If are integrable indp. i.d. r.v. show that

Any help would be much apreciated, i can post what i've done so far and where i am stuck, would that help?

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- March 28th 2010, 03:18 PMmabrukaSum of i.i .d. r.v.
If are integrable indp. i.d. r.v. show that

Any help would be much apreciated, i can post what i've done so far and where i am stuck, would that help? - March 28th 2010, 03:37 PMMoo
Hello,

This is a very common problem. I'm copying it from something I typed before, so don't worry if the are transformed into X.

(set of permutations) because the random variables are iid.

Considering the permutation that inverts 1 and k for some k, we obtain that has the same distribution as .

So , the follow the same distribution.

Now consider the function :

and apply it to the vectors .

It follows that are identically distributed for all k.

In particular, it follows that .

Hence - March 28th 2010, 05:15 PMmabruka
Wow what an interesting approach there!

The only thing that remains unclear to me (maybe i am forgetting a fundamental result) is why using that function on Y_k yields that (X_k, S_n) are i.d. for all k?

Is it true that if Z and W are r.v. i.d. then f(Z) is i.d. to f(W)

for any function ? (does this function needs to be continuous? or maybe just measurable? )

Also, where is the independence hypothesis being used implicitly?

thank you - March 28th 2010, 05:33 PMmabruka
Oh well i think this answer my question:

Z , Y are i.d. iff E f(Z) = E f(Y) for every borel measurable f

So the only thing it has yet to be proved is that the function

is measurable

right?

(i am betting it is continuous..) - March 28th 2010, 06:45 PMLaurent
That's a nice post! Actually I wouldn't have bothered going into such details... I would have just said: by symmetry, have same distribution.

Or: Let be a bounded measurable function. We have, by symmetry, , hence . Since is a measurable function of , we have checked the definition of the conditional expectation: . - March 28th 2010, 07:03 PMmabrukaQuote:

In particular, it follows that .

Thank you - March 29th 2010, 12:35 PMMoo
Because the follow the same distribution for all k ! So if you apply the function f to each of them, the results will all have the same distribution.

Quote:

Is it true that if Z and W are r.v. i.d. then f(Z) is i.d. to f(W)

for any function ? (does this function needs to be continuous? or maybe just measurable? )

thank you[/QUOTE]

Not quite, it's rather that for every bounded measurable h, E(h(Z)=E(h(Y)), and then let h=gof, where g is any bounded measurable function. And f, defined as above.

Quote:

So the only thing it has yet to be proved is that the function

is measurable

right?

Just recall that the projections (the functions that return the i-th coordinate) are measurable, and so is the sum.

Quote:

Why does this follows?

There's a sort of equivalence between having a conditional expectation X|Z, than having the expectation of X multiplied by any function of Z. - March 29th 2010, 12:38 PMMooQuote:

Actually I wouldn't have bothered going into such details...

So I copied what my own teacher did on that forum, and then here (the second time is always less painful) :P - March 29th 2010, 02:13 PMmabruka
True! the measurability of f is not a problem now =)

Quote:

Quote:

Why does this follows?

Read Laurent's answer about this.

There's a sort of equivalence between having a conditional expectation X|Z, than having the expectation of X multiplied by any function of Z.

What i know is that if we check that

for every

then it follows what we need:

Alternatively in order to get that , it suffices to show that

for every which to tell you the truth seems a little bit hard !

This last step is what i am not convinced yet.

thank you

Sorry if i am being stubborn but as you wrote earlier usually teachers dont give rigorous proofs but in order for one to learn something one must know the do's and dont's :) - March 30th 2010, 02:18 AMLaurent
- March 30th 2010, 11:05 AMmabruka
Thank you it is all clear now !

(Rofl)