- #1

- 12

- 0

[itex]CDF(Z) = Prob(Z < z)[/itex]

[itex]CDF(Y) = Prob(Y < y)[/itex] where y = f(z)

[itex]PDF(Z) = \frac{d(CDF(Z))}{dz}[/itex]

[itex]PDF(Y) = \frac{d(CDF(Y))}{df(z)}[/itex]

Now, it is known from various internet sources and wikipedia that:

[itex]E(f(z))= \int_{-\infty}^{\infty}{f(z) PDF(z) }dz [/itex] - (1)

Also, since z is a random variable, f(z) is also a random variable, hence:

[itex]E(f(z))= \int_{-\infty}^{\infty}{f(z) PDF(f(z)) }df(z) [/itex] - (2)

From (1) and (2),

[itex]PDF(z)dz = PDF(f(z)) df(z)[/itex]

From this doesn't it follow that:

[itex] CDF(z) = CDF(f(z)) + const.[/itex]