Revision as of 02:12, 4 December 2015 by Yan115 (Talk | contribs)


ECE Ph.D. Qualifying Exam

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.


Back to QE CS question 1, August 2015

Back to ECE Qualifying Exams (QE) page

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva