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Given $F(x)$ is the value of the distribution function of the continuous random variable $X$ at $x$, what should be probability density of $Y=F(X)$?

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can anyone show me how to get the ans fY(y)=1? – johnny Jan 1 '12 at 9:22
up vote 3 down vote accepted

Let $A=\{x:f_X(x) \gt 0\}$ and $B = \{y: y = g(x) ~\text{for some}~ x \in A\}$ . In general if $X$ has distribution function $F_X(x)$ and $Y=g(X)$, then we have the following:
a. $F_Y(y) = F_X(g^{-1}(y))$ for $y\in B$ if $g$ is an increasing function.
b. $F_Y(y) = 1 - F_X(g^{-1}(y))$ for $y\in B$ if $g$ is a decreasing function and $X$ is a continuous random variable.

Differentiating $F_Y(y)$ gives you the density function of $Y$.


In your case we have that $F_Y(y)= F(F^{-1}(y))=y$. So $F^~{'}_Y(y)=1.$ Hence $f_Y(y) =1.$

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i cant figure out why the professor said f Y (y)=1,do you get the same answer? – johnny Jan 1 '12 at 9:14
you mean F(F^{-1}(y))=y? – johnny Jan 1 '12 at 9:26
@johnny: Yes, that's what I mean. – Nana Jan 1 '12 at 9:28
thx very much,i think i knew what's wrong with my concern – johnny Jan 1 '12 at 9:29
you are welcome. – Nana Jan 1 '12 at 9:31

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