Practice Problem Set 4 – Correlation Coefficient

This post provides practice problems to reinforce the concept of correlation coefficient discussed in this
post in a companion blog. The post in the companion blog shows how to evaluate the covariance \text{Cov}(X,Y) and the correlation coefficient \rho of two continuous random variables X and Y. It also discusses the connection between \rho and the regression curve E[Y \lvert X=x] and the least squares regression line.

The structure of the practice problems found here is quite simple. Given a joint density function for a pair of random variables X and Y (with an appropriate region in the xy-plane as support), determine the following four pieces of information.

  • The covariance \text{Cov}(X,Y)
  • The correlation coefficient \rho
  • The regression curve E[Y \lvert X=x]
  • The least squares regression line y=a+b x

The least squares regression line y=a+bx whose slope b and y-intercept a are given by:

    \displaystyle b=\rho \ \frac{\sigma_Y}{\sigma_X}

    \displaystyle a=\mu_Y-b \ \mu_X

where \mu_X=E[X], \sigma_X^2=Var[X], \mu_Y=E[Y] and \sigma_Y^2=Var[Y].

.

For some of the problems, the regression curves E[Y \lvert X=x] coincide with the least squares regression lines. When the regression curve is in a linear form, it coincides with the least squares regression line.

As mentioned, the practice problems are to reinforce the concepts discussed in this post.

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Practice Problem 4-A
    \displaystyle f(x,y)=\frac{3}{4} \ (2-y) \ \ \ \ \ \ \ 0<x<y<2

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Practice Problem 4-B
    \displaystyle f(x,y)=\frac{1}{2} \ \ \ \ \ \ \ \ \ \ \ \ 0<x<y<2

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Practice Problem 4-C
    \displaystyle f(x,y)=\frac{1}{8} \ (x+y) \ \ \ \ \ \ \ \ \ 0<x<2, \ 0<y<2

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Practice Problem 4-D
    \displaystyle f(x,y)=\frac{1}{2 \ x^2} \ \ \ \  \ \ \ \ \ \ \ \ \ 0<x<2, \ 0<y<x^2

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Practice Problem 4-E
    \displaystyle f(x,y)=\frac{1}{2} \ (x+y) \ e^{-x-y} \ \ \ \  \ \ \ \ \ \ \ \ \ x>0, \ y>0

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Practice Problem 4-F
    \displaystyle f(x,y)=\frac{3}{8} \ x \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ 0<y<x<2

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Practice Problem 4-G
    \displaystyle f(x,y)=\frac{1}{2} \ xy \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ 0<y<x<2

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Practice Problem 4-H
    \displaystyle f(x,y)=\frac{3}{14} \ (xy +x) \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ 0<y<x<2

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Practice Problem 4-I
    \displaystyle f(x,y)=\frac{3}{32} \ (x+y) \ xy \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ 0<x<2, \ 0<y<2

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Practice Problem 4-J
    \displaystyle f(x,y)=\frac{3y}{(x+1)^6} \ \ e^{-y/(x+1)} \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ x>0, \ y>0

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Practice Problem 4-K
    \displaystyle f(x,y)=\frac{y}{(x+1)^4} \ \ e^{-y/(x+1)} \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ x>0, \ y>0

For this problem, only work on the regression curve E[Y \lvert X=x]. Note that E[X] and Var[X] do not exist.

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Problem ………..Answer
4-A
  • \displaystyle \text{Cov}(X,Y)=\frac{1}{10}
  • \displaystyle \rho=\sqrt{\frac{1}{3}}=0.57735
  • \displaystyle E[Y \lvert X=x]=\frac{2 (4-3 x^2+x^3)}{3 (4- 4x+x^2)}=\frac{2 (2+x-x^2)}{3 (2-x)} \ \ \ \ \ 0<x<2
  • \displaystyle y=\frac{2}{3} \ (x+1)
4-B
  • \displaystyle \text{Cov}(X,Y)=\frac{1}{9}
  • \displaystyle \rho=\frac{1}{2}
  • \displaystyle E[Y \lvert X=x]=1+\frac{1}{2} x \ \ \ \ \ 0<x<2
  • \displaystyle y=1+\frac{1}{2} x
4-C
  • \displaystyle \text{Cov}(X,Y)=-\frac{1}{36}
  • \displaystyle \rho=-\frac{1}{11}
  • \displaystyle E[Y \lvert X=x]=\frac{x+\frac{4}{3}}{x+1} \ \ \ \ \ 0<x<2
  • \displaystyle y=\frac{14}{11}-\frac{1}{11} x
4-D
  • \displaystyle \text{Cov}(X,Y)=\frac{1}{3}
  • \displaystyle \rho=\frac{1}{2} \ \sqrt{\frac{15}{7}}=0.7319
  • \displaystyle E[Y \lvert X=x]=\frac{x^2}{2} \ \ \ \ \ 0<x<2
  • \displaystyle y=-\frac{1}{3}+ x
4-E
  • \displaystyle \text{Cov}(X,Y)=-\frac{1}{4}
  • \displaystyle \rho=-\frac{1}{7}=-0.1429
  • \displaystyle E[Y \lvert X=x]=\frac{x+2}{x+1} \ \ \ \ \ x>0
  • \displaystyle y=\frac{12}{7}-\frac{1}{7} x
4-F
  • \displaystyle \text{Cov}(X,Y)=-\frac{3}{40}
  • \displaystyle \rho=\frac{3}{\sqrt{19}}=0.3974
  • \displaystyle E[Y \lvert X=x]=\frac{x}{2} \ \ \ \ \ 0<x<2
  • \displaystyle y=\frac{x}{2}
4-G
  • \displaystyle \text{Cov}(X,Y)=\frac{16}{225}
  • \displaystyle \rho=\frac{4}{\sqrt{66}}=0.4924
  • \displaystyle E[Y \lvert X=x]=\frac{2}{3} x \ \ \ \ \ 0<x<2
  • \displaystyle y=\frac{2}{3} x
4-H
  • \displaystyle \text{Cov}(X,Y)=\frac{298}{3675}
  • \displaystyle \rho=\frac{149}{3 \sqrt{12259}}=0.4486
  • \displaystyle E[Y \lvert X=x]=\frac{x (2x+3)}{3x+6}  \ \ \ \ \ 0<x<2
  • \displaystyle y=-\frac{2}{41}+\frac{149}{246} x
4-I
  • \displaystyle \text{Cov}(X,Y)=-\frac{1}{144}
  • \displaystyle \rho=-\frac{5}{139}=-0.03597
  • \displaystyle E[Y \lvert X=x]=\frac{4x+6}{3x+4}  \ \ \ \ \ 0<x<2
  • \displaystyle y=\frac{204}{139}-\frac{5}{139} x
4-J
  • \displaystyle \text{Cov}(X,Y)=\frac{3}{2}
  • \displaystyle \rho=\frac{1}{\sqrt{3}}=0.57735
  • \displaystyle E[Y \lvert X=x]=2 (x+1) \ \ \ \ \ x>0
  • \displaystyle y=2 (x+1)
4-K
  • \displaystyle E[Y \lvert X=x]=2 (x+1) \ \ \ \ \ x>0

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2 thoughts on “Practice Problem Set 4 – Correlation Coefficient

  1. […] Practice Problem Set 4 – Correlation Coefficient | Probability and Statistics Problem Solve (22:00:15) : […]

  2. […] The concept of covariance and correlation coefficient is given detailed treatment in this post in a companion blog. Practice problems are available here. […]

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