Revision as of 18:36, 3 November 2014

Part 2

Let $ X_1,X_2,... $ be a sequence of jointly Gaussian random variables with covariance

$ Cov(X_i,X_j) = \left\{ \begin{array}{ll} {\sigma}^2, & i=j\\ \rho{\sigma}^2, & |i-j|=1\\ 0, & otherwise \end{array} \right. $

Suppose we take 2 consecutive samples from this sequence to form a vector $ X $, which is then linearly transformed to form a 2-dimensional random vector $ Y=AX $. Find a matrix $ A $ so that the components of $ Y $ are independent random variables You must justify your answer.

Solution 1

Suppose

$ A=\left(\begin{array}{cc} a & b\\ c & d \end{array} \right) $.

Then the new 2-D random vector can be expressed as

$ Y=\left(\begin{array}{c}Y_1 \\ Y_2\end{array} \right)=A\left(\begin{array}{c}X_i \\ X_j\end{array} \right)=\left(\begin{array}{c}aX_i+bX_j \\ cX_i+dX_j\end{array} \right) $

Therefore,

$ \begin{array}{l}Cov(Y_1,Y_2)=E[(aX_i+bX_j-E(aX_i+bX_j))(cX_i+dX_j-E(cX_i+dX_j))] \\ =E[(aX_i+bX_j-aE(X_i)-bE(X_j))(cX_i+dX_j-cE(X_i)-dE(X_j))] \\ =E[acX_i^2+adX_iX_j-acX_iE(X_i)-adX_iE(X_j)+bcX_iX_j+bdX_j^2-bcX_jE(X_i)\\ -bdX_jE(X_j)-acX_iE(X_i)-adX_jE(X_i)+acE(X_i)^2+adE(X_i)E(X_j)\\ -bcX_iE(X_j)-bdX_jE(X_j)+bcE(X_i)E(X_j)+bdE(X_i)^2]\\ =E(ac(X_i-E(X_i))^2+(ad+bc)(X_i-E(X_i)(X_j-E(X_j))+bd(X_j-E(X_j))^2]\\ =(ac)Cov(X_i,X_i)+(ad+bc)Cov(X-i,X_j)+(bd)Cov(X_j,X_j)\\ =ac\sigma^2+(ad+bc)\rho\sigma^2+bd\sigma^2 \end{array} $

Let the above formula equal to 0 and $ a=b=d=1 $, we get $ c=-1 $.

Therefore, a solution is

$ A=\left(\begin{array}{cc} 1 & 1\\ -1 & 1 \end{array} \right) $.

Solution 2

For $ n $ flips, there are $ n-1 $ changeovers at most. Assume random variable $ k_i $ for changeover,

$ p({k_i}=1)=p(1-p)+(1-p)p=2p(1-p) $

$ E(k)=\sum_{i_1}^{n-1}p(k_i=1)=2(n-1)p(p-1) $

Critique on Solution 2:

It might be better to claim the changeover as a Bernoulli random variable so the logic is easier to understand.

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