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<math> | <math> | ||
| − | E[X_iX_j]=\int_{-\infty}^{+\infty} | + | E[X_iX_j]=\int_{-\infty}^{+\infty}x_ix_jf_{x_ix_j}(x_i,x_j)dx_idx_j |
</math> | </math> | ||
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Revision as of 12:08, 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[X_iX_j]=\int_{-\infty}^{+\infty}x_ix_jf_{x_ix_j}(x_i,x_j)dx_idx_j $
