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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)=E(X_i-\frac{1}{n}\sum_k^n X_k) =E(X_i)-E(\frac{1}{n}\sum_k^n X_k)=0 | ||
</math> | </math> | ||
| + | |||
| + | <math> | ||
| + | E((X_i-S_n)S_n)=E(X_iS_n-S_n^2)=E(X_i-\frac{1}{n}\sum_k^n X_k) | ||
| + | </math> | ||
| + | |||
| + | As for any <math> i,j\in \{1,2,...,n\} </math>, we have <math>E(X_i\cdot X_j) = E(X_i)E(X_j)=0 </math> | ||
---- | ---- | ||
[[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 02:07, 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)=E(X_i-\frac{1}{n}\sum_k^n X_k) $
As for any $ i,j\in \{1,2,...,n\} $, we have $ E(X_i\cdot X_j) = E(X_i)E(X_j)=0 $
