/*! 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} Q90E Suppose a randomly chosen indivi... [FREE SOLUTION] | 91影视

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Chapter 5: Q90E (page 245)

Suppose a randomly chosen individual's verbal score \({\rm{X}}\)and quantitative score \({\rm{Y}}\)on a nationally administered aptitude examination have a joint pdf

\({\rm{f(x,y) = }}\left\{ {\begin{array}{*{20}{c}}{\frac{{\rm{2}}}{{\rm{5}}}{\rm{(2x + 3y)}}}&{{\rm{0拢 x拢 1,0拢 y拢 1}}}\\{\rm{0}}&{{\rm{ otherwise }}}\end{array}} \right.\)

You are asked to provide a prediction \({\rm{t}}\)of the individual's total score\({\rm{X + Y}}\). The error of prediction is the mean squared error\({\rm{E}}\left( {{{{\rm{(X + Y - t)}}}^{\rm{2}}}} \right)\). What value of \({\rm{t}}\)minimizes the error of prediction?

Short Answer

Expert verified

The \({\rm{2}}{\rm{.333}}\)value of t minimizes the error of prediction.

Step by step solution

01

Definition

A linear combination is an expression made up of a series of terms multiplied by a constant and then added together. Linear combinations are a fundamental idea in linear algebra and related branches of mathematics.

02

Calculating the value of t

In order to minimize the error of prediction

\({\rm{E}}\left( {{{{\rm{(X + Y - t)}}}^{\rm{2}}}} \right)\)

Because we are minimizing \({\rm{t}}\), one should take the derivative with regard to \({\rm{t}}\) and equate it with zero (typical minimizing technique).

The

Expected Value

(Mean value) of a random variable\({\rm{g(X,Y)}}\), where \({\rm{g( \times )}}\) is a function, denoted as \({\rm{E(g(X,Y))}}\) is given by

\(E(g(X,Y)) = \left\{ {\begin{array}{*{20}{l}}{\sum\limits_x {\sum\limits_y g } (x,y) \cdot p(x,y)}&{,X{\rm{ and }}Y{\rm{ discrete }}}\\{\int_{ - \infty }^\infty {\int_{ - \infty }^\infty g } (x,y) \cdot f(x,y)dxdy}&{,X{\rm{ and }}Y{\rm{ continuous}}{\rm{. }}}\end{array}} \right.\)

where \({\rm{p(x,y)}}\) is \({\rm{pmf}}\)and \({\rm{f(x,y)}}\) pdf.

Therefore, for\({\rm{g(X,Y) = (X + Y - t}}{{\rm{)}}^{\rm{2}}}\), the error of prediction is

\({\rm{E}}\left( {{{{\rm{(X + Y - t)}}}^{\rm{2}}}} \right){\rm{ = }}\int_{\rm{0}}^{\rm{1}} {\int_{\rm{0}}^{\rm{1}} {{{{\rm{(x + y - t)}}}^{\rm{2}}}} } {\rm{f(x,y)dxdy}}\)

and the derivative is

\(\begin{array}{l}\frac{{\rm{d}}}{{{\rm{dt}}}}\int_{\rm{0}}^{\rm{1}} {\int_{\rm{0}}^{\rm{1}} {{{{\rm{(x + y - t)}}}^{\rm{2}}}} } {\rm{f(x,y)dxdy}}\\{\rm{ = }}\int_{\rm{0}}^{\rm{1}} {\int_{\rm{0}}^{\rm{1}} {\frac{{\rm{d}}}{{{\rm{dt}}}}} } \left( {{{{\rm{(x + y - t)}}}^{\rm{2}}}} \right){\rm{f(x,y)dxdy}}\\{\rm{ = }}\int_{\rm{0}}^{\rm{1}} {\int_{\rm{0}}^{\rm{1}} {\rm{2}} } {\rm{((x + y - t)) \times ( - 1)f(x,y)dxdy}}\\{\rm{ = - 2}}\int_{\rm{0}}^{\rm{1}} {\int_{\rm{0}}^{\rm{1}} {{\rm{(x + y)}}} } {\rm{f(x,y)dxdy + t}}\underbrace {\int_{\rm{0}}^{\rm{1}} {\int_{\rm{0}}^{\rm{1}} {\rm{f}} } {\rm{(x,y)dxdy}}}_{\rm{0}}\\{\rm{ = - 2E(X + Y) + t}}\end{array}\)

03

Calculating the value of t

The value \({\rm{t}}\)for which \({\rm{E}}\left( {{{{\rm{(X + Y - t)}}}^{\rm{2}}}} \right)\)is minimum can be calculated from

\({\rm{ - 2E(X + Y) + t = 0}}\)

or equally

\({\rm{t = 2E(X + Y)}}{\rm{.}}\)

The expected value of random variable \({\rm{X + Y}}\) is

\(\begin{aligned}E(X + Y) &= \int_{\rm{0}}^{\rm{1}} {\int_{\rm{0}}^{\rm{1}} {{\rm{(x + y)}}} } {\rm{f(x,y)dxdy}}\\& =\frac{{\rm{2}}}{{\rm{5}}}\int_{\rm{0}}^{\rm{1}} {\int_{\rm{0}}^{\rm{1}} {{\rm{(x + y)}}} } {\rm{(2x + 3y)dxdy}}\\&= \frac{{\rm{2}}}{{\rm{5}}}\int_{\rm{0}}^{\rm{1}} {\int_{\rm{0}}^{\rm{1}} {\left( {{\rm{2}}{{\rm{x}}^{\rm{2}}}{\rm{ + 3yx + 2xy + 3}}{{\rm{y}}^{\rm{2}}}} \right)} } {\rm{dxdy}}\\ &= \frac{{\rm{2}}}{{\rm{5}}}\int_{\rm{0}}^{\rm{1}} {\left( {\left. {{\rm{2}}\frac{{{{\rm{x}}^{\rm{3}}}}}{{\rm{3}}}} \right|_{\rm{0}}^{\rm{1}}{\rm{ + }}\left. {{\rm{5y}}\frac{{{{\rm{x}}^{\rm{2}}}}}{{\rm{2}}}} \right|_{\rm{0}}^{\rm{1}}{\rm{ + }}\left. {{\rm{3}}{{\rm{y}}^{\rm{2}}}{\rm{x}}} \right|_{\rm{0}}^{\rm{1}}} \right)} {\rm{dy}}\\&= \frac{{\rm{2}}}{{\rm{5}}}\int_{\rm{0}}^{\rm{1}} {\left( {{\rm{2}}\frac{{{{\rm{1}}^{\rm{3}}}}}{{\rm{3}}}{\rm{ + 5y}}\frac{{{{\rm{1}}^{\rm{2}}}}}{{\rm{2}}}{\rm{ + 3}}{{\rm{y}}^{\rm{2}}}{\rm{ \times 1}}} \right)} {\rm{dy}}\\&= \frac{{\rm{2}}}{{\rm{5}}}\left( {\left. {\frac{{\rm{2}}}{{\rm{3}}}{\rm{y}}} \right|_{\rm{0}}^{\rm{1}}{\rm{ + }}\left. {\frac{{\rm{5}}}{{\rm{2}}}\frac{{{{\rm{y}}^{\rm{2}}}}}{{\rm{2}}}} \right|_{\rm{0}}^{\rm{1}}{\rm{ + }}\left. {{\rm{3}}\frac{{{{\rm{y}}^{\rm{3}}}}}{{\rm{3}}}} \right|_{\rm{0}}^{\rm{1}}} \right)\\&= \frac{{\rm{2}}}{{\rm{5}}}\left( {\frac{{\rm{2}}}{{\rm{3}}}{\rm{ + }}\frac{{\rm{5}}}{{\rm{4}}}{\rm{ + 1}}} \right)\\&= 1{\rm{.167}}\end{aligned}\)

Finally, the value \({\rm{t}}\)which minimizes the mean squared error is\({\rm{t = 2}}{\rm{.333}}{\rm{.}}\)

Therefore, the value of t is \({\rm{2}}{\rm{.333}}\).

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Most popular questions from this chapter

Exercise introduced random variables X and Y, the number of cars and buses, respectively, carried by ferry on a single trip. The joint pmf of X and Y is given in the table in Exercise. It is readily verified that X and Y are independent.

a. Compute the expected value, variance, and standard deviation of the total number of vehicles on a single trip.

b. If each car is charged\(\$ {\bf{3}}\)and each bus\(\$ {\bf{10}}\), compute the expected value, variance, and standard deviation of the revenue resulting from a single trip.

Suppose the proportion of rural voters in a certain state who favor a particular gubernatorial candidate is\(.{\bf{45}}\)and the proportion of suburban and urban voters favouring the candidate is\(.{\bf{60}}\). If a sample of\({\bf{200}}\)rural voters and\({\bf{300}}\)urban and suburban voters is obtained, what is the approximate probability that at least\(\;{\bf{250}}\)of these voters favour this candidate?

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

A more accurate approximation to \({\rm{E}}\left( {{\rm{h}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}} \right)} \right)\) in Exercise 95 is

\(h\left( {{\mu _1}, \ldots ,{\mu _n}} \right) + \frac{1}{2}\sigma _1^2\left( {\frac{{{\partial ^2}h}}{{\partial x_1^2}}} \right) + \cdots + \frac{1}{2}\sigma _n^2\left( {\frac{{{\partial ^2}h}}{{\partial x_n^2}}} \right)\)

Compute this for \({\rm{Y = h}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{,}}{{\rm{X}}_{\rm{3}}}{\rm{,}}{{\rm{X}}_{\rm{4}}}} \right)\)given in Exercise 93 , and compare it to the leading term \({\rm{h}}\left( {{{\rm{\mu }}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{\mu }}_{\rm{n}}}} \right)\).

Compute the correlation coefficient \({\rm{\rho }}\) for \({\rm{X}}\) and \({\rm{Y}}\)(the covariance has already been computed).
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