Is this derivation correct?

1. May 9, 2013

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:

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

2. May 9, 2013

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.