Question about continuous random variable?

Let X be a continuous random variable taking values in (a,b) with c.d.f. F, strictly increasing on (a,b). Show that Y = F(X) has a uniform distribution on (0,1).





hello, i'm very stuck on this. i've tried using integrals of cdfs and stuff but i'm not getting anywhere.

Question about continuous random variable?
There might be another way to show this, but I first thought of using the change of variables formula. In case you're unfamiliar (and to introduce my notation), we have two random variables Y and X, where Y=g(X), and g has an inverse function h; so X=h(Y). Then if f_X and f_Y are the pdf's of X and Y respectively, we have





f_Y(y) = f_X( h(Y) ) * |d/dy h(Y)|.





Now to our case. Notice first that d/dx(F(x))=f(x), the pdf of X. Also note that since F is strictly increasing, the inverse of F exists, and d/dy(F^-1(y)) = 1/[d/dx(F(x)] = 1/f(x). Then we have





f_Y(y) = f( F^-1(y) ) * |1/f(x)|.





But x=F^-1(y) and f(x) is positive for all x in (a,b), so the two terms are identical, and f_Y(y)=1. The only thing left to consider is the support for Y, but it's fairly obvious that it's (0,1) since the cdf takes on all values from 0 to 1...or does it?





We've hit the heart of the problem, showing that the pdf of Y is 1 and that the values for Y cannot be outside of the interval [0,1]. I'm not, however, convinced that Y can take on all values in this interval. The conditions on this problem and on the change of variables formula might make sure it does, but without spending some time looking into all of the consequences, I'm not certain...