Hallo,

$\displaystyle X_1, X_2, X_3$ are iid random variables with a lognormal distribution.

Am I right, that

$\displaystyle f(X_1),f(X_2),f(X_3)$ where $\displaystyle f(x)=(1+x)^\gamma, 0< \gamma < 1$

are also independent?

Thanks in advance!

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- Jun 25th 2011, 01:40 PMJujufunction of iid random variables also independent?
Hallo,

$\displaystyle X_1, X_2, X_3$ are iid random variables with a lognormal distribution.

Am I right, that

$\displaystyle f(X_1),f(X_2),f(X_3)$ where $\displaystyle f(x)=(1+x)^\gamma, 0< \gamma < 1$

are also independent?

Thanks in advance! - Jun 25th 2011, 02:13 PMMooRe: function of iid random variables also independent?
Hello,

Yes, just use the definition of independence : X and Y are independent iff for any *correct* functions f,g, E[f(X)g(Y)]=E[f(X)]E[g(Y)]

*correct* in the sense that the expectations are defined for such functions. That would be measurable and bounded for the simplest situation I guess... - Jun 25th 2011, 02:32 PMJujuRe: function of iid random variables also independent?
Thank you!

I didn't know this definition. - Jun 25th 2011, 09:20 PMGuyRe: function of iid random variables also independent?
You can also prove that

$\displaystyle P[f_1(X_1) \in B_1, ..., f_n (X_n) \in B_n] = P[f_1 (X_1) \in B_1] \cdots P[f_n (X_n) \in B_n]$

where the $\displaystyle B_i$ are arbitrary 1d Borel sets. This implies independence from a different starting place; all my references use a different definition of independence than Moo's, but they are equivalent. - Jun 25th 2011, 09:55 PMJujuRe: function of iid random variables also independent?
Thank you. So if I show it this way, I just need the inverse function and use the independence of$\displaystyle X_1,X_2,X_3$ and then it follows.

Am I right?

Doesn't it matter that my function $\displaystyle f(x)=(1+x)^\gamma,0<\gamma<1, $ is just defined for $\displaystyle x\geq -1$ and the inverse function for $\displaystyle x \geq 0$? - Jun 26th 2011, 07:12 AMGuyRe: function of iid random variables also independent?
- Jun 26th 2011, 10:37 AMJujuRe: function of iid random variables also independent?
my function is continous (on $\displaystyle x \geq -1$) and thus measurable. right?

- Jun 26th 2011, 10:45 AMGuyRe: function of iid random variables also independent?
Heh, you have to try much harder than that to cook up a function that isn't measurable :) Continuity is more than enough.