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Two variates $x$ and y are uncorrelated and have standard deviations $σ_{x}$ and $σ_{y}$ respectively. What is the correlation coefficient between $x + y$ and $- y$ ?

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\frac{σ_{x} σ_{y}}{{σ_{x}}^{2} + {σ_{y}}^{2}}

\frac{σ_{x} + σ_{y}}{2 σ_{x} σ_{y}}

\frac{{σ_{x}}^{2} - {σ_{y}}^{2}}{{σ_{x}}^{2} + {σ_{y}}^{2}}

\frac{σ_{y} - σ_{x}}{σ_{x} σ_{y}}

answer is C.

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Detailed Solution

Let

p = x + y

and q =

x - y

.
We know that if three variates are such that

z = x + y

, then

\underset{̲}{z} = \underset{̲}{x} + \underset{̲}{y}

.
Therefore, since

p = x + y

and q =

x - y

, we get

\underset{̲}{p} = \underset{̲}{x} + \underset{̲}{y}

and

\underset{̲}{q} = \underset{̲}{x} - \underset{̲}{y}

The covariance between two variates

x

and

y

is given by the formula

cov (x, y) = E (p - \underset{̲}{p}) (q - \underset{̲}{q})

, where

\underset{̲}{x}

is the mean of the variate

x

, and

\underset{̲}{y}

is the mean of the variate y.
Therefore, we get
⇒

cov (p, q) = E (p - \underset{̲}{p}) (q - \underset{̲}{q})

Substituting

p = x + y

, q =

x - y

\underset{̲}{p} = \underset{̲}{x} + \underset{̲}{y}

, and

\underset{̲}{q} = \underset{̲}{x} - \underset{̲}{y}

in the expression, we get
⇒

cov (p, q) = E [(x + y) - (\underset{̲}{x} + \underset{̲}{y})] [(x - y) - (\underset{̲}{x} - \underset{̲}{y})]

Rewriting the expressions, we get
⇒

cov (p, q) = E [x + y - \underset{̲}{x} - \underset{̲}{y}] [x - y - \underset{̲}{x} + \underset{̲}{y}]

⇒

cov (p, q) = E

[

(x - \underset{̲}{x}) + (y - \underset{̲}{y})

]

[(x - \underset{̲}{x}) - (y - \underset{̲}{y})]

The product of the sum and difference of two numbers is given by the algebraic identity

(a - b) (a + b) = a^{2} - b^{2}

.
Therefore, the equation becomes

cov (p, q) = E [{(x - \underset{̲}{x})}^{2} - {(y - \underset{̲}{y})}^{2}]

The expression

E (x + y)

can be written as

E (x) + E (y)

.
Therefore, we get
⇒

cov (p, q) = E {(x - \underset{̲}{x})}^{2} - E {(y - \underset{̲}{y})}^{2}

The variance of a variate z is given by the formula

E (z - z^{2})

Therefore, we get
Variance of

x

E {(x - \underset{̲}{x})}^{2}

Variance of y =

E {(y - \underset{̲}{y})}^{2}

It is given that the two variates x and y have standard deviations σx and σy respectively.
Variance is the square of the standard deviation.
Therefore, we get the equations

{σx}^{2} = E {(x - \underset{̲}{x})}^{2}

and

{σy}^{2} = E {(y - \underset{̲}{y})}^{2}

Substituting

{σx}^{2} = E {(x - \underset{̲}{x})}^{2}

and

{σy}^{2} = E {(y - \underset{̲}{y})}^{2}

in the equation

cov (p, q) = E [{(x - \underset{̲}{x})}^{2} - {(y - \underset{̲}{y})}^{2}]

, we get
⇒

cov (p, q) = {σx}^{2} - {σy}^{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 {(z - \underset{̲}{z})}^{2}

Therefore, we get
⇒ Variance of p =

E {(p - \underset{̲}{p})}^{2}

⇒Variance of q =

E {(q - \underset{̲}{q})}^{2}

Substituting

p = x + y

, q =

x - y

\underset{̲}{p} = \underset{̲}{x} + \underset{̲}{y}

and

\underset{̲}{q} = \underset{̲}{x} - \underset{̲}{y}

, in the expressions, we get
⇒Variance of p =

E {[(x + y) - (\underset{̲}{x} + \underset{̲}{y})]}^{2}

⇒Variance of q =

E {[(x - y) - (\underset{̲}{x} - \underset{̲}{y})]}^{2}

Rewriting the expressions, we get
⇒Variance of p =

E {[(x - \underset{̲}{x}) + (y - \underset{̲}{y})]}^{2}

⇒Variance of q =

E {[(x - \underset{̲}{x}) - (y - \underset{̲}{y})]}^{2}

Expanding the terms using algebraic identity,

{(a + b)}^{2} = a^{2} + b^{2} + 2 ab

, we get
⇒Variance of p =

E {(x - \underset{̲}{x})}^{2} + E {(y - \underset{̲}{y})}^{2} + 2 E (x - \underset{̲}{x}) (y - \underset{̲}{y})

⇒Variance of q =

E {(x - \underset{̲}{x})}^{2} + E {(y - \underset{̲}{y})}^{2} - 2 E (x - \underset{̲}{x}) (y - \underset{̲}{y})

Since the two variates x and y are uncorrelated, we get

E [(x - \underset{̲}{x}) (y - \underset{̲}{y})]

=0.
Therefore, the equations become
⇒Variance of p =

E {(x - \underset{̲}{x})}^{2} + E {(y - \underset{̲}{y})}^{2}

⇒Variance of q =

E {(x - \underset{̲}{x})}^{2} + E {(y - \underset{̲}{y})}^{2}

Thus, we get
⇒Variance of p =

{σx}^{2} + {σy}^{2}

⇒Variance of q =

{σx}^{2} + {σy}^{2}

Taking the square roots on both the sides, we get the standard deviations as
⇒Standard deviation of p =

\sqrt {σx}^{2} + {σy}^{2}

⇒Standard deviation of q =

{σx}^{2} + {σy}^{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 =

\frac{cov (p, q)}{s . d . (p) s . d . (q)}

Substituting

s . d . (p) = \sqrt {σx}^{2} + {σy}^{2}

, and cov(p,q)=

{σx}^{2} - {σy}^{2}

in the expression, we get
⇒Coefficient of correlation =

\frac{{σx}^{2} - {σy}^{2}}{\sqrt {σx}^{2} + {σy}^{2} \times \sqrt {σx}^{2} + {σy}^{2}}

Rewriting the expression, we get
⇒Coefficient of correlation =

\frac{{σx}^{2} - {σy}^{2}}{\sqrt{({σx}^{2} + {σy}^{2}) ({σx}^{2} + {σy}^{2})}}

Simplifying the expression, we get
Coefficient of correlation =

\frac{{σx}^{2} - {σy}^{2}}{{σx}^{2} + {σy}^{2}}

Therefore, the coefficient of correlation of

x + y

and

x - y

\frac{{σx}^{2} - {σy}^{2}}{{σx}^{2} + {σy}^{2}}

Thus, the correct option is option (3).

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Two variates x and y are uncorrelated and have standard deviations σx and σy respectively. What is the correlation coefficient between x+y and -y ?

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Two variates $x$ and y are uncorrelated and have standard deviations $σ_{x}$ and $σ_{y}$ respectively. What is the correlation coefficient between $x + y$ and $- y$ ?