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 PM #1

- March 28th 2010, 03:37 PM #2
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 PM #3
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 PM #4

- March 28th 2010, 06:45 PM #5

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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 PM #6

- March 29th 2010, 12:35 PM #7
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.

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.

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.

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 PM #8Actually 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 PM #9
True! the measurability of f is not a problem now =)

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 AM #10

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- March 30th 2010, 11:05 AM #11