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Math Help - Finding distribution of new random variable

  1. #1
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    Finding distribution of new random variable

    If X is continous with distribution function F(x) and pdf f(x). It is given
    Y=2X, finding the distribution of Y will be:
    Fy(a)= P(Y< a)
    = P(2X< a)
    = P(X< a/2)
    = Fx(a/2)
    Differentiation will give me:
    fy(a) = 1/2fx(a/2)

    What if the relationship is Y= 2 F(X), where F(X) is the cumulative distribution of X. I can't seem to make X the subject of the above equation.
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  2. #2
    MHF Contributor matheagle's Avatar
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    I'll aussme F_X(x) has an inverse.
    But we all know what happens when we ass/u/me.

    F_Y(y)=P(Y\le y)=P(2F_X(x)\le y)=P(F_X(x)\le y/2)

    =P(X\le F^{-1}_X (y/2))=F_X(F^{-1}_X (y/2))=y/2.

    This seems right and 0<y<2.

    So Y\sim U(0,2).

    Or f_Y(y)=f_X(x)|dx/dy|.

    With y=2F_X(x) we have dy/dx=2f_X(x)

    giving us f_Y(y)=f_X(x){1\over 2f_X(x)}=1/2.
    Last edited by matheagle; March 16th 2009 at 06:11 PM.
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  3. #3
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    Normal distribution

    Hey really thanks a lot.
    I think the case when F_X(F^{-1}_X (y/2))=y/2 really help a lot!

    By the way, assuming now X is a continous normal distribution with cumulative density φ(X), will this still work? For example may i use
    F_X(Phi^{-1}_X (y/2)) when in this case, Y= 2φ(X) rather then Y= 2F(x)
    Will the answer be still be y/2 ??
    Last edited by noob mathematician; March 17th 2009 at 04:12 AM.
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