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Math Help - transformation of a random variable

  1. #1
    saoricapital
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    transformation of a random variable

    let Y = aX + b, a and b are constants

    Fx(x) and fx(x) are the cdf and pdf of X

    so fy(y) = 1/|a| fx(y-b/a), why is that?

    in other words why is dFx(y-b/a) / dy = 1/|a| fx(y-b/a)?
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  2. #2
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    Quote Originally Posted by saoricapital View Post
    let Y = aX + b, a and b are constants

    Fx(x) and fx(x) are the cdf and pdf of X

    so fy(y) = 1/|a| fx(y-b/a), why is that?

    in other words why is dFx(y-b/a) / dy = 1/|a| fx(y-b/a)?
    Suppose that a>0:
    Let F_Y(y) be the cdf of f_Y(y). Now F_Y(y) = P(y\leq Y) = P(y\leq aX+b) = P(X \geq a^{-1}(y-b)) = 1 - P(X\leq a^{-1}(y-b)) = 1- F_X(a^{-1}(y-b))
    That means,
    f_Y(y) = F'_Y(y) = 1 - F_X'(a^{-1}(y-b)) = -\frac{1}{a}f_x(a^{-1}(y-b)).
    It seems your formula is wrong.
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