Expectation of expansion of sum of independent random variables raised to some power.

[Win_odd Dhamnekar]

Suppose $X_1, X_2, \dots$ are independent random variables with $\mathbb{P}[X_i=1]=\frac13$ and $\mathbb{P}[X_i= -1] =\frac23$.
Let $$S_n = X_1 +\dots +X_n$$ and let $\mathcal{F}_n$ denote the information in $X_1, \dots, X_n$

1.Find$$\mathbb{E}[S_n], \mathbb{E}[S^2_n], \mathbb{E}[S^3_n]$$2. If m < n Find $$E[S_n| \mathcal{F}_m], E[S^2_n|\mathcal{F}_m], E[S^3_n|\mathcal{F}_m]$$3. If m < n , Find $$E[X_m|S_n]$$
How to answer all these questions?

I am working on these questions. Any math help will be accepted.

 

[BigBeachBanana]

1.
$$E(S_n) = E(X_1+\dots+X_n)=E(nX_i)=nE(X_i)\\ E(S_n^2)=\dots = n^2E(X_i^2)\\ E(S_n^3)=\dots = n^3E(X_i^3)$$

 

[Win_odd Dhamnekar]

1.
$$E(S_n) = E(X_1+\dots+X_n)=E(nX_i)=nE(X_i)\\ E(S_n^2)=\dots = n^2E(X_i^2)\\ E(S_n^3)=\dots = n^3E(X_i^3)$$

I computed the answers as $$\mathbb{E}[S_n]=-\frac{n}{3}$$

Are these answers correct?

 

[BigBeachBanana]

I computed the answers as $$\mathbb{E}[S_n]=-\frac{n}{3}$$View attachment 35170

Are these answers correct?

Agree with the first.

Didn't have time to check the other 2, but it doesn't feel right. Try with small cases like n=2, 3 to see if your formulas work.

 

[Win_odd Dhamnekar]

Agree with the first.

Didn't have time to check the other 2, but it doesn't feel right. Try with small cases like n=2, 3 to see if your formulas work.

Correct answers are $$\mathbb{E}[S^2_n]= n + \frac{n(n-1)}{2}, \mathbb{E}[S^3_n]= -\frac{n}{3}-n(n-1)- \frac{n(n-1)}{9}$$

 

[blamocur]

1.
$$E(S_n) = E(X_1+\dots+X_n)=E(nX_i)=nE(X_i)\\ E(S_n^2)=\dots = n^2E(X_i^2)\\ E(S_n^3)=\dots = n^3E(X_i^3)$$

I don't agree that $E(S_n^2) = n^2 E(X_i^2)$.

 

[blamocur]

Correct answers are $$\mathbb{E}[S^2_n]= n + \frac{n(n-1)}{2}, \mathbb{E}[S^3_n]= -\frac{n}{3}-n(n-1)- \frac{n(n-1)}{9}$$

I get a different answer for $E(S_n^2)$. Where did you get yours?

 

[BigBeachBanana]

I assumed the RVs are symmetric above.

$\displaystyle E(X_i) = -\dfrac{1}{3}$

$\displaystyle E(X_i^2) = 1$

$\displaystyle E(X_i^3) = -\dfrac{1}{3}$

---

$\displaystyle E(S_n) = nE(X_i) = \boxed{-\dfrac{n}{3}}$

$\displaystyle E(S_n^2) = nE(X_i^2) + n(n-1)E(X_i)^2 = \boxed{n + \dfrac{n(n-1)}{9}}$

$\displaystyle E(S_n^3) = nE(X_i^3) + 3n(n-1)E(X_i^2)E(X_i) +n(n-1)(n-2)E(X_i)^3 = \boxed{-\dfrac{n}{3} - n(n-1) -\dfrac{ n(n-1)(n-2)}{27}}$

 

[blamocur]

I assumed the RVs are symmetric above.

$\displaystyle E(X_i) = -\dfrac{1}{3}$

$\displaystyle E(X_i^2) = 1$

$\displaystyle E(X_i^3) = -\dfrac{1}{3}$

---

$\displaystyle E(S_n) = nE(X_i) = \boxed{-\dfrac{n}{3}}$

$\displaystyle E(S_n^2) = nE(X_i^2) + n(n-1)E(X_i)^2 = \boxed{n + \dfrac{n(n-1)}{9}}$

$\displaystyle E(S_n^3) = nE(X_i^3) + 3n(n-1)E(X_i^2)E(X_i) +n(n-1)(n-2)E(X_i)^3 = \boxed{-\dfrac{n}{3} - n(n-1) -\dfrac{ n(n-1)(n-2)}{27}}$

I like this version of $E(S_n^2)$ better

 

[Win_odd Dhamnekar]

I assumed the RVs are symmetric above.

$\displaystyle E(X_i) = -\dfrac{1}{3}$

$\displaystyle E(X_i^2) = 1$

$\displaystyle E(X_i^3) = -\dfrac{1}{3}$

---

$\displaystyle E(S_n) = nE(X_i) = \boxed{-\dfrac{n}{3}}$

$\displaystyle E(S_n^2) = nE(X_i^2) + n(n-1)E(X_i)^2 = \boxed{n + \dfrac{n(n-1)}{9}}$

$\displaystyle E(S_n^3) = nE(X_i^3) + 3n(n-1)E(X_i^2)E(X_i) +n(n-1)(n-2)E(X_i)^3 = \boxed{-\dfrac{n}{3} - n(n-1) -\dfrac{ n(n-1)(n-2)}{27}}$

$E[S_n | \mathcal{F}_n] = E[S_m | \mathcal{F}_n] +E[S_n - S_m|\mathcal{F}_n] = S_m + \mathbb{E}[S_n-S_m] = S_m + \mathbb{E}[X_j] (n -m) = S_m - \frac13 (n-m)$

We know $E[X_j]=-\frac13 , \mathbb{E}[X^2_j]= 1$ for each j.
Then if m < n ,
$$E[S^2_n| \mathcal{F}_m] = E([S_m + (S_n - S_m)]^2 | \mathcal{F}_m)$$$$E[S^2_n |\mathcal{F}_m] = E[ S^2_m |\mathcal{F}_m] + 2 E [S_m (S_n -S_m)| \mathcal{F}_m] + E [(S_n -S_m)^2 | \mathcal{F}_m ]$$
Since $S_m$ is $\mathcal{F}_n$ -measurable and $S_n - S_m$ is independent of $\mathcal{F}_m$

$E[S^2_m|\mathcal{F}_m] = S^2_m$
$E[S_m(S_n -S_m) | \mathcal{F}_m]= S_m E[ S_n - S_m |\mathcal{F}_m]= S_m E[S_n - S_m] = -\frac{S_m}{3}$

$E[(S_n -S_m)^2 |\mathcal{F}_m] = \mathbb{E}[(S_n -S_m)^2] = Var (S_n - S_m) = (E[X^2_j] - E[X_j])(n-m) = \frac43 (n-m)$

and hence,
$$E[S^2_n |\mathcal{F}_m] = S^2_m -\frac{S_m}{3} +\frac43(n-m)$$
Are these above answers correct?

Note : Above answers are computed by referring to G.F. Lawler's Book "Stochastic Calculus: An Introduction with Applications"

How to compute $E[ S^3_n|\mathcal{F}_m]?$

 

[blamocur]

$E[S_n | \mathcal{F}_n] = E[S_m | \mathcal{F}_n] +E[S_n - S_m|\mathcal{F}_n] = S_m + \mathbb{E}[S_n-S_m] = S_m + \mathbb{E}[X_j] (n -m) = S_m - \frac13 (n-m)$

We know $E[X_j]=-\frac13 , \mathbb{E}[X^2_j]= 1$ for each j.
Then if m < n ,
$$E[S^2_n| \mathcal{F}_m] = E([S_m + (S_n - S_m)]^2 | \mathcal{F}_m)$$$$E[S^2_n |\mathcal{F}_m] = E[ S^2_m |\mathcal{F}_m] + 2 E [S_m (S_n -S_m)| \mathcal{F}_m] + E [(S_n -S_m)^2 | \mathcal{F}_m ]$$
Since $S_m$ is $\mathcal{F}_n$ -measurable and $S_n - S_m$ is independent of $\mathcal{F}_m$

$E[S^2_m|\mathcal{F}_m] = S^2_m$
$E[S_m(S_n -S_m) | \mathcal{F}_m]= S_m E[ S_n - S_m |\mathcal{F}_m]= S_m E[S_n - S_m] = -\frac{S_m}{3}$

$E[(S_n -S_m)^2 |\mathcal{F}_m] = \mathbb{E}[(S_n -S_m)^2] = Var (S_n - S_m) = (E[X^2_j] - E[X_j])(n-m) = \frac43 (n-m)$

and hence,
$$E[S^2_n |\mathcal{F}_m] = S^2_m -\frac{S_m}{3} +\frac43(n-m)$$
Are these above answers correct?

Note : Above answers are computed by referring to G.F. Lawler's Book "Stochastic Calculus: An Introduction with Applications"

How to compute $E[ S^3_n|\mathcal{F}_m]?$

Are you done with $E(S_n^2)$ ?

 

[Win_odd Dhamnekar]

Are you done with $E(S_n^2)$ ?

$E[S^2_n]= n +\displaystyle\frac{n(n-1)}{9}$

In #8 , all the answers are correctly stated for question 1 considering multinomial theorem for $( x_1 + \dots + x_k)^n$

 

[Win_odd Dhamnekar]

Errata $[S^2_n|\mathcal{F}_m] = S^2_m -\frac23S_m + (n- m)$



$E[S^3_n|\mathcal{F}_m]= E([ S_m + (S_n - S_m)]^3|\mathcal{F}_m)$

$E[S^3_n|\mathcal{F}_m] = E[S^3_n|\mathcal{F}_m] + 3 E[S^2_m(S_n-S_m)|\mathcal{F}_m] + 3E[ S_m(S_n - S_m)^2 | \mathcal{F}_m] + E[(S_n -S_m)^3| \mathcal{F}_m]$

Since$S_m$ is $\mathcal{F}_m$ -measurable and $S_n - S_m$ is independent of $\mathcal{F}_m$

$$E[S^3_m|\mathcal{F}_m] = S^3_m$$
$E[S^2_m(S_n -S_m )|\mathcal{F}_m] = S^2_m E[(S_n -S_m)|\mathcal{F}_m] = S^2_m \mathbb{E}[ S_n -S_m] = -\displaystyle\frac{S^2_m}{3}$
=
$E[S_m )(S_n -S_m)^2|\mathcal{F}_m] = S_m E[(S_n -S_m)^2 |\mathcal{F}_m]= S_m E[X^2_j] = S_m (1) = S_m$

$E[(S_n -S_m)^3 |\mathcal{F}_m]= E[X^3_j]= -\displaystyle\frac13 (n-m)$

$E[S^3_n|\mathcal{F}_m] = S^3_m - S^2_m + 3 S_m - \displaystyle\frac13 (n-m )$

3.How to find $E[X_m | S_n]$ given that m < n.

 

[Win_odd Dhamnekar]

Suppose $X_1, X_2, \dots$ are i.i.d. random variables. We will compute $E[X_1| S_n]$
Note that the information contained in the one data point $S_n$ is less than the information contained in $X_1, \dots , X_n.$ However, since the random variables are independent and identically distributed, it must be the case that$$E[X_1|S_n ] = E[X_m| S_n ] = E[X_n |S_n]$$
Linearity implies that
$$m E[X_1|S_n] =\displaystyle\sum_{j=1}^m E[X_j|S_n] = E[ X_1 +\dots + X_n| S_n] = E[S_m |S_n] = S_m$$
$$\therefore E[X_m|S_n] =\frac{S_m}{m}$$