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June 18th, 2014, 04:05 AM  #1 
Newbie Joined: Jun 2014 From: Italy Posts: 6 Thanks: 0  Expected mean squared error
Could anyone help me to solve this exercise? Suppose the estimation errors e_i are independent and identically distributed random variables that takes values 1 and 1 with an equal probability. X is a uniform random variable that takes on 100 possible values (P(X=x)=1/100 , x=1,2,..100). Since x is a number that can take on different values with different probabilities, we should talk about the expected MSE, where the expectation E x is the averaging procedure over the probability distribution of x. Calculate E_AE and E_EA with the formulas below for N=100 and tell if the relationship between E_AE and E_EA hold. Please show me how to solve this exercise 
June 18th, 2014, 12:17 PM  #2  
Global Moderator Joined: May 2007 Posts: 6,754 Thanks: 695  Quote:
Since the e_i are independent and have mean 0, the only terms that survive in E_EA are the $\displaystyle e_i^2$ terms, so the expressions are almost the same, the difference being a factor of 1/N.  
June 18th, 2014, 01:55 PM  #3 
Newbie Joined: Jun 2014 From: Italy Posts: 6 Thanks: 0 
I try to explain better. We are talking about the problem of averaging a crowd. we say the number that a crowd of N people wants to guess is x, and each person in the crowd makes a guess yi : yi = x+ei (x) i.e. the true value plus some error ei ... then the book explains that we are interested to compare the following two quantities: 1.the average of expected meansquared error (E_AE) 2.the expected meansquared error of the average (E_EA) Since x is a number that can take on different values with different probabilities, we should talk about the expected MSE, where the expectation E x is the averaging procedure over the probability distribution of x. if the estimates yi are independent, we have E_EA=(1/N)E_AE, instead if the estimates are completely dependent then the averaged estimate is just the same as each estimate,so the error is the same and we have E_EA=E_AE 
June 19th, 2014, 01:35 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,754 Thanks: 695 
Mathematically you seem to be correct. I have trouble understanding what point you are trying to make.

June 20th, 2014, 12:52 PM  #5 
Newbie Joined: Jun 2014 From: Italy Posts: 6 Thanks: 0 
I simply would understand how to solve the exercise. Calculate E_AE and E_EA with the formulas below for N=100 considering the different probabilities of X. I don't know how to apply these data in the two formulas. At the end I think E_AE and E_EA should have a value. Which is?

June 21st, 2014, 12:13 PM  #6 
Global Moderator Joined: May 2007 Posts: 6,754 Thanks: 695 
$\displaystyle E(e_i^2) = 1, E(e_ie_j) = 0$ for all i, and j ≠ i. Net result: $\displaystyle E_{AE}=1, E_{EA} = \frac{1}{N}$ 

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