# Sum of i.i .d. r.v.

• Mar 28th 2010, 03:18 PM
mabruka
Sum of i.i .d. r.v.
If $\displaystyle \xi_1,\ldots,\xi_n$ are integrable indp. i.d. r.v. show that

$\displaystyle E\left[\xi_1 | \xi_1+\cdots+\xi_n\right] =\frac{\xi_1+\cdots+\xi_n}{n}$

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

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

$\displaystyle (X_1,\dots,X_n)\stackrel{distribution}{=} (X_{\sigma(1)},\dots,X_{\sigma(n)}) ~,~ \forall \sigma\in\mathfrak{S}_n$ (set of permutations) because the random variables are iid.

Considering the permutation $\displaystyle \sigma$ that inverts 1 and k for some k, we obtain that $\displaystyle Y_k=(X_k,X_2,\dots,X_{k-1},X_1,X_{k+1},\dots,X_n)$ has the same distribution as $\displaystyle (X_1,\dots,X_n)$.
So $\displaystyle \forall k=1,\dots,n$, the $\displaystyle Y_k$ follow the same distribution.

Now consider the function :

\displaystyle \begin{aligned} f ~:~ & \mathbb{R}^n &\to &\mathbb{R}^2 \\ & (x_1,\dots,x_n) & \mapsto & (x_1,\sum_{i=1}^n x_i) \end{aligned}

and apply it to the vectors $\displaystyle Y_k$.

It follows that $\displaystyle (X_k,S_n)$ are identically distributed for all k.

In particular, it follows that $\displaystyle \forall j,k, ~ E[X_j|S_n]=E[X_k|S_n]$.

Hence $\displaystyle S_n=E[S_n|S_n]=E\left[\sum_{k=1}^n X_k\bigg|S_n\right]=\sum_{k=1}^n E[X_k|S_n]=n E[X_1|S_n]$
• Mar 28th 2010, 05:15 PM
mabruka
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 $\displaystyle f:\mathbb R^n \longrightarrow \mathbb R^m$ ? (does this function needs to be continuous? or maybe just measurable? )

Also, where is the independence hypothesis being used implicitly?
thank you
• Mar 28th 2010, 05:33 PM
mabruka
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

\displaystyle \begin{aligned} f ~:~ & \mathbb{R}^n &\to &\mathbb{R}^2 \\ & (x_1,\dots,x_n) & \mapsto & (x_1,\sum_{i=1}^n x_i) \end{aligned}

is measurable

right?

(i am betting it is continuous..)
• Mar 28th 2010, 06:45 PM
Laurent
Quote:

Originally Posted by Moo
This is a very common problem. I'm copying it from something I typed before, so don't worry if the $\displaystyle \xi$ are transformed into X.

That's a nice post! Actually I wouldn't have bothered going into such details... I would have just said: by symmetry, $\displaystyle (X_1,S_n),\ldots,(X_n,S_n)$ have same distribution.

Or: Let $\displaystyle f$ be a bounded measurable function. We have, by symmetry, $\displaystyle E[X_1f(S_n)]=\cdots=E[X_n f(S_n)]$, hence $\displaystyle E[X_1 f(S_n)]=\frac{1}{n}E[(X_1+\cdots+X_n)f(S_n)]=E[\frac{S_n}{n}f(S_n)]$. Since $\displaystyle \frac{S_n}{n}$ is a measurable function of $\displaystyle S_n$, we have checked the definition of the conditional expectation: $\displaystyle E[X_1|S_n]=\frac{S_n}{n}$.
• Mar 28th 2010, 07:03 PM
mabruka
Quote:

In particular, it follows that $\displaystyle \forall j,k, ~ E[X_j|S_n]=E[X_k|S_n]$.
Why does this follows?

Thank you
• Mar 29th 2010, 12:35 PM
Moo
Quote:

Originally Posted by mabruka
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?

Because the $\displaystyle Y_k$ follow the same distribution for all k ! So if you apply the function f to each of them, the results $\displaystyle (X_k,S_n)$ 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 $\displaystyle f:\mathbb R^n \longrightarrow \mathbb R^m$ ? (does this function needs to be continuous? or maybe just measurable? )
Also, where is the independence hypothesis being used implicitly?
thank you[/QUOTE]

Quote:

Originally Posted by mabruka
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

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

\displaystyle \begin{aligned} f ~:~ & \mathbb{R}^n &\to &\mathbb{R}^2 \\ & (x_1,\dots,x_n) & \mapsto & (x_1,\sum_{i=1}^n x_i) \end{aligned}

is measurable

right?
Well yeah, it has to be done...
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.
• Mar 29th 2010, 12:38 PM
Moo
Quote:

Actually I wouldn't have bothered going into such details...
Lol, the story is that on les-maths.net, someone was talking about the probably little number of teachers who did it the rigorous way.
So I copied what my own teacher did on that forum, and then here (the second time is always less painful) :P
• Mar 29th 2010, 02:13 PM
mabruka
True! the measurability of f is not a problem now =)

Quote:

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.
What definition of conditional expected value is he using?

What i know is that if we check that

$\displaystyle E(\frac{S_n}{n}1_A) = E(X_11_A)$ for every $\displaystyle A\in \sigma(S_n)$

then it follows what we need: $\displaystyle E[X_1|S_n]=\frac{S_n}{n}$

Alternatively in order to get that $\displaystyle E[X_k|S_n]=E[X_j|S_n]$ , it suffices to show that

$\displaystyle E(X_k 1_A)=E(X_j 1_A)$ for every $\displaystyle A\in\sigma(S_n)$ 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 :)
• Mar 30th 2010, 02:18 AM
Laurent
Quote:

Originally Posted by mabruka
What definition of conditional expected value is he using?

What i know is that if we check that

$\displaystyle E(\frac{S_n}{n}1_A) = E(X_11_A)$ for every $\displaystyle A\in \sigma(S_n)$

then it follows what we need: $\displaystyle E[X_1|S_n]=\frac{S_n}{n}$

Alternatively in order to get that $\displaystyle E[X_k|S_n]=E[X_j|S_n]$ , it suffices to show that

$\displaystyle E(X_k 1_A)=E(X_j 1_A)$ for every $\displaystyle A\in\sigma(S_n)$ which to tell you the truth seems a little bit hard !

Read my post again by replacing $\displaystyle f(S_n)$ by $\displaystyle {\bf 1}_{\{S_n\in B\}}$, where $\displaystyle B$ is any measurable subset of $\displaystyle \mathbb{R}$. Indeed, an event $\displaystyle A\in\sigma(S_n)$ can be written $\displaystyle A=\{S_n\in B\}$ where $\displaystyle B$ is a measurable subset of $\displaystyle \mathbb{R}$.
• Mar 30th 2010, 11:05 AM
mabruka
Thank you it is all clear now !

(Rofl)