Thread: function of iid random variables also independent?

Hallo,

are iid random variables with a lognormal distribution.

Am I right, that
where
are also independent?

Thanks in advance!

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...

Thank you!
I didn't know this definition.

You can also prove that



where the 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.

Thank you. So if I show it this way, I just need the inverse function and use the independence of and then it follows.
Am I right?

Doesn't it matter that my function is just defined for and the inverse function for ?

Originally Posted by Juju

Thank you. So if I show it this way, I just need the inverse function and use the independence of and then it follows.
Am I right?

Doesn't it matter that my function is just defined for and the inverse function for ?

You don't need inverse function. Inverse image works and is always there for you. You need the to be measurable to make sure that is measurable.

my function is continous (on ) and thus measurable. right?

Heh, you have to try much harder than that to cook up a function that isn't measurable Continuity is more than enough.