Is the Derivation of the Relationship Between CDF and PDF Correct?

[RRraskolnikov]

No, this is not for a homework. Please don't delete the thread.

CDF(Z) = Prob(Z < z)
CDF(Y) = Prob(Y < y) where y = f(z)
PDF(Z) = \frac{d(CDF(Z))}{dz}
PDF(Y) = \frac{d(CDF(Y))}{df(z)}

Now, it is known from various internet sources and wikipedia that:
E(f(z))= \int_{-\infty}^{\infty}{f(z) PDF(z) }dz - (1)

Also, since z is a random variable, f(z) is also a random variable, hence:
E(f(z))= \int_{-\infty}^{\infty}{f(z) PDF(f(z)) }df(z) - (2)

From (1) and (2),
PDF(z)dz = PDF(f(z)) df(z)

From this doesn't it follow that:

CDF(z) = CDF(f(z)) + const.

 

[Simon Bridge]

OK - check by seeing if you can find some function f(z) where the CDF of f(z) differs from CDF of z by something other than a constant.