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</math> | </math> | ||
| − | Thus <math>E(X_i-S_n | + | Thus <math>E(X_i-S_n)E(S_n)=E((X_i-S_n)S_n)</math>, <math>S_n</math> and <math>X_i-S_n</math> are uncorrelated. |
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[[ECE-QE_CS1-2015|Back to QE CS question 1, August 2015]] | [[ECE-QE_CS1-2015|Back to QE CS question 1, August 2015]] | ||
[[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]] | [[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]] | ||
Revision as of 01:13, 4 December 2015
Communication, Networking, Signal and Image Processing (CS)
Question 1: Probability and Random Processes
August 2015
$ E(S_n)=E(\frac{1}{n}\sum_i^n X_i) =\frac{1}{n}\sum_i^n E(X_i)=0 $
$ E(X_i-S_n)=E(X_i-\frac{1}{n}\sum_k^n X_k) =E(X_i)-E(\frac{1}{n}\sum_k^n X_k)=0 $
$ E((X_i-S_n)S_n)=E(X_iS_n-S_n^2) $
As for any $ i,j\in \{1,2,...,n\} $, we have $ E(X_i\cdot X_j) = E(X_i)E(X_j)=0 $
$ E(X_iS_n-S_n^2) = E(X_iS_n)-E(S_n^2)\\ =E(\sum_k^nX_iX_K) - E(\sum_i^n\sum_k^nX_iX_K)\\ =\sum_k^nE(X_iX_K) - \sum_i^n\sum_k^nE(X_iX_K) \\ =0 $
Thus $ E(X_i-S_n)E(S_n)=E((X_i-S_n)S_n) $, $ S_n $ and $ X_i-S_n $ are uncorrelated.
