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Math Help - Statistics help needed

  1. #1
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    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

    Thanks in advance!
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  2. #2
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    Quote Originally Posted by bill2010 View Post
    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 ...?
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  3. #3
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    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!
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  4. #4
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    Quote Originally Posted by bill2010 View Post
    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}.
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    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.
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  6. #6
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    Thanks again Mr F
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