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(Y≤y)=P(FX(X)≤y)=P(P(X≤X)≤y)
The chance that X≤X is always 1, right ? Therefore I would say that:
FY(y)=P(1≤y)=1[1,∞)(y)
But this is wrong... Why is this ?
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