Standard Deviation of transformed R.V.

Hello all,

Suppose you have some random variable x that is distributed according to a pareto distribution f(x) = k m^{k} x^{-k-1}. And then you have a transformation y = c x where c is a constant. I want to find out the standard deviation of the natural log of y. I have been trying for several hours but have had no success. I have tried to do it the standard way E[x^2]-E[x]^2 but am quickly running into a mess that I can't deal with. Does anybody here have any idea how I can approach this problem? I really appreciate the help.

Cheers,

N

Re: Standard Deviation of transformed R.V.

ive got two suggestions for you:

**Simplify**

since y = log c + log x; the sd of y does not depend on the value of c.

so without loss of generality, consider the case c=1.

**Use MGFs**

with C=1, consider the MGF of y=logx

$\displaystyle M_y(t) = E \left(e^{ty} \right) = E \left(e^{t \ln x} \right) = E \left(e^{\ln x^t} \right) = E \left(x^t} \right)$

$\displaystyle E(x^t)$ is just the underlying (non central) moment of X. since x follows a standard distribution, you can look this up.

Since you now have a working expression for $\displaystyle M_y(t)$ you can use standard methods to obtain the moments of Y from $\displaystyle M_y(t)$. (assuming you can differenciate $\displaystyle E(x^t)$ at t=0.)

PS: i haven't actually done this, but i assume it is tractable.

PPS: if this post makes no sense to you, then you probably have not learned "moment generating functions" yet, and youll have to solve your problem the old fashioned way.