## 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 $\sigma_Y^2=Var[Y]$.

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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

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

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

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

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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

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

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

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

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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
• $\displaystyle y=\frac{2}{3} \ (x+1)$
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
• $\displaystyle y=1+\frac{1}{2} x$
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
• $\displaystyle y=\frac{14}{11}-\frac{1}{11} x$
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
• $\displaystyle y=-\frac{1}{3}+ x$
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
• $\displaystyle y=\frac{x}{2}$
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
• $\displaystyle y=\frac{2}{3} x$
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
• $\displaystyle y=-\frac{2}{41}+\frac{149}{246} x$
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
• $\displaystyle y=\frac{204}{139}-\frac{5}{139} x$
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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