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Thread: random variable function

  1. #1
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    random variable function

    If $\displaystyle X \sim N( \mu , {\sigma}^2)$, then how do you derive the density of $\displaystyle Y = |X| $?
    This is what I did:

    $\displaystyle F_{y}(y) = P(Y < y) $
    $\displaystyle F_{y}(y) = P(|X| < y) $
    $\displaystyle F_{y}(y) = \left\{\begin{array}{cc}P(X< y),&\mbox{ if }
    x > 0 \\ P(-X <y), & \mbox{ if } x<0\end{array}\right. $

    Is this right so far?
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  2. #2
    MHF Contributor matheagle's Avatar
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    The last line is off, you don't want to change the RV X, you want to play with the constant y.

    $\displaystyle F_Y(y) = P(Y < y) = P(|X| < y) $

    $\displaystyle = P(-y<X < y)=F_X(y)-F_X(-y) $

    You can now differentiate wrt y, and obtain

    $\displaystyle f_Y(y) =f_X(y)+f_X(-y) $
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  3. #3
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    I got
    $\displaystyle \frac{exp(\frac{-y^2}{2 \sigma^2})}{\sigma \sqrt{2 \pi}}$
    I'm pretty sure this is right, but can someone vouch for me? Thanks.
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  4. #4
    MHF Contributor matheagle's Avatar
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    Did you really add the two densities?
    AND do note the domain of Y.
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  5. #5
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    Ahh, there's a typo on my original question. If $\displaystyle \mu =0 $, then am I right?
    y>0
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  6. #6
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by ezong View Post
    Ahh, there's a typo on my original question. If $\displaystyle \mu =0 $, then am I right?
    y>0

    I would like to see your addition.
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