| Line 21: | Line 21: | ||
</center> | </center> | ||
---- | ---- | ||
| + | ===Solution 1=== | ||
<math> | <math> | ||
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 | ||
| Line 43: | Line 44: | ||
Thus <math>E(X_i-S_n)E(S_n)=E((X_i-S_n)S_n)</math>, <math>S_n</math> and <math>X_i-S_n</math> are uncorrelated. | Thus <math>E(X_i-S_n)E(S_n)=E((X_i-S_n)S_n)</math>, <math>S_n</math> and <math>X_i-S_n</math> are uncorrelated. | ||
| + | |||
| + | ===Solution 2=== | ||
| + | <math> S_n=\frac{1}{n}\sum_{j=1}{n}X_j </math>, note: in the problem statement, it should be <math>\frac{1}{n}, because <math>S_n</math> is the sample mean. | ||
| + | |||
| + | <math> | ||
| + | E[S_n]=E[\frac{1}{n}\sum_{j=1}{n}X_j] = \frac{1}{n}\sum_{j=1}{n}E[X_j ] = \frac{1}{n}\sum_{j=1}{n} \mu = 0\\ | ||
| + | E[(X_i-\mu)^2]=E[X_i^2]=\sigma^2 | ||
| + | </math> | ||
| + | |||
| + | <math> | ||
| + | E[S_n]=E[\frac{1}{n}\sum_{j=1}{n}X_j] = \frac{1}{n}\sum_{j=1}{n}E[X_j ] = \frac{1}{n}\sum_{j=1}{n} \mu = 0\\ | ||
| + | E[(X_i-\mu)^2]=E[X_i^2]=\sigma^2 | ||
| + | </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 13:05, 7 December 2015
Communication, Networking, Signal and Image Processing (CS)
Question 1: Probability and Random Processes
August 2015
Solution 1
$ 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.
Solution 2
$ S_n=\frac{1}{n}\sum_{j=1}{n}X_j $, note: in the problem statement, it should be $ \frac{1}{n}, because <math>S_n $ is the sample mean.
$ E[S_n]=E[\frac{1}{n}\sum_{j=1}{n}X_j] = \frac{1}{n}\sum_{j=1}{n}E[X_j ] = \frac{1}{n}\sum_{j=1}{n} \mu = 0\\ E[(X_i-\mu)^2]=E[X_i^2]=\sigma^2 $
$ E[S_n]=E[\frac{1}{n}\sum_{j=1}{n}X_j] = \frac{1}{n}\sum_{j=1}{n}E[X_j ] = \frac{1}{n}\sum_{j=1}{n} \mu = 0\\ E[(X_i-\mu)^2]=E[X_i^2]=\sigma^2 $
