Is this derivation correct?

  • #1
No, this is not for a homework. Please don't delete the thread.

[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]
 

Answers and Replies

  • #2
Simon Bridge
Science Advisor
Homework Helper
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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.
 

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