# Thread: function of iid random variables also independent?

1. ## function 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?

2. ## 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...

3. ## Re: function of iid random variables also independent?

Thank you!
I didn't know this definition.

4. ## Re: 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.

5. ## 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$\displaystyle X_1,X_2,X_3$ 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 $\displaystyle x \geq 0$?

6. ## Re: function of iid random variables also independent?

Originally Posted by Juju
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 $\displaystyle x \geq 0$?
You don't need inverse function. Inverse image works and is always there for you. You need the $\displaystyle f_i$ to be measurable to make sure that $\displaystyle f_i ^{-1} (B_i)$ is measurable.

7. ## Re: function of iid random variables also independent?

my function is continous (on $\displaystyle x \geq -1$) and thus measurable. right?

8. ## 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.