Thursday, 19 March 2015

probability theory - Let X a random variable with a strictly increasing distrubution function FX. Show that Y=FX(X)simhom(0,1) distrubution.

Let X a random variable with a strictly increasing distrubution function FX. Show that the random variable Y=FX(X) has a hom(0,1) distrubution.



Here is what I thought:



FY(y)=P(Yy)=P(FX(X)y)=P(P(XX)y)



The chance that XX is always 1, right ? Therefore I would say that:



FY(y)=P(1y)=1[1,)(y)



But this is wrong... Why is this ?

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