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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 | + | E(S_n)=E(\frac{1}{n}\sum_i^n X_i) =\frac{1}{n}\sum_i^n E(X_i)=0 |
</math> | </math> | ||
Revision as of 02:00, 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 $
