# function of iid random variables also independent?

• Jun 25th 2011, 02:40 PM
Juju
function of iid random variables also independent?
Hallo,

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

Am I right, that
$f(X_1),f(X_2),f(X_3)$ where $f(x)=(1+x)^\gamma, 0< \gamma < 1$
are also independent?

• Jun 25th 2011, 03:13 PM
Moo
Re: 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, 03:32 PM
Juju
Re: function of iid random variables also independent?
Thank you!
I didn't know this definition.
• Jun 25th 2011, 10:20 PM
Guy
Re: function of iid random variables also independent?
You can also prove that

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

where the $B_i$ 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, 10:55 PM
Juju
Re: 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 $X_1,X_2,X_3$ and then it follows.
Am I right?

Doesn't it matter that my function $f(x)=(1+x)^\gamma,0<\gamma<1,$ is just defined for $x\geq -1$ and the inverse function for $x \geq 0$?
• Jun 26th 2011, 08:12 AM
Guy
Re: function of iid random variables also independent?
Quote:

Originally Posted by Juju
Thank you. So if I show it this way, I just need the inverse function and use the independence of $X_1,X_2,X_3$ and then it follows.
Am I right?

Doesn't it matter that my function $f(x)=(1+x)^\gamma,0<\gamma<1,$ is just defined for $x\geq -1$ and the inverse function for $x \geq 0$?

You don't need inverse function. Inverse image works and is always there for you. You need the $f_i$ to be measurable to make sure that $f_i ^{-1} (B_i)$ is measurable.
• Jun 26th 2011, 11:37 AM
Juju
Re: function of iid random variables also independent?
my function is continous (on $x \geq -1$) and thus measurable. right?
• Jun 26th 2011, 11:45 AM
Guy
Re: 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.