Stochastic Calculus: Conditional Expectation

  • #1
WMDhamnekar
MHB
359
28
Homework Statement:
Suppose ##X_1,X_2 \dots ## are independent random variables with ##\mathbb{P}[X_j= 1] =1- \mathbb{P}[X_j=-1]=\frac13## Let ## S_n = X_1 +\dots +X_n## Find ## \mathbb{E}[S_n], \mathbb{E}[S^2_n], \mathbb{E}[S^3_n]##
Relevant Equations:
Not applicable
Are my following answers correct?


1677864912441.png

1677864978342.png

1677877283450.png
 
Last edited:

Answers and Replies

  • #2
mjc123
Science Advisor
Homework Helper
1,294
699
Have you tried the simple expedient of substituting some values of n and seeing if you get the right answer? Try n=1 to start.
You are making some careless mistakes
e.g. the expression for E(S2) should be n(1) + n(n-1)(1/9)
Your statement about the product of 3 random variables makes no sense; they are certainly not equal to zero. Each of the 3 random variables may independently take either of the values 1 or -1.
 
  • Like
  • Informative
Likes WMDhamnekar and WWGD
  • #3
WWGD
Science Advisor
Gold Member
6,320
8,379
Homework Statement:: Suppose ##X_1,X_2 \dots ## are independent random variables with ##\mathbb{P}[X_j= 1] =1- \mathbb{P}[X_j=-1]=\frac13## Let ## S_n = X_1 +\dots +X_n## Find ## \mathbb{E}[S_n], \mathbb{E}[S^2_n], \mathbb{E}[S^3_n]##
Relevant Equations:: Not applicable

Are my following answers correct?


View attachment 323141
View attachment 323142
View attachment 323157
I agree with mjc -- hey, it rhymes!

You can't treat a multinomial ## ( X_1+X_2+....+X_n)^2## as a standard binomial ##( X_1+X_2)^2##. Look up multinomial coefficients.
 
  • #4
WMDhamnekar
MHB
359
28
Have you tried the simple expedient of substituting some values of n and seeing if you get the right answer? Try n=1 to start.
You are making some careless mistakes
e.g. the expression for E(S2) should be n(1) + n(n-1)(1/9)
Your statement about the product of 3 random variables makes no sense; they are certainly not equal to zero. Each of the 3 random variables may independently take either of the values 1 or -1.
So, taking into consideration your this reply, I correct my amswers as follows:
##\mathbb{E}[S_n]=-\displaystyle\frac{n}{3}, \mathbb{E}[S^2_n]= n +\displaystyle\frac{n(n-1)}{9}, \mathbb{E}[S^3_n ] = -\displaystyle\frac{n}{3}-\frac{n(n-1)}{2} -\displaystyle\frac{n(n-1)}{9}##

Now, are these above answers correct?
 
Last edited:
  • #5
WMDhamnekar
MHB
359
28
Correct answers are
1677954533684.png
 
  • #6
mjc123
Science Advisor
Homework Helper
1,294
699
No, they are not. The answer for E(Sn2) is wrong for n = 2, and that for E(Sn3) is wrong for n = 3.

The answer you gave in post #4 for E(Sn2) is right, but the answer for E(Sn3) is not. You didn't show your working, but I suspect the mistake may lie in enumerating the terms of different kinds.
There are n terms of the form Xi3, each of which has expectation -1/3.
There are 3n(n-1) terms of the form Xi2Xj, each of which has expectation -1/3*1.
There are n(n-1)(n-2) terms of the form XiXjXk, each of which has expectation -1/27.
(Check that the total number of terms is n3.)
 
  • Like
Likes WMDhamnekar
  • #7
WWGD
Science Advisor
Gold Member
6,320
8,379
Sorry to insist on this, but I suggest you check multinomial coefficients , to determine how to expand multinomials ##( x_1+x_2+...+x_k)^n##. I suspect the errors may be partially due to this.
 
  • Like
Likes WMDhamnekar

Suggested for: Stochastic Calculus: Conditional Expectation

Replies
9
Views
154
Replies
5
Views
191
Replies
29
Views
604
Replies
4
Views
841
Replies
4
Views
354
  • Last Post
Replies
16
Views
300
Replies
2
Views
549
Replies
5
Views
2K
  • Last Post
Replies
28
Views
912
  • Last Post
Replies
8
Views
755
Top