# Statistics help needed

• Jan 7th 2009, 09:16 AM
bill2010
Statistics help needed
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

• Jan 7th 2009, 02:23 PM
mr fantastic
Quote:

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

What is the actual problem you have here ...?
• Jan 7th 2009, 03:23 PM
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.

• Jan 8th 2009, 07:05 PM
mr fantastic
Quote:

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.

Have you tried starting from the definition: $\rho = \frac{Cov(X, Y)}{\sigma_X \sigma_Y}$.
• Jan 8th 2009, 09:35 PM
mr fantastic
Quote:

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 $\rho$: $\rho = \frac{-a \sigma_X^2}{a \sigma_X^2} = -1$.
• Jan 9th 2009, 10:51 AM
bill2010
Thanks again Mr F(Bow)