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91Ó°ÊÓ

a. Compute the covariance between \({\rm{X}}\) and\({\rm{Y}}\).

b. Compute the correlation coefficient \({\rm{\rho }}\) for this \({\rm{X}}\) and \({\rm{Y}}\).

Short Answer

Expert verified

a) The covariance is \({\mathop{\rm Cov}\nolimits} (X,Y) = - 0.1619\).

b) The correlation coefficient is \({\rho _{X,Y}} = - 0.0024\).

Step by step solution

01

Definition

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

02

Calculating covariance

(a):

Covariance

between \({\rm{X}}\) and \({\rm{Y}}\) is

\({\rm{Cov(X,Y) = E((X - E(X))(Y - E(Y)))}}\)

or equally

\({\rm{Cov(X,Y) = }}\left\{ {\begin{aligned}{*{20}{l}}{\sum\limits_{\rm{x}} {\sum\limits_{\rm{y}}^{\rm{¥}} {{\rm{(x - E(}}} } {\rm{X))(y - E(Y)) \times p(x,y)}}}&{{\rm{,X and Y discrete, }}}\\{\int_{{\rm{ - ¥}}}^{\rm{¥}} {\int_{{\rm{ - ¥}}}^{\rm{¥}} {{\rm{(x - E(}}} } {\rm{X))(y - E(Y)) \times f(x,y)dxdy}}}&{{\rm{,X and Y continuous}}{\rm{. }}}\end{aligned}} \right.\)

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

However, it requires more time to compute the covariance by the definition, this is why we will use the following proposition.

Proposition:

The following holds

\({\rm{Cov(X,Y) = E(XY) - E(X) \times E(Y)}}\)

From the exercise \({\rm{9}}\) we can see how does the marginal pdf's look like, therefore we can compute the expectations.

The following holds

\(\begin{aligned}{\rm{E(Y) = E(X)}}\\{\rm{ = }}\int_{{\rm{20}}}^{{\rm{30}}} {\rm{x}} {{\rm{f}}_{\rm{X}}}{\rm{(x)dx}}\\{\rm{ = }}\int_{{\rm{20}}}^{{\rm{30}}} {\rm{x}} \left( {{\rm{10K}}{{\rm{x}}^{\rm{2}}}{\rm{ + 0}}{\rm{.05}}} \right){\rm{dx}}\\{\rm{ = }}\left. {{\rm{10K}}\frac{{{{\rm{x}}^{\rm{3}}}}}{{\rm{3}}}} \right|_{{\rm{20}}}^{{\rm{30}}}{\rm{ + }}\left. {{\rm{0}}{\rm{.05}}\frac{{{{\rm{x}}^{\rm{2}}}}}{{\rm{2}}}} \right|_{{\rm{20}}}^{{\rm{30}}}\\{\rm{ = 25}}{\rm{.33}}{\rm{.}}\end{aligned}\)

We also need the following expectation

\(\begin{aligned}{\rm{E(XY) = }}\int_{{\rm{20}}}^{{\rm{30}}} {\int_{{\rm{20}}}^{{\rm{30}}} {\rm{x}} } {\rm{y \times K}}\left( {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} \right){\rm{dxdy}}\\{\rm{ = K}}\int_{{\rm{20}}}^{{\rm{30}}} {\int_{{\rm{20}}}^{{\rm{30}}} {{{\rm{x}}^{\rm{3}}}} } {\rm{ydydx + K}}\int_{{\rm{20}}}^{{\rm{30}}} {\int_{{\rm{20}}}^{{\rm{30}}} {{{\rm{y}}^{\rm{3}}}} } {\rm{xdxdy}}\\{\rm{ = K}}\int_{{\rm{20}}}^{{\rm{30}}} {{{\rm{x}}^{\rm{3}}}} \left( {\left. {\frac{{{{\rm{y}}^{\rm{2}}}}}{{\rm{2}}}} \right|_{{\rm{20}}}^{{\rm{30}}}} \right){\rm{dx + K}}\int_{{\rm{20}}}^{{\rm{30}}} {{{\rm{y}}^{\rm{3}}}} \left( {\left. {\frac{{{{\rm{x}}^{\rm{2}}}}}{{\rm{2}}}} \right|_{{\rm{20}}}^{{\rm{30}}}} \right){\rm{dy}}\\{\rm{ = }}\left. {\left( {\left. {\frac{{{{\rm{y}}^{\rm{2}}}}}{{\rm{2}}}} \right|_{{\rm{20}}}^{{\rm{30}}}} \right){\rm{K \times }}\frac{{{{\rm{x}}^{\rm{4}}}}}{{\rm{4}}}} \right|_{{\rm{20}}}^{{\rm{30}}}{\rm{ + }}\left. {\left( {\left. {\frac{{{{\rm{x}}^{\rm{2}}}}}{{\rm{2}}}} \right|_{{\rm{20}}}^{{\rm{30}}}} \right){\rm{K \times }}\frac{{{{\rm{y}}^{\rm{4}}}}}{{\rm{4}}}} \right|_{{\rm{20}}}^{{\rm{30}}}\\{\rm{ = 641}}{\rm{.447}}\end{aligned}\)

Now we can compute the covariance as follows

\(\begin{aligned}{\rm{Cov(X,Y) = E(XY) - E(X) \times E(Y)}}\\{\rm{ = 641}}{\rm{.447 - 25}}{\rm{.33 \times 25}}{\rm{.33}}\\{\rm{ = - 0}}{\rm{.1619}}{\rm{.}}\end{aligned}\)

Therefore, the covariance is \({\mathop{\rm Cov}\nolimits} (X,Y) = - 0.1619\).

03

Calculating correlation coefficient

(b):

The

correlation coefficient

of \({\rm{X}}\) and \({\rm{Y}}\) is

\({{\rm{\rho }}_{{\rm{X,Y}}}}{\rm{ = }}\frac{{{\rm{Cov(X,Y)}}}}{{{{\rm{\sigma }}_{\rm{X}}}{\rm{ \times }}{{\rm{\sigma }}_{\rm{Y}}}}}\)

We need to compute the variances in order to determine the correlation coefficient.

The Variance of \({\rm{X}}\), denoted by \({\rm{V(X)}}\left( {{\rm{\sigma }}_{\rm{X}}^{\rm{2}}} \right.\) or \(\left. {{{\rm{\sigma }}^{\rm{2}}}} \right)\) is

\(\begin{aligned}{\rm{V(X) = }}{{\rm{\sigma }}_{\rm{X}}}\\{\rm{ = E}}\left( {{{{\rm{(X - E(X))}}}^{\rm{2}}}} \right)\\{\rm{ = E}}\left( {{{\rm{X}}^{\rm{2}}}} \right){\rm{ - (E(X)}}{{\rm{)}}^{\rm{2}}}\end{aligned}\)

The Standard Deviation of \({\rm{X}}\) is

\({{\rm{\sigma }}_{\rm{X}}}{\rm{ = }}\sqrt {{\rm{\sigma }}_{\rm{X}}^{\rm{2}}} \)

The expectation of random variables \({\rm{X}}\) and \({\rm{Y}}\) has been calculated. The following is true

\(\begin{aligned}{c}{\rm{E}}\left( {{{\rm{Y}}^{\rm{2}}}} \right){\rm{ = E}}\left( {{{\rm{X}}^{\rm{2}}}} \right)\\{\rm{ = }}\int_{{\rm{20}}}^{{\rm{30}}} {\rm{x}} {{\rm{f}}_{\rm{X}}}{\rm{(x)dx}}\\{\rm{ = }}\int_{{\rm{20}}}^{{\rm{30}}} {{{\rm{x}}^{\rm{2}}}} \left( {{\rm{10K}}{{\rm{x}}^{\rm{2}}}{\rm{ + 0}}{\rm{.05}}} \right){\rm{dx}}\\{\rm{ = }}\left. {{\rm{10K}}\frac{{{{\rm{x}}^{\rm{4}}}}}{{\rm{4}}}} \right|_{{\rm{20}}}^{{\rm{30}}}{\rm{ + }}\left. {{\rm{0}}{\rm{.05}}\frac{{{{\rm{x}}^{\rm{3}}}}}{{\rm{3}}}} \right|_{{\rm{20}}}^{{\rm{30}}}\\{\rm{ = 649}}{\rm{.825}}{\rm{.}}\end{aligned}\)

The variance is

\(\begin{aligned}{\rm{V(Y) = V(X)}}\\{\rm{ = E}}\left( {{{\rm{X}}^{\rm{2}}}} \right){\rm{ - (E(X)}}{{\rm{)}}^{\rm{2}}}\\{\rm{ = 649}}{\rm{.825 - (25}}{\rm{.33}}{{\rm{)}}^{\rm{2}}}\\{\rm{ = 8}}{\rm{.2161}}\end{aligned}\)

Finally, the correlation coefficient is

\(\begin{aligned}{{\rm{\rho }}_{{\rm{X,Y}}}}{\rm{ = }}\frac{{{\rm{Cov(X,Y)}}}}{{{{\rm{\sigma }}_{\rm{X}}}{\rm{ \times }}{{\rm{\sigma }}_{\rm{Y}}}}}\\{\rm{ = }}\frac{{{\rm{ - 0}}{\rm{.1619}}}}{{\sqrt {{\rm{8}}{\rm{.2161}}} \sqrt {{\rm{8}}{\rm{.2161}}} }}\\{\rm{ = - 0}}{\rm{.0024}}{\rm{.}}\end{aligned}\)

Therefore, the correlation coefficient is \({\rho _{X,Y}} = - 0.0024\).

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Most popular questions from this chapter

Let X1, X2, and X3 represent the times necessary to perform three successive repair tasks at a certain service facility. Suppose they are independent, normal rv’s with expected values \({\mu _1}, {\mu _2}, and {\mu _3}\)and variances \(\sigma _1^2 , \sigma _2^2, and \sigma _3^2 \), respectively. a. If \(\mu = {\mu _2} = {\mu _3} = 60\)and\(\sigma _1^2 = \sigma _2^2 = \sigma _3^2 = 15\), calculate \(P\left( {{T_0} \le 200} \right)\)and\(P\left( {150 \le {T_0} \le 200} \right)\)? b. Using the \(\mu 's and \sigma 's\)given in part (a), calculate both \(P\left( {55 \le X} \right)\)and \(P\left( {58 \le X \le 62} \right)\).c. Using the \(\mu 's and \sigma 's\)given in part (a), calculate and interpret\(P\left( { - 10 \le {X_1} - .5{X_2} - .5{X_3} \le 5} \right)\). d. If\({\mu _1} = 40, {\mu _1} = 50, {\mu _1} = 60,\),\( \sigma _1^2 = 10, \sigma _2^2 = 12, and \sigma _3^2 = 14\) calculate \(P\left( {{X_1} + {X_2} + {X_3} \le 160} \right)\)and also \(P\left( {{X_1} + {X_2} \ge 2{X_3}} \right).\)

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\(f(x,y) = \left\{ {\begin{array}{*{20}{c}}{kxy\;\;\;\;\;\;\;x \ge 0,\;y \ge 0,\;20 \le x + y \le 30}\\{0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;otherwise}\end{array}} \right\}\)

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f. What is the variance of the total amount of grain on hand?

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