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Question: The correlation coefficient between two variables \(X\) and \(Y\) is found to be 0.6. All the observ...

The correlation coefficient between two variables XX and YY is found to be 0.6. All the observations on XX and YY are transformed using the transformations U=23XU=2-3X andV=4Y+1V=4Y+1. The correlation coefficient between the transformed variables UU and VV will be: $$$$
A. -0.5$$$$$ B. +0.5 C. $-0.6
D. +0.6$$$$$ E. -0.5$$$$$

Explanation

Solution

We recall the definition of linear operator, expectation, covariance, variance and correlation coefficient. We show that covariance and variance are linear operators as they are defined on expectation E(X)E\left( X \right) and expectation is a linear operator. We use the property of linear operators (addition, scalar multiplication) to convert the correlation coefficient of U,VU, V in terms of the correlation coefficient of X,YX, Y.$$$$

Complete step-by-step solution:
We know that a linear map or operator is defined between vector spaces V,WV,W such that f:VWf:V\to W then it has to satisfy the property of addition that is for some x,yVx,y\in V
f(x+y)=f(x+y)f\left( x+y \right)=f\left( x+y \right)
The operator also has to satisfy property of scalar multiplication that is for some real number aacalled scalar
f(ax)=af(x)f\left( ax \right)=af\left( x \right)
We know that expectation of any random variable XX represented by E(X)E\left( X \right) follows the property of linear operator which means for some scalars aa and bb and two variables X,YX,Y

& E\left( X+Y \right)=E\left( X+Y \right) \\\ & E\left( ax \right)=aE\left( X \right) \\\ \end{aligned}$$ We know that the expectation of constant say $c$ is zero which means $$E\left( c \right)=0$$ The covariance of random variables $X$ and $Y$ measures the amount of joint variability between $X$ and $Y$ who vary depending on each other. It is defined as the expected value (or mean) of the product of their deviations from their individual expected values It is denoted by COV and given by the formula $$\text{COV}\left( X,Y \right)=E\left[ \left( X-E\left( X \right) \right)\left( Y-E\left( Y \right) \right) \right]$$ We see that covariance is also an expectation and will follow the property of linear operator which means $$\begin{aligned} & \text{COV}\left( X+a,Y+c \right)=\text{COV}\left( X,Y \right) \\\ & \text{COV}\left( aX,cY \right)=ab\text{COV}\left( X,Y \right) \\\ \end{aligned}$$ We know variance is the expectation of the squared deviation of a random variable from its mean. The variance of random variable $X$ is, $$\text{Var}\left( X \right)=E\left[ {{\left( X-E\left( X \right) \right)}^{2}} \right]$$ We see that covariance is also an expectation and will follow the property of linear operator which means $$\begin{aligned} & \text{Var}\left( X+Y \right)=\text{Var}\left( X \right)+\text{Var}\left( Y \right) \\\ & \text{Var}\left( aX \right)={{a}^{2}}X \\\ \end{aligned}$$ So we have $$\text{COV}\left( c \right)=0,\text{Var}\left( c \right)=0$$ We know that the correlation coefficient measures the degree of relationship between the random variables $X,Y$ .It is denoted by $\rho $ and given by $$\rho \left( X,Y \right)=\dfrac{\text{COV}\left( X,Y \right)}{\sqrt{\text{Var}\left( X \right)\text{Var}\left( Y \right)}}$$ We are given the question that the correlation coefficient between two variables $X$ and $Y$ is 0.6 which means $\rho \left( X,Y \right)=-0.6$ . All the observations on $X$ and $Y$ are transformed using the transformations $U=2-3X$ and$V=4Y+1$. The correlation coefficient between the transformed variables $U$ and $V$ is $$\begin{aligned} & \rho \left( U,V \right)=\dfrac{\text{COV}\left( U,V \right)}{\sqrt{\text{Var}\left( U \right)\text{Var}\left( V \right)}} \\\ & \Rightarrow \rho \left( U,V \right)=\dfrac{\text{COV}\left( 2-3X,4Y+1 \right)}{\sqrt{\text{Var}\left( 2-3X \right)\text{Var}\left( 4Y+1 \right)}} \\\ \end{aligned}$$ We use the property of linear operator of addition and have $$\Rightarrow \rho \left( U,V \right)=\dfrac{\text{COV}\left( -3X,4Y \right)}{\sqrt{\text{Var}\left( -3X \right)\text{Var}\left( 4Y \right)}}$$ We see that in the above step $\text{Var}\left( -2 \right)=0,\text{Var}\left( 1 \right)=0$. We proceed to use scalar multiplication property and have, $$\begin{aligned} & \rho \left( U,V \right)=\dfrac{\left( -3\times 4 \right)\text{COV}\left( X,Y \right)}{\sqrt{{{\left( -3 \right)}^{2}}\text{Var}\left( U \right){{\left( 4 \right)}^{2}}\text{Var}\left( V \right)}} \\\ & \Rightarrow \rho \left( U,V \right)=\dfrac{-12\text{COV}\left( X, \right)}{12\sqrt{\text{Var}\left( X \right)\text{Var}\left( Y \right)}} \\\ & \Rightarrow \rho \left( U,V \right)=-1\times \dfrac{\text{COV}\left( X, \right)}{\sqrt{\text{Var}\left( X \right)\text{Var}\left( Y \right)}}=-\rho \left( X,Y \right)=-0.6 \\\ \end{aligned}$$ **So the correct option is C.** **Note:** We note that the square root of variance $\text{SD}=\sqrt{\text{Var}\left( X \right)}$ is always positive quantity and that is why we have rejected negative values for square root. The correlation coefficient always lies between $-1$ to 1. If the value is closer to 1 there is a high positive relation, if closer to $-1$ there is a negative relationship and if closer to 0 there is no relation as it is the slope of the line $Y=aX+b$.