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| + | E(S_n)=E(\frac{1}{n}\sum_i^n X_i) =frac{1}{n}\sum_i^n E(X_i)=0 | ||
| + | </math> | ||
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| + | <math> | ||
| + | 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> | ||
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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 00:59, 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 $
