Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
can anyone show me how to get the ans fY(y)=1? –  johnny Jan 1 '12 at 9:22

1 Answer 1

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 $Y$.

Edit:

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

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.