Two variates \[x\] and \[y\] are uncorrelated and have standard deviations \[{\sigma _x}\] and \[{\sigma _y}\] respectively. What is the correlation coefficient between \[x + y\] and \[x - y\]?
(a) \[\dfrac{{{\sigma _x}{\sigma _y}}}{{\sigma _x^2 + \sigma _y^2}}\]
(b) \[\dfrac{{{\sigma _x} + {\sigma _y}}}{{2{\sigma _x}{\sigma _y}}}\]
(c) \[\dfrac{{\sigma _x^2 - \sigma _y^2}}{{\sigma _x^2 + \sigma _y^2}}\]
(d) \[\dfrac{{{\sigma _y} - {\sigma _x}}}{{{\sigma _x}{\sigma _y}}}\]
Answer
Verified558.3k+ views
Hint: We need to find the correlation coefficient between \[x + y\] and \[x - y\]. Let \[p = x + y\] and \[q = x - y\]. We will find the variance of these variates. Then, we will calculate the covariance between them. Finally, we will find the correlation coefficient using the calculated variances and covariance. The correlation coefficient between two variates is the ratio of the covariance between them, and the product of their standard deviations.
Formula Used:
We will use the following formulas:
1.The covariance between two variates \[x\] and \[y\] is given by the formula \[{\mathop{\rm cov}} \left( {x,y} \right) = E\left( {x - \overline x } \right)\left( {y - \overline y } \right)\], where \[\overline x \] is the mean of the variate \[x\], and \[\overline y \] is the mean of the variate \[y\].
2.The product of the sum and difference of two numbers is given by the algebraic identity \[\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}\].
3.The variance of a variate \[z\] is given by the formula \[E{\left( {z - \overline z } \right)^2}\].
Complete step-by-step answer:
Let \[p = x + y\] and \[q = x - y\].
We know that if three variates are such that \[z = x + y\], then \[\overline z = \overline x + \overline y \].
Therefore, since \[p = x + y\] and \[q = x - y\], we get
\[\overline p = \overline x + \overline y \] and \[\overline q = \overline x - \overline y \]
The covariance between two variates \[x\] and \[y\] is given by the formula \[{\mathop{\rm cov}} \left( {x,y} \right) = E\left( {x - \overline x } \right)\left( {y - \overline y } \right)\], where \[\overline x \] is the mean of the variate \[x\], and \[\overline y \] is the mean of the variate \[y\].
Therefore, we get
\[ \Rightarrow {\mathop{\rm cov}} \left( {p,q} \right) = E\left( {p - \overline p } \right)\left( {q - \overline q } \right)\]
Substituting \[p = x + y\], \[q = x - y\], \[\overline p = \overline x + \overline y \], and \[\overline q = \overline x - \overline y \] in the expression, we get
\[ \Rightarrow {\mathop{\rm cov}} \left( {p,q} \right) = E\left[ {\left( {x + y} \right) - \left( {\overline x + \overline y } \right)} \right]\left[ {\left( {x - y} \right) - \left( {\overline x - \overline y } \right)} \right]\]
Rewriting the expressions, we get
\[\begin{array}{l} \Rightarrow {\mathop{\rm cov}} \left( {p,q} \right) = E\left[ {x + y - \overline x - \overline y } \right]\left[ {x - y - \overline x + \overline y } \right]\\ \Rightarrow {\mathop{\rm cov}} \left( {p,q} \right) = E\left[ {\left( {x - \overline x } \right) + \left( {y - \overline y } \right)} \right]\left[ {\left( {x - \overline x } \right) - \left( {y - \overline y } \right)} \right]\end{array}\]
The product of the sum and difference of two numbers is given by the algebraic identity \[\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}\].
Therefore, the equation becomes
\[ \Rightarrow {\mathop{\rm cov}} \left( {p,q} \right) = E\left[ {{{\left( {x - \overline x } \right)}^2} - {{\left( {y - \overline y } \right)}^2}} \right]\]
The expression \[E\left( {x + y} \right)\] can be written as \[E\left( x \right) + E\left( y \right)\].
Therefore, we get
\[ \Rightarrow {\mathop{\rm cov}} \left( {p,q} \right) = E{\left( {x - \overline x } \right)^2} - E{\left( {y - \overline y } \right)^2}\]
The variance of a variate \[z\] is given by the formula \[E{\left( {z - \overline z } \right)^2}\].
Therefore, we get
Variance of \[x\] \[ = E{\left( {x - \overline x } \right)^2}\]
Variance of \[y\] \[ = E{\left( {y - \overline y } \right)^2}\]
It is given that the two variates \[x\] and \[y\] have standard deviations \[{\sigma _x}\] and \[{\sigma _y}\] respectively.
Variance is the square of the standard deviation.
Therefore, we get the equations
\[\sigma _x^2 = E{\left( {x - \overline x } \right)^2}\] and \[\sigma _y^2 = E{\left( {y - \overline y } \right)^2}\]
Substituting \[\sigma _x^2 = E{\left( {x - \overline x } \right)^2}\] and \[\sigma _y^2 = E{\left( {y - \overline y } \right)^2}\] in the equation \[{\mathop{\rm cov}} \left( {p,q} \right) = E{\left( {x - \overline x } \right)^2} - E{\left( {y - \overline y } \right)^2}\], we get
\[ \Rightarrow {\mathop{\rm cov}} \left( {p,q} \right) = \sigma _x^2 - \sigma _y^2\]
Now, we will find the variance of the variates \[p = x + y\] and \[q = x - y\].
The variance of a variate \[z\] is given by the formula \[E{\left( {z - \overline z } \right)^2}\].
Therefore, we get
\[ \Rightarrow \] Variance of \[p\] \[ = E{\left( {p - \overline p } \right)^2}\]
\[ \Rightarrow \] Variance of \[q\] \[ = E{\left( {q - \overline q } \right)^2}\]
Substituting \[p = x + y\], \[q = x - y\], \[\overline p = \overline x + \overline y \], and \[\overline q = \overline x - \overline y \] in the expressions, we get
\[ \Rightarrow \] Variance of \[p\] \[ = E{\left[ {\left( {x + y} \right) - \left( {\overline x + \overline y } \right)} \right]^2}\]
\[ \Rightarrow \] Variance of \[q\] \[ = E{\left[ {\left( {x - y} \right) - \left( {\overline x - \overline y } \right)} \right]^2}\]
Rewriting the expressions, we get
\[ \Rightarrow \]Variance of \[p\] \[ = E{\left[ {\left( {x - \overline x } \right) + \left( {y - \overline y } \right)} \right]^2}\]
\[ \Rightarrow \]Variance of \[q\] \[ = E{\left[ {\left( {x - \overline x } \right) - \left( {y - \overline y } \right)} \right]^2}\]
Expanding the terms using algebraic identity, \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\], we get
\[ \Rightarrow \] Variance of \[p\] \[ = E{\left[ {\left( {x - \overline x } \right)} \right]^2} + E{\left[ {\left( {y - \overline y } \right)} \right]^2} + 2E\left[ {\left( {x - \overline x } \right)\left( {y - \overline y } \right)} \right]\]
\[ \Rightarrow \]Variance of \[q\] \[ = E{\left[ {\left( {x - \overline x } \right)} \right]^2} + E{\left[ {\left( {y - \overline y } \right)} \right]^2} - 2E\left[ {\left( {x - \overline x } \right)\left( {y - \overline y } \right)} \right]\]
Since the two variates \[x\] and \[y\] are uncorrelated, we get \[E\left[ {\left( {x - \overline x } \right)\left( {y - \overline y } \right)} \right] = 0\].
Therefore, the equations become
\[ \Rightarrow \]Variance of \[p\] \[ = E{\left[ {\left( {x - \overline x } \right)} \right]^2} + E{\left[ {\left( {y - \overline y } \right)} \right]^2}\]
\[ \Rightarrow \] Variance of \[q\] \[ = E{\left[ {\left( {x - \overline x } \right)} \right]^2} + E{\left[ {\left( {y - \overline y } \right)} \right]^2}\]
Thus, we get
\[ \Rightarrow \] Variance of \[p\] \[ = \sigma _x^2 + \sigma _y^2\]
\[ \Rightarrow \] Variance of \[q\] \[ = \sigma _x^2 + \sigma _y^2\]
Taking the square roots on both the sides, we get the standard deviations as
\[ \Rightarrow \] Standard deviation of \[p\] \[ = \sqrt {\sigma _x^2 + \sigma _y^2} \]
\[ \Rightarrow \] Standard deviation of \[q\] \[ = \sqrt {\sigma _x^2 + \sigma _y^2} \]
Finally, we will find the coefficient of correlation of the two variates \[p = x + y\] and \[q = x - y\].
Therefore, we get
Coefficient of correlation \[ = \dfrac{{{\mathop{\rm cov}} \left( {p,q} \right)}}{{s.d.\left( p \right)s.d.\left( q \right)}}\]
Substituting \[s.d.\left( p \right) = \sqrt {\sigma _x^2 + \sigma _y^2} \], \[s.d.\left( q \right) = \sqrt {\sigma _x^2 + \sigma _y^2} \], and \[{\mathop{\rm cov}} \left( {p,q} \right) = \sigma _x^2 - \sigma _y^2\] in the expression, we get
\[ \Rightarrow \] Coefficient of correlation \[ = \dfrac{{\sigma _x^2 - \sigma _y^2}}{{\sqrt {\sigma _x^2 + \sigma _y^2} \times \sqrt {\sigma _x^2 + \sigma _y^2} }}\]
Rewriting the expression, we get
\[ \Rightarrow \] Coefficient of correlation \[ = \dfrac{{\sigma _x^2 - \sigma _y^2}}{{\sqrt {\left( {\sigma _x^2 + \sigma _y^2} \right)\left( {\sigma _x^2 + \sigma _y^2} \right)} }} = \dfrac{{\sigma _x^2 - \sigma _y^2}}{{\sqrt {{{\left( {\sigma _x^2 + \sigma _y^2} \right)}^2}} }}\]
Simplifying the expression, we get
Coefficient of correlation \[ = \dfrac{{\sigma _x^2 - \sigma _y^2}}{{\sigma _x^2 + \sigma _y^2}}\]
Therefore, the coefficient of correlation of \[x + y\] and \[x - y\] is \[\dfrac{{\sigma _x^2 - \sigma _y^2}}{{\sigma _x^2 + \sigma _y^2}}\].
Thus, the correct option is option (c).
Note: We used the coefficient of correlation in the solution. The correlation coefficient is donated by the greek letter ‘rho’, that is \[\rho \]. Here we have found out the variance and standard deviation and substituted in the formula of the coefficient of correlation. Standard deviation measures the spread of a statistical data and it is denoted by \[\sigma \]. The standard deviation can never be negative and therefore its smallest value is 0. Variance measures how much each data of a set from its average value. It is always positive and it is denoted by \[{\sigma ^2}\].
Formula Used:
We will use the following formulas:
1.The covariance between two variates \[x\] and \[y\] is given by the formula \[{\mathop{\rm cov}} \left( {x,y} \right) = E\left( {x - \overline x } \right)\left( {y - \overline y } \right)\], where \[\overline x \] is the mean of the variate \[x\], and \[\overline y \] is the mean of the variate \[y\].
2.The product of the sum and difference of two numbers is given by the algebraic identity \[\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}\].
3.The variance of a variate \[z\] is given by the formula \[E{\left( {z - \overline z } \right)^2}\].
Complete step-by-step answer:
Let \[p = x + y\] and \[q = x - y\].
We know that if three variates are such that \[z = x + y\], then \[\overline z = \overline x + \overline y \].
Therefore, since \[p = x + y\] and \[q = x - y\], we get
\[\overline p = \overline x + \overline y \] and \[\overline q = \overline x - \overline y \]
The covariance between two variates \[x\] and \[y\] is given by the formula \[{\mathop{\rm cov}} \left( {x,y} \right) = E\left( {x - \overline x } \right)\left( {y - \overline y } \right)\], where \[\overline x \] is the mean of the variate \[x\], and \[\overline y \] is the mean of the variate \[y\].
Therefore, we get
\[ \Rightarrow {\mathop{\rm cov}} \left( {p,q} \right) = E\left( {p - \overline p } \right)\left( {q - \overline q } \right)\]
Substituting \[p = x + y\], \[q = x - y\], \[\overline p = \overline x + \overline y \], and \[\overline q = \overline x - \overline y \] in the expression, we get
\[ \Rightarrow {\mathop{\rm cov}} \left( {p,q} \right) = E\left[ {\left( {x + y} \right) - \left( {\overline x + \overline y } \right)} \right]\left[ {\left( {x - y} \right) - \left( {\overline x - \overline y } \right)} \right]\]
Rewriting the expressions, we get
\[\begin{array}{l} \Rightarrow {\mathop{\rm cov}} \left( {p,q} \right) = E\left[ {x + y - \overline x - \overline y } \right]\left[ {x - y - \overline x + \overline y } \right]\\ \Rightarrow {\mathop{\rm cov}} \left( {p,q} \right) = E\left[ {\left( {x - \overline x } \right) + \left( {y - \overline y } \right)} \right]\left[ {\left( {x - \overline x } \right) - \left( {y - \overline y } \right)} \right]\end{array}\]
The product of the sum and difference of two numbers is given by the algebraic identity \[\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}\].
Therefore, the equation becomes
\[ \Rightarrow {\mathop{\rm cov}} \left( {p,q} \right) = E\left[ {{{\left( {x - \overline x } \right)}^2} - {{\left( {y - \overline y } \right)}^2}} \right]\]
The expression \[E\left( {x + y} \right)\] can be written as \[E\left( x \right) + E\left( y \right)\].
Therefore, we get
\[ \Rightarrow {\mathop{\rm cov}} \left( {p,q} \right) = E{\left( {x - \overline x } \right)^2} - E{\left( {y - \overline y } \right)^2}\]
The variance of a variate \[z\] is given by the formula \[E{\left( {z - \overline z } \right)^2}\].
Therefore, we get
Variance of \[x\] \[ = E{\left( {x - \overline x } \right)^2}\]
Variance of \[y\] \[ = E{\left( {y - \overline y } \right)^2}\]
It is given that the two variates \[x\] and \[y\] have standard deviations \[{\sigma _x}\] and \[{\sigma _y}\] respectively.
Variance is the square of the standard deviation.
Therefore, we get the equations
\[\sigma _x^2 = E{\left( {x - \overline x } \right)^2}\] and \[\sigma _y^2 = E{\left( {y - \overline y } \right)^2}\]
Substituting \[\sigma _x^2 = E{\left( {x - \overline x } \right)^2}\] and \[\sigma _y^2 = E{\left( {y - \overline y } \right)^2}\] in the equation \[{\mathop{\rm cov}} \left( {p,q} \right) = E{\left( {x - \overline x } \right)^2} - E{\left( {y - \overline y } \right)^2}\], we get
\[ \Rightarrow {\mathop{\rm cov}} \left( {p,q} \right) = \sigma _x^2 - \sigma _y^2\]
Now, we will find the variance of the variates \[p = x + y\] and \[q = x - y\].
The variance of a variate \[z\] is given by the formula \[E{\left( {z - \overline z } \right)^2}\].
Therefore, we get
\[ \Rightarrow \] Variance of \[p\] \[ = E{\left( {p - \overline p } \right)^2}\]
\[ \Rightarrow \] Variance of \[q\] \[ = E{\left( {q - \overline q } \right)^2}\]
Substituting \[p = x + y\], \[q = x - y\], \[\overline p = \overline x + \overline y \], and \[\overline q = \overline x - \overline y \] in the expressions, we get
\[ \Rightarrow \] Variance of \[p\] \[ = E{\left[ {\left( {x + y} \right) - \left( {\overline x + \overline y } \right)} \right]^2}\]
\[ \Rightarrow \] Variance of \[q\] \[ = E{\left[ {\left( {x - y} \right) - \left( {\overline x - \overline y } \right)} \right]^2}\]
Rewriting the expressions, we get
\[ \Rightarrow \]Variance of \[p\] \[ = E{\left[ {\left( {x - \overline x } \right) + \left( {y - \overline y } \right)} \right]^2}\]
\[ \Rightarrow \]Variance of \[q\] \[ = E{\left[ {\left( {x - \overline x } \right) - \left( {y - \overline y } \right)} \right]^2}\]
Expanding the terms using algebraic identity, \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\], we get
\[ \Rightarrow \] Variance of \[p\] \[ = E{\left[ {\left( {x - \overline x } \right)} \right]^2} + E{\left[ {\left( {y - \overline y } \right)} \right]^2} + 2E\left[ {\left( {x - \overline x } \right)\left( {y - \overline y } \right)} \right]\]
\[ \Rightarrow \]Variance of \[q\] \[ = E{\left[ {\left( {x - \overline x } \right)} \right]^2} + E{\left[ {\left( {y - \overline y } \right)} \right]^2} - 2E\left[ {\left( {x - \overline x } \right)\left( {y - \overline y } \right)} \right]\]
Since the two variates \[x\] and \[y\] are uncorrelated, we get \[E\left[ {\left( {x - \overline x } \right)\left( {y - \overline y } \right)} \right] = 0\].
Therefore, the equations become
\[ \Rightarrow \]Variance of \[p\] \[ = E{\left[ {\left( {x - \overline x } \right)} \right]^2} + E{\left[ {\left( {y - \overline y } \right)} \right]^2}\]
\[ \Rightarrow \] Variance of \[q\] \[ = E{\left[ {\left( {x - \overline x } \right)} \right]^2} + E{\left[ {\left( {y - \overline y } \right)} \right]^2}\]
Thus, we get
\[ \Rightarrow \] Variance of \[p\] \[ = \sigma _x^2 + \sigma _y^2\]
\[ \Rightarrow \] Variance of \[q\] \[ = \sigma _x^2 + \sigma _y^2\]
Taking the square roots on both the sides, we get the standard deviations as
\[ \Rightarrow \] Standard deviation of \[p\] \[ = \sqrt {\sigma _x^2 + \sigma _y^2} \]
\[ \Rightarrow \] Standard deviation of \[q\] \[ = \sqrt {\sigma _x^2 + \sigma _y^2} \]
Finally, we will find the coefficient of correlation of the two variates \[p = x + y\] and \[q = x - y\].
Therefore, we get
Coefficient of correlation \[ = \dfrac{{{\mathop{\rm cov}} \left( {p,q} \right)}}{{s.d.\left( p \right)s.d.\left( q \right)}}\]
Substituting \[s.d.\left( p \right) = \sqrt {\sigma _x^2 + \sigma _y^2} \], \[s.d.\left( q \right) = \sqrt {\sigma _x^2 + \sigma _y^2} \], and \[{\mathop{\rm cov}} \left( {p,q} \right) = \sigma _x^2 - \sigma _y^2\] in the expression, we get
\[ \Rightarrow \] Coefficient of correlation \[ = \dfrac{{\sigma _x^2 - \sigma _y^2}}{{\sqrt {\sigma _x^2 + \sigma _y^2} \times \sqrt {\sigma _x^2 + \sigma _y^2} }}\]
Rewriting the expression, we get
\[ \Rightarrow \] Coefficient of correlation \[ = \dfrac{{\sigma _x^2 - \sigma _y^2}}{{\sqrt {\left( {\sigma _x^2 + \sigma _y^2} \right)\left( {\sigma _x^2 + \sigma _y^2} \right)} }} = \dfrac{{\sigma _x^2 - \sigma _y^2}}{{\sqrt {{{\left( {\sigma _x^2 + \sigma _y^2} \right)}^2}} }}\]
Simplifying the expression, we get
Coefficient of correlation \[ = \dfrac{{\sigma _x^2 - \sigma _y^2}}{{\sigma _x^2 + \sigma _y^2}}\]
Therefore, the coefficient of correlation of \[x + y\] and \[x - y\] is \[\dfrac{{\sigma _x^2 - \sigma _y^2}}{{\sigma _x^2 + \sigma _y^2}}\].
Thus, the correct option is option (c).
Note: We used the coefficient of correlation in the solution. The correlation coefficient is donated by the greek letter ‘rho’, that is \[\rho \]. Here we have found out the variance and standard deviation and substituted in the formula of the coefficient of correlation. Standard deviation measures the spread of a statistical data and it is denoted by \[\sigma \]. The standard deviation can never be negative and therefore its smallest value is 0. Variance measures how much each data of a set from its average value. It is always positive and it is denoted by \[{\sigma ^2}\].
Recently Updated Pages
Master Class 8 Maths: Engaging Questions & Answers for Success
Class 8 Question and Answer - Your Ultimate Solutions Guide
Master Class 7 Maths: Engaging Questions & Answers for Success
Class 7 Question and Answer - Your Ultimate Solutions Guide
Master Class 6 Maths: Engaging Questions & Answers for Success
Class 6 Question and Answer - Your Ultimate Solutions Guide
Trending doubts
Why is there a time difference of about 5 hours between class 10 social science CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE
What is the median of the first 10 natural numbers class 10 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Which of the following does not have a fundamental class 10 physics CBSE
State and prove converse of BPT Basic Proportionality class 10 maths CBSE