\({\rm{h(X,Y)( = max(X,Y))}}\).

Short Answer

Expert verified

a) The formula \({\rm{V(h(X,Y)) = E}}\left( {{\rm{h(X,Y}}{{\rm{)}}^{\rm{2}}}} \right){\rm{ - (E(h(X,Y))}}{{\rm{)}}^{\rm{2}}}\)that would be appropriate for computing the variance of a function \({\rm{h(X,Y)}}\) of two random variables

b) The variance of the recorded score is \({\rm{V(max(X,Y)) = 13}}{\rm{.34}}\).

Step by step solution

01

Definition

Probability simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes鈥攈ow likely they are鈥攚hen we're unclear about the result of an event. Statistics is the study of occurrences guided by probability.

02

Formula that appropriate for calculating the variance

(a):

Only discrete random variables will be proved. Simply substitute sum with integral and pmf with podf to gain evidence for the continuous case.

The Variance of\({\rm{X}}\), where \({\rm{X}}\) is a discrete random variable \({\rm{X}}\) with set of possible values \({\rm{S}}\) and pmf\({\rm{p(x)}}\), denoted by \({\rm{V(X)}}\) (\({\rm{\sigma }}_{\rm{X}}^{\rm{2}}\)or\({{\rm{\sigma }}^{\rm{2}}}\)) is

\({\rm{V(X) = }}\sum\limits_{{\rm{xI S}}} {{{{\rm{(x - \mu )}}}^{\rm{2}}}} {\rm{ \times p(x) = E}}\left( {{{{\rm{(X - \mu )}}}^{\rm{2}}}} \right)\)

The Expected Value (mean value) of any function\({\rm{g(X)}}\), where X is a discrete random variable \({\rm{X}}\) with set of possible values \({\rm{S}}\) and pmf\({\rm{p(x)}}\), denoted as\({\rm{E(g(X))}}\left( {{{\rm{\mu }}_{{\rm{g(X)}}}}} \right)\), is

\({\rm{E(g(X)) = }}{{\rm{\mu }}_{{\rm{g(X)}}}}{\rm{ = }}\sum\limits_{{\rm{xI S}}} {\rm{g}} {\rm{(x) \times p(x)}}\)

03

Calculation

Although \({\rm{h(X,Y)}}\) is a two-dimensional random variable, the previous definitions and assertions apply equally to two-dimensional random variables (just replace pmf with joint pmf).

Therefore, the following holds

\(\begin{aligned}{\rm{V(h(X,Y)) = E}}\left( {{{{\rm{(h(X,Y) - E(h(X,Y)))}}}^{\rm{2}}}} \right)\\{\rm{ = }}\sum\limits_{\rm{x}} {\sum\limits_{\rm{y}} {\left( {{{{\rm{(h(X,Y) - E(h(X,Y)))}}}^{\rm{2}}}} \right)} } {\rm{p(x,y)}}\\{\rm{ = }}\sum\limits_{\rm{x}} {\sum\limits_{\rm{y}} {\left( {{\rm{h(X,Y}}{{\rm{)}}^{\rm{2}}}{\rm{p(x,y)}}} \right)} } {\rm{ - }}\sum\limits_{\rm{x}} {\sum\limits_{\rm{y}} {\rm{2}} } {\rm{h(X,Y)E(h(X,Y))p(x,y) + (E(h(X,Y))}}{{\rm{)}}^{\rm{2}}}\\{\rm{ = }}\sum\limits_{\rm{x}} {\sum\limits_{\rm{y}} {\left( {{\rm{h(X,Y}}{{\rm{)}}^{\rm{2}}}{\rm{p(x,y)}}} \right)} } {\rm{ - 2E(h(X,Y))}}\sum\limits_{\rm{x}} {\sum\limits_{\rm{y}} {\rm{h}} } {\rm{(X,Y)p(x,y) + }}\left( {{\rm{E(h(X,Y)}}{{\rm{)}}^{\rm{2}}}} \right.\\{\rm{ = }}\sum\limits_{\rm{x}} {\sum\limits_{\rm{y}} {\left( {{\rm{h(X,Y}}{{\rm{)}}^{\rm{2}}}{\rm{p(x,y)}}} \right)} } {\rm{ - 2E(h(X,Y)) \times E(h(X,Y)) + }}\left( {{\rm{E(h(X,Y)}}{{\rm{)}}^{\rm{2}}}} \right.\\{\rm{ = E}}\left( {{\rm{h(X,Y}}{{\rm{)}}^{\rm{2}}}} \right){\rm{ - (E(h(X,Y))}}{{\rm{)}}^{\rm{2}}}\end{aligned}\)

Therefore, the formula is \({\rm{V(h(X,Y)) = E}}\left( {{\rm{h(X,Y}}{{\rm{)}}^{\rm{2}}}} \right){\rm{ - (E(h(X,Y))}}{{\rm{)}}^{\rm{2}}}\).

04

Calculating the variance

(b):

We are given

\({\rm{h(X,Y) = max(X,Y)}}\)

We are interested in expectation of random variable\({\rm{h(X,Y) = max(X,Y)}}\). Therefore,

\(\begin{aligned}{\rm{E(max(X,Y)) = }}\sum\limits_{\rm{x}} {\sum\limits_{\rm{y}} {\rm{h}} } {\rm{(x,y) \times p(x,y)}}\\{\rm{ = max(0,0) \times 0}}{\rm{.02 + max(0,5) \times 0}}{\rm{.06 + \ldots + max(10,10) \times 0}}{\rm{.14 + max(10,15) \times 0}}{\rm{.01}}\\{\rm{ = 0 \times 0}}{\rm{.02 + 5 \times 0}}{\rm{.06 + \ldots + \ldots 10 \times 0}}{\rm{.14 + 15 \times 0}}{\rm{.01}}\\{\rm{ = 9}}{\rm{.6}}{\rm{.}}\end{aligned}\)

Similarly,

\(\begin{aligned}{\rm{E}}\left( {{{{\rm{(max(X,Y))}}}^{\rm{2}}}} \right){\rm{ = }}\sum\limits_{\rm{x}} {\sum\limits_{\rm{y}} {\rm{h}} } {\rm{(x,y) \times p(x,y)}}\\{\rm{ = max(0,0}}{{\rm{)}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.02 + max(0,5}}{{\rm{)}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.06 + \ldots + max(10,10}}{{\rm{)}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.14 + max(10,15}}{{\rm{)}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.01}}\\{\rm{ = }}{{\rm{0}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.02 + }}{{\rm{5}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.06 + \ldots + \ldots 1}}{{\rm{0}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.14 + 1}}{{\rm{5}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.01}}\\{\rm{ = 105}}{\rm{.5}}{\rm{.}}\end{aligned}\)

Using the formula, we have

\(\begin{aligned}{\rm{V(h(X,Y)) = V(max(X,Y))}}\\{\rm{ = E}}\left( {{\rm{h(X,Y}}{{\rm{)}}^{\rm{2}}}} \right){\rm{ - (E(h(X,Y))}}{{\rm{)}}^{\rm{2}}}\\{\rm{ = 105}}{\rm{.5 - 9}}{\rm{.}}{{\rm{6}}^{\rm{2}}}\\{\rm{ = 13}}{\rm{.34}}{\rm{.}}\end{aligned}\)

Therefore, the variance is \({\rm{V(max(X,Y)) = 13}}{\rm{.34}}\).

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Most popular questions from this chapter

Let \({\rm{X}}\) be the number of packages being mailed by a randomly selected customer at a certain shipping facility. Suppose the distribution of \({\rm{X}}\) is as follows:

a. Consider a random sample of size \({\rm{n = 2}}\) (two customers), and let \({\rm{\bar X}}\) be the sample mean number of packages shipped. Obtain the probability distribution of\({\rm{\bar X}}\).

b. Refer to part (a) and calculate\({\rm{P(\bar X\pounds2}}{\rm{.5)}}\).

c. Again consider a random sample of size\({\rm{n = 2}}\), but now focus on the statistic \({\rm{R = }}\) the sample range (difference between the largest and smallest values in the sample). Obtain the distribution of\({\rm{R}}\). (Hint: Calculate the value of \({\rm{R}}\) for each outcome and use the probabilities from part (a).)

d. If a random sample of size \({\rm{n = 4}}\) is selected, what is \({\rm{P(\bar X\pounds1}}{\rm{.5)}}\) ? (Hint: You should not have to list all possible outcomes, only those for which\({\rm{\bar x\pounds1}}{\rm{.5}}\).)

Consider a random sample of size n from a continuous distribution having median \({\rm{0}}\)so that the probability of any one observation being positive is \({\rm{.5}}\). Disregarding the signs of the observations, rank them from smallest to largest in absolute value, and let \({\rm{w = }}\)the sum of the ranks of the observations having positive signs. For example, if the observations are \({\rm{ - }}{\rm{.3, + }}{\rm{.7, + 2}}{\rm{.1 and - 2}}{\rm{.5 }}\) then the ranks of positive observations are \({\rm{ 2and 3 }}\), so \({\rm{w = 5}}\). In Chapter \({\rm{15}}\), W will be called Wilcoxon鈥檚 signed-rank statistic. \({\rm{w}}\)can be represented as follows:

where the Yi 鈥檚 are independent Bernoulli rv鈥檚, each with \({\rm{ p = }}{\rm{.5 }}\) (\({{\rm{Y}}_{\rm{1}}}{\rm{ = 1}}\)corresponds to the observation with rank \({\rm{i }}\)being positive)

a. Determine \({\rm{E}}\left( {{{\rm{Y}}_{\rm{i}}}} \right){\rm{ }}\)and then \({\rm{E(W)}}\)using the equation for \({\rm{W}}\). (Hint: The first n positive integers sum to \({\rm{n(n + 1)/2}}{\rm{.)}}\)

b. Determine \({\rm{V(}}{{\rm{Y}}_{\rm{1}}}{\rm{)}}\)and then \({\rm{V(W)}}\). (Hint: The sum of the squares of the first n positive integers can be expressed as \({\rm{n(n + 1)(2n + 1)/6}}{\rm{.)}}\)

Suppose that you have ten lightbulbs, that the lifetime of each is independent of all the other lifetimes, and that each lifetime has an exponential distribution with parameter\({\rm{\lambda }}\).

a. What is the probability that all ten bulbs fail before time \({\rm{t}}\)?

b. What is the probability that exactly \({\rm{k}}\) of the ten bulbs fail before time \({\rm{t}}\)?

c. Suppose that nine of the bulbs have lifetimes that are exponentially distributed with parameter \({\rm{\lambda }}\) and that the remaining bulb has a lifetime that is exponentially distributed with parameter \({\rm{\theta }}\) (it is made by another manufacturer). What is the probability that exactly five of the ten bulbs fail before time\({\rm{t}}\)?

Suppose that for a certain individual, calorie intake at breakfast is a random variable with an expected value of \({\bf{500}}\)and standard deviation of \({\bf{50}}\), calorie intake at lunch is random with an expected value of \({\bf{900}}\) and standard deviation of\(100\), and calorie intake at dinner is a random variable with expected value \({\bf{2000}}\)and standard deviation\({\bf{180}}\). Assuming that intakes at different meals are independent of one another, what is the probability that the average calorie intake per day over the next (\({\bf{365}}\)- day) year is at most\({\bf{3500}}\)?

A company maintains three offices in a certain region, each staffed by two employees. Information concerning yearly salaries (\({\rm{1000}}\)s of dollars) is as follows:

\(\begin{array}{*{20}{c}}{{\rm{ Office }}}&{\rm{1}}&{\rm{1}}&{\rm{2}}&{\rm{2}}&{\rm{3}}&{\rm{3}}\\{{\rm{ Employee }}}&{\rm{1}}&{\rm{2}}&{\rm{3}}&{\rm{4}}&{\rm{5}}&{\rm{6}}\\{{\rm{ Salary }}}&{{\rm{29}}{\rm{.7}}}&{{\rm{33}}{\rm{.6}}}&{{\rm{30}}{\rm{.2}}}&{{\rm{33}}{\rm{.6}}}&{{\rm{25}}{\rm{.8}}}&{{\rm{29}}{\rm{.7}}}\end{array}\)

a. Suppose two of these employees are randomly selected from among the six (without replacement). Determine the sampling distribution of the sample mean salary\({\rm{\bar X}}\).

b. Suppose one of the three offices is randomly selected. Let\({{\rm{X}}_{\rm{1}}}\)and\({{\rm{X}}_{\rm{2}}}\)denote the salaries of the two employees. Determine the sampling distribution of\({\rm{\bar X}}\).

c. How does \({\rm{E(\bar X)}}\) from parts (a) and (b) compare to the population mean salary\({\rm{\mu }}\)?
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