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Chapter 5: Q30E (page 220)

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

b. Compute \({\rm{\rho }}\) for \({\rm{X}}\) and \({\rm{Y}}\).

Short Answer

Expert verified

(a) The covariance for \({\rm{X}}\) and \({\rm{Y}}\) is \({\rm{Cov(X,Y) = - 3}}{\rm{.2025}}\).

(b) The \({\rm{\rho }}\)for \({\rm{X}}\) and\({\rm{Y}}\)\({\rm{\rho = Corr(X,Y) = - 0}}{\rm{.2074}}\).

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

Calculating covariance

Given:

(a) The expected value (or mean) \({\rm{\mu }}\)is the sum of the product of each possibility \({\rm{x}}\) with its probability\({\rm{P(x)}}\):

Note: The marginal probability distributions \({{\rm{P}}_{\rm{X}}}\) and \({{\rm{P}}_{\rm{Y}}}\) are the column totals \({\rm{(Y)}}\) and row totals\({\rm{(X)}}\).

\(\begin{aligned}{l}{\rm{E(X) = }}\sum {\rm{x}} {{\rm{P}}_{\rm{X}}}{\rm{(x) = 0(0}}{\rm{.02 + 0}}{\rm{.06 + 0}}{\rm{.02 + 0}}{\rm{.10) + 5(0}}{\rm{.04 + 0}}{\rm{.15 + 0}}{\rm{.20 + 0}}{\rm{.10) + 10(0}}{\rm{.01 + 0}}{\rm{.15 + 0}}{\rm{.14 + 0}}{\rm{.01) = 5}}{\rm{.55}}\\{\rm{E(Y) = }}\sum {\rm{y}} {{\rm{P}}_{\rm{Y}}}{\rm{(x) = 0(0}}{\rm{.02 + 0}}{\rm{.04 + 0}}{\rm{.01) + 5(0}}{\rm{.06 + 0}}{\rm{.15 + 0}}{\rm{.15) + 10(0}}{\rm{.02 + 0}}{\rm{.20 + 0}}{\rm{.14) + 15(0}}{\rm{.10 + 0}}{\rm{.10 + 0}}{\rm{.01) = 8}}{\rm{.55}}\\{\rm{E(XY) = }}\sum {\rm{x}} {\rm{yp(x,y) = 0(0)(0}}{\rm{.02) + 0(5)(0}}{\rm{.06) + 0(10)(0}}{\rm{.02) + 0(15)(0}}{\rm{.10) + 5(0)(0}}{\rm{.04) + 5(5)(0}}{\rm{.15) + 5(10)(0}}{\rm{.20) + 5(15)(0}}{\rm{.10)}}\\{\rm{ + 10(0)(0}}{\rm{.01) + 10(5)(0}}{\rm{.15) + 10(10)(0}}{\rm{.14) + 10(15)(0}}{\rm{.01) = 44}}{\rm{.25}}\end{aligned}\)

Using the property\({\rm{Cov(X,Y) = E(XY) - E(X) - E(Y)}}\), we can then determine the covariance:

\({\rm{Cov(X,Y) = E(XY) - E(X) - E(Y) = 44}}{\rm{.25 - 5}}{\rm{.55(8}}{\rm{.55) = - 3}}{\rm{.2025}}\)

Therefore, the covariance for \({\rm{X}}\) and \({\rm{Y}}\) is \({\rm{Cov(X,Y) = - 3}}{\rm{.2025}}\).

03

Calculating \({\rm{\rho }}\)

(b) The variance is the expected value of the squared deviation from the mean:

\(\begin{aligned}{l}{\rm{\sigma }}_{\rm{X}}^{\rm{2}}{\rm{ = }}\sum {{{{\rm{(x - \mu )}}}^{\rm{2}}}} {\rm{P(x) = (0 - 5}}{\rm{.55}}{{\rm{)}}^{\rm{2}}}{\rm{ \times (0}}{\rm{.02 + 0}}{\rm{.06 + 0}}{\rm{.02 + 0}}{\rm{.10) + (5 - 5}}{\rm{.55}}{{\rm{)}}^{\rm{2}}}{\rm{ \times (0}}{\rm{.04 + 0}}{\rm{.15 + 0}}{\rm{.20 + 0}}{\rm{.10)}}\\{\rm{ + (10 - 5}}{\rm{.55}}{{\rm{)}}^{\rm{2}}}{\rm{ \times (0}}{\rm{.01 + 0}}{\rm{.15 + 0}}{\rm{.14 + 0}}{\rm{.01) = 12}}{\rm{.4475}}\\{\rm{\sigma }}_{\rm{Y}}^{\rm{2}}{\rm{ = }}\sum {{{{\rm{(y - \mu )}}}^{\rm{2}}}} {\rm{P(y) = (0 - 8}}{\rm{.55}}{{\rm{)}}^{\rm{2}}}{\rm{ \times (0}}{\rm{.02 + 0}}{\rm{.04 + 0}}{\rm{.01) + (5 - 8}}{\rm{.55}}{{\rm{)}}^{\rm{2}}}{\rm{ \times (0}}{\rm{.06 + 0}}{\rm{.15 + 0}}{\rm{.15)}}\\{\rm{ + (10 - 8}}{\rm{.55}}{{\rm{)}}^{\rm{2}}}{\rm{ \times (0}}{\rm{.02 + 0}}{\rm{.20 + 0}}{\rm{.14) + (15 - 8}}{\rm{.55}}{{\rm{)}}^{\rm{2}}}{\rm{ \times (0}}{\rm{.10 + 0}}{\rm{.10 + 0}}{\rm{.01) = 19}}{\rm{.1475}}\end{aligned}\)

The standard deviation is the square root of the variance:

\(\begin{aligned}{l}{{\rm{\sigma }}_{\rm{X}}}{\rm{ = }}\sqrt {{\rm{\sigma }}_{\rm{X}}^{\rm{2}}} {\rm{ = }}\sqrt {{\rm{12}}{\rm{.4475}}} \\{{\rm{\sigma }}_{\rm{Y}}}{\rm{ = }}\sqrt {{\rm{\sigma }}_{\rm{Y}}^{\rm{2}}} {\rm{ = }}\sqrt {{\rm{19}}{\rm{.1475}}} \end{aligned}\)

The correlation \({\rm{\rho }}\) is then the covariance divided by the standard deviation of each variable:

\(\begin{aligned}{l}{\rm{\rho = Corr(X,Y) = }}\frac{{{\rm{Cov(X,Y)}}}}{{{{\rm{\sigma }}_{\rm{X}}}{{\rm{\sigma }}_{\rm{Y}}}}}\\{\rm{ = }}\frac{{{\rm{ - 3}}{\rm{.2025}}}}{{\sqrt {{\rm{12}}{\rm{.4475}}} \sqrt {{\rm{19}}{\rm{.1475}}} }}\\{\rm{\gg - 0}}{\rm{.2074}}\end{aligned}\)

Therefore, the \({\rm{\rho }}\)for \({\rm{X}}\) and\({\rm{Y}}\)\({\rm{\rho = Corr(X,Y) = - 0}}{\rm{.2074}}\).


(b) The variance is the expected value of the squared deviation from the mean:

\(\begin{aligned}{l}{\rm{\sigma }}_{\rm{X}}^{\rm{2}}{\rm{ = }}\sum {{{{\rm{(x - \mu )}}}^{\rm{2}}}} {\rm{P(x) = (0 - 5}}{\rm{.55}}{{\rm{)}}^{\rm{2}}}{\rm{ \times (0}}{\rm{.02 + 0}}{\rm{.06 + 0}}{\rm{.02 + 0}}{\rm{.10) + (5 - 5}}{\rm{.55}}{{\rm{)}}^{\rm{2}}}{\rm{ \times (0}}{\rm{.04 + 0}}{\rm{.15 + 0}}{\rm{.20 + 0}}{\rm{.10)}}\\{\rm{ + (10 - 5}}{\rm{.55}}{{\rm{)}}^{\rm{2}}}{\rm{ \times (0}}{\rm{.01 + 0}}{\rm{.15 + 0}}{\rm{.14 + 0}}{\rm{.01) = 12}}{\rm{.4475}}\\{\rm{\sigma }}_{\rm{Y}}^{\rm{2}}{\rm{ = }}\sum {{{{\rm{(y - \mu )}}}^{\rm{2}}}} {\rm{P(y) = (0 - 8}}{\rm{.55}}{{\rm{)}}^{\rm{2}}}{\rm{ \times (0}}{\rm{.02 + 0}}{\rm{.04 + 0}}{\rm{.01) + (5 - 8}}{\rm{.55}}{{\rm{)}}^{\rm{2}}}{\rm{ \times (0}}{\rm{.06 + 0}}{\rm{.15 + 0}}{\rm{.15)}}\\{\rm{ + (10 - 8}}{\rm{.55}}{{\rm{)}}^{\rm{2}}}{\rm{ \times (0}}{\rm{.02 + 0}}{\rm{.20 + 0}}{\rm{.14) + (15 - 8}}{\rm{.55}}{{\rm{)}}^{\rm{2}}}{\rm{ \times (0}}{\rm{.10 + 0}}{\rm{.10 + 0}}{\rm{.01) = 19}}{\rm{.1475}}\end{aligned}\)

The standard deviation is the square root of the variance:

\(\begin{aligned}{l}{{\rm{\sigma }}_{\rm{X}}}{\rm{ = }}\sqrt {{\rm{\sigma }}_{\rm{X}}^{\rm{2}}} {\rm{ = }}\sqrt {{\rm{12}}{\rm{.4475}}} \\{{\rm{\sigma }}_{\rm{Y}}}{\rm{ = }}\sqrt {{\rm{\sigma }}_{\rm{Y}}^{\rm{2}}} {\rm{ = }}\sqrt {{\rm{19}}{\rm{.1475}}} \end{aligned}\)

The correlation \({\rm{\rho }}\) is then the covariance divided by the standard deviation of each variable:

\(\begin{aligned}{l}{\rm{\rho = Corr(X,Y) = }}\frac{{{\rm{Cov(X,Y)}}}}{{{{\rm{\sigma }}_{\rm{X}}}{{\rm{\sigma }}_{\rm{Y}}}}}\\{\rm{ = }}\frac{{{\rm{ - 3}}{\rm{.2025}}}}{{\sqrt {{\rm{12}}{\rm{.4475}}} \sqrt {{\rm{19}}{\rm{.1475}}} }}\\{\rm{\gg - 0}}{\rm{.2074}}\end{aligned}\)

Therefore, the \({\rm{\rho }}\)for \({\rm{X}}\) and\({\rm{Y}}\)\({\rm{\rho = Corr(X,Y) = - 0}}{\rm{.2074}}\).


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