Conditional expectation given ##\mathcal{F}_m##

In summary, the conversation discusses the correct calculations for E(Sn-Sm) and E[(Sn-Sm)^2] as well as the incorrect calculation for E[(Sn-Sm)^3]. The correct calculation for E(Xm|Sn) is also discussed.
  • #1
WMDhamnekar
MHB
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Homework Statement
Suppose ## X_1, X_2, \dots ## are independent random variables with ## \mathbb{P}[X_j =1] = 1- \mathbb{P}[X_j =-1] =\displaystyle\frac13## Let ##S_n = X_1 +\dots + X_n## and let
##\mathcal{F}_n## denote the information contained in ## X_1 , \dots , X_n ##
1.If m < n Find ## E[S_n |\mathcal{F}_m], E[S^2_n| \mathcal{F}_m], E[S^3_n |\mathcal{F}_m]##

2. If m < n Find ## E [X_m| S_n ] ##
Relevant Equations
Not applicable
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Are these above answers correct?
 
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  • #2
You're getting careless again. In the first line you correctly say E(Sn-Sm) = (-1/3)(n-m).
So when later on you say SmE(Sn-Sm) = -Sm/3, that's wrong, isn't it?
And E[(Sn-Sm)2] is not E(Xj2)(n-m). We worked this out in another thread, didn't we? Similarly with the cube.
 
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  • #3
mjc123 said:
You're getting careless again. In the first line you correctly say E(Sn-Sm) = (-1/3)(n-m).
So when later on you say SmE(Sn-Sm) = -Sm/3, that's wrong, isn't it?
And E[(Sn-Sm)2] is not E(Xj2)(n-m). We worked this out in another thread, didn't we? Similarly with the cube.
E(Sn - Sm)= -1/3 (n-m)
So, SmE[Sn -Sm ] = -Sm(n-m)/3

Hence ##E[S^2_n|\mathcal{F}_m] = S^2_m -\frac23 S_m (n-m) + (n-m) + \frac{(n-m)(n-m-1)}{9}##

##E[S^3_n|\mathcal{F}_m] = S^3_m - S^2_m(n-m) + S_m(n-m)\frac{(n-m)(n-m-1)}{3} -\frac{n-m}{3} - (n-m)(n-m-1)- \frac{(n-m)(n-m-1)(n-m-2)}{27}##
Are these above answer correct?
Note:

Now, do you mean to say E[(Sn - Sm)2] = (n- m) +##\frac{(n-m)(n-m-1)}{9}##
and E[(Sn - Sm)3] = ##-\frac{n-m}{3} - (n-m)(n-m-1)- \frac{(n-m)(n-m-1)(n-m-2)}{27}##
 
Last edited:
  • #4
I think that's correct, except that the Sm term in the cube should be
3Sm{(n-m) + (n-m)(n-m-1)/9}
You multiplied when you should have added.
 
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  • #5
mjc123 said:
I think that's correct, except that the Sm term in the cube should be
3Sm{(n-m) + (n-m)(n-m-1)/9}
You multiplied when you should have added.
Is my computed E(Xm| Sn] correct?
 
  • #6
You want E[Xm|Sn]. Why do you bring in Sm? You don't know that (in the terms of the question). I would say it's Sn/n.
Sn can take the values n-2k (0≤k≤n). If there are k values of -1 in the string, the probability of any one value being -1 is k/n, and the probability of it being 1 is (n-k)/n. The expectation of any Xi is then
1*(n-k)/n - 1*k/n = (n-2k)/n = Sn/n.
 
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FAQ: Conditional expectation given ##\mathcal{F}_m##

What is conditional expectation given ##\mathcal{F}_m##?

Conditional expectation given ##\mathcal{F}_m##, denoted as ##\mathbb{E}[X | \mathcal{F}_m]##, is the expected value of a random variable ##X## given the information contained in the sigma-algebra ##\mathcal{F}_m##. It provides a way to update our expectations based on new information.

How is conditional expectation given ##\mathcal{F}_m## calculated?

Conditional expectation given ##\mathcal{F}_m## is calculated by averaging the values of the random variable ##X## over the sets in the sigma-algebra ##\mathcal{F}_m##. Mathematically, it is defined as the ##\mathcal{F}_m##-measurable function that satisfies the property: for any set ##A \in \mathcal{F}_m##, the integral of ##X## over ##A## equals the integral of ##\mathbb{E}[X | \mathcal{F}_m]## over ##A##.

What are the properties of conditional expectation given ##\mathcal{F}_m##?

Conditional expectation given ##\mathcal{F}_m## has several important properties:1. Linearity: ##\mathbb{E}[aX + bY | \mathcal{F}_m] = a\mathbb{E}[X | \mathcal{F}_m] + b\mathbb{E}[Y | \mathcal{F}_m]## for constants ##a## and ##b##.2. Taking out what is known: If ##Y## is ##\mathcal{F}_m##-measurable, then ##\mathbb{E}[YX | \mathcal{F}_m] = Y\mathbb{E}[X | \mathcal{F}_m]##.3. Law of total expectation: ##\mathbb{E}[X] = \mathbb{E}[\mathbb{E}[X | \mathcal{F}_m]]##.4. Tower property: If ##\mathcal{F}_n \subseteq \mathcal{F}_m##, then ##\mathbb{E}[\mathbb{E}[X | \mathcal{F}_m] | \mathcal{F}_n] = \mathbb{E}[X | \mathcal{F}_n]##.5. Non-negativity: If ##X \geq 0##, then ##\mathbb{E}[X | \mathcal{F}_m] \geq 0##.

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