Statistics help needed

**bill2010** · January 7th 2009, 09:16 AM

Im stuck with understanding course material on probability distributions.
I need some help working out and understanding :-

if there is an exact negative linear relationship between Y and X -

Y = a - bX,

a and b are positive-valued parameters, and X and Y are R.V.s

The value of the correlation coefficient corresponding to X and Y is equal to –1

Thanks in advance!

**mr fantastic** · January 7th 2009, 02:23 PM

Originally Posted by bill2010

Im stuck with understanding course material on probability distributions.
I need some help working out and understanding :-

if there is an exact negative linear relationship between Y and X -

Y = a - bX,

a and b are positive-valued parameters, and X and Y are R.V.s

The value of the correlation coefficient corresponding to X and Y is equal to –1

Thanks in advance!

What is the actual problem you have here ...?

**bill2010** · January 7th 2009, 03:23 PM

Its not a very clear question.

I think its saying that if there is a negative linear relationship between X and Y, that it must have a correllation coefficient of -1.
Basically it has to just be shown.

Thanks for your response!

**mr fantastic** · January 8th 2009, 07:05 PM

Originally Posted by bill2010

Its not a very clear question.

I think its saying that if there is a negative linear relationship between X and Y, that it must have a correllation coefficient of -1.
Basically it has to just be shown.

Thanks for your response!

Have you tried starting from the definition: $\rho = \frac{Cov(X, Y)}{\sigma_X \sigma_Y}$ .

**mr fantastic** · January 8th 2009, 09:35 PM

Originally Posted by mr fantastic

Have you tried starting from the definition: $\rho = \frac{Cov(X, Y)}{\sigma_X \sigma_Y}$ .

$Y = b - aX$

$\Rightarrow E(Y) = b - a E(X)$ .... (1)

and

$\sigma_Y = a \sigma_X$ .... (2) since $a > 0$ is given.

$Cov(X, Y) = E(XY) - E(X)E(Y) = E(X (b - aX)) - E(X) \cdot [b - a E(X)]$

using (1)

$= E(bX - aX^2) - b E(X) + a [E(X)]^2 = b E(X) - a E(X^2) - b E(X) + a [E(X)]^2$

$= a ([E(X)]^2 - E(X^2)) = - a Var(X) = -a \sigma^2_X$ .... (3)

Substitute (2) and (3) into the formula for $\rho$ : $\rho = \frac{-a \sigma_X^2}{a \sigma_X^2} = -1$ .

**bill2010** · January 9th 2009, 10:51 AM

Thanks again Mr F

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