Difference between revisions of "ECE PhD QE CNSIP 2015 Problem1.4" - Rhea

Revision as of 16:59, 26 January 2016

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=\int_{-\infty}^{+\infty}x_if_{X_i}(x_i)x_jf_{X_j}(x_j)dx_idx_j=E[X_i]E[X_j]=\mu\cdot\mu=0 $

$ E[X_i-S_n]=E[X_i]-E[S_n]=0-0=0 $

$ E[X_i\cdot S_n]=E[\frac{1}{n}\sum_{j=1}^{n}X_j\cdot X_i]=\frac{1}{n}\sum_{j=1}^{n}E[X_j\cdot X_i]=\frac{1}{n}\cdot \sigma^2 $

$ E[S_n^2]=E[\frac{1}{n^2}\sum_{j=1}^{n}\sum_{i=1}^{n}X_j\cdot X_i]=\frac{1}{n^2}\sum_{j=1}^{n}E[X_i^2]+\frac{1}{n^2}\sum_{j=1}^{n}\sum_{i=1}^{n}E[X_i\cdot X_j]=\frac{1}{n^2}\cdot (n\cdot \sigma^2) + \frac{1}{n^2}\cdot 0 = \frac{\sigma^2}{n} $

Therefore,

$ E[(S_n-0)(X_i-S_n-0)]=E[S_nX_i-S_n^2]= E[S_nX_i]-E[S_n^2]=\frac{\sigma^2}{n}-\frac{\sigma^2}{n}=0 $

So $ r = \frac{cov(S_n,X_i-S_n)}{\sigma_{S_n}\sigma_{X_i-S_n}}=0 $

Thus $ S_n $ and $ X_i-S_n $ are uncorrelated.

Solution 3

Our goal is to show that $ S_n $ and $ X_i $ - $ S_n $ are uncorrelated $ \forall i \in 1, 2, ..., n $. If we can show that the covariance between $ S_n $ and $ X_i - S_n $ is equal to 0 $ \forall i $, then we will have shown the aforementioned property. Recalling that

$ cov(X,Y) = E[XY] - E[X]E[Y], $

we aim to show that

$ E[(S_n)(X_i -S_n)] - E[S_n]E[X_i-S_n] = 0. $

Let us consider the LHS of the above equation. This can be written as

$ E[(S_n)(X_i -S_n)] - (E[S_n]E[X_i]-E[S_n]E[S_n]). $

We are given that $ E[X_i] = 0 \forall i $, and since $ S_n \triangleq \frac{1}{n}\sum^n_{j=1}X_j $ it is easy to show that $ E[S_n] = 0 $ as well. Thus the above becomes

$ E[(S_n)(X_i -S_n)] $

or

$ E[S_n X_i] - E[S_n^2]. $

Recalling that the above expression must be equal to 0, our problem has reduced to showing that

$ E[S_n X_i] = E[S_n^2]. $

Let us first examine $ E[S_n X_i] $. This can be rewritten as

$ E[S_n X_i] = E\left[\left(\frac{1}{n}\sum^n_{j = 1}X_j \right)\cdot X_i\right] = \frac{1}{n}E[X_1 X_i + X_2 Xi + ... + X_i X_i + ... + X_n X_i]. $

Since the sequence $ X_1, X_2, ..., X_n $ is i.i.d, if $ i\neq j, E[X_i X_j] = 0 $. Thus, the above becomes

$ \frac{1}{n}\left(0 + 0 + ...\, E[X_i^2] + ... + 0 + 0\right). $

Since $ X_i $ is zero-mean $ \forall i $, we know that $ E[X_i^2] = var(X_i) = \sigma^2 $ for any $ i $, and that $ E[S_n X_i] = \frac{\sigma^2}{n} $. Now let us examine $ E[S_n^2] $. This can be rewritten as

$ E[S_n^2] = E\left[\frac{1}{n}\left(X_1 + X_2 + ... + X_n\right)\frac{1}{n}\left(X_1 + X_2 + ... + X_n\right)\right] = \frac{1}{n^2}E\left[\left(X_1 + X_2 + ... + X_n\right)^2\right] $.

Squaring out the expression inside the expectation operator will result in an expression with $ n $ square terms and $ \sum^n_{j-1}(j-1) $ cross terms. Recall that the expectation values of the cross terms will all be zero since $ X_i $ is zero-mean $ \forall i $. Then we have

$ \frac{1}{n^2}E\left[\left(X_1 + X_2 + ... + X_n\right)^2\right] = \frac{1}{n^2}\left(E[X_1^2] + E[X_2^2] + ... + E[X_n^2]\right) = \frac{1}{n}(n\sigma^2) = \frac{\sigma^2}{n}. $

Thus we have shown that $ E[S_n X_i] = E[S_n^2] = \frac{\sigma^2}{n} $, and we are done.

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