Revision as of 01:07, 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)=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 $


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