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If the random vector $\mathbf{X}$ is transformed according to \begin{align*} Y_1 &= X_1\\ Y_2 &= X_1 + X_2 \end{align*} and has a covariance matrix $$ \mathbf{C}_X = \begin{bmatrix} 2 & 1\\ 1 & 2 \end{bmatrix} $$ find the covariance matrix for $\mathbf{Y} = [Y_1 Y_2]^{\intercal}$.


My Solution

Let $\mathbf{Y} = \mathbf{A}\mathbf{X}$. Then by definition, $\mathbf{C}_Y = \mathbf{V}^{\intercal}\mathbf{C}_X\mathbf{V} = \Lambda$ where $\mathbf{A} = \mathbf{V}^{\intercal}$, $\mathbf{V}$ is the matrix of eigenvectors for the eigenvalues $\lambda_i$ for $\mathbf{C}_x$, and $\Lambda$ is a diagonal matrix. First, we need to find the eignvalues and eigenvectors for $\mathbf{C}_X$. The characteristic equation for a $2\times 2$ matrix is $$ p(\lambda) = \lambda^2 - \lambda\text{tr}(\mathbf{M}) + \det(\mathbf{M}) $$ where $\mathbf{M}$ is the matrix of interest and tr is the trace of the matrix. $$ p(\lambda) = \lambda^2 - 4\lambda + 3 = (\lambda - 3)(\lambda - 1) = 0 $$ Therefore, the eigenvalues are $\lambda = 1, 3$. The eigenvector for $\lambda = 1$ can be found by $$ \begin{bmatrix} 1 & 1\\ 1 & 1 \end{bmatrix} \Rightarrow \begin{bmatrix} 1 & 1\\ 0 & 0 \end{bmatrix} $$ This tells us that $x_1 = -x_2$ and $x_2$ is a free variable. In other words, $$ \mathbf{v}_{\lambda = 1} = \begin{bmatrix} -x_2\\ x_2 \end{bmatrix} = x_2 \begin{bmatrix} -1\\ 1 \end{bmatrix} $$ For $\lambda = 3$, we have $$ \mathbf{v}_{\lambda = 3} = x_2 \begin{bmatrix} 1\\ 1 \end{bmatrix} $$ Finally, we have that our matrix $\mathbf{V}$ is $$ \mathbf{V} = \begin{bmatrix} -1 & 1\\ 1 & 1 \end{bmatrix} $$ so $$ \mathbf{C}_Y = \mathbf{V}^{\intercal}\mathbf{C}_x\mathbf{V} = \begin{bmatrix} -1 & 1\\ 1 & 1 \end{bmatrix} \begin{bmatrix} 2 & 1\\ 1 & 2 \end{bmatrix} \begin{bmatrix} -1 & 1\\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 0\\ 0 & 6 \end{bmatrix}. $$


The book's answers

The book's solution is $$ \begin{bmatrix} 2 & 3\\ 3 & 6 \end{bmatrix}. $$ Why is this? By definition, the covariance matrix of $Y$ is diagonal.

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The entries of the covariance matrix of $\vec{Y}$ are $$C^Y_{ij}=Cov(Y_i, Y_j)$$ for $i=1,2$. Moreover the matrix is symmetric since $$C^Y_{ij}=Cov(Y_i, Y_j)=Cov(Y_j,Y_i)=C^Y_{ji}$$ Thus $$\begin{align*}C^Y_{11}&=Cov(Y_1,Y_1)=Var(Y_1)=Var(X_1)=Cov(X_1,X_1)=C^{X}_{11}=2\\C^Y_{12}&=Cov(Y_1,Y_2)=Cov(X_1,X_1+X_2)=Cov(X_1,X_1)+Cov(X_1,X_2)=\\&=C^{X}_{11}+C^{X}_{12}=2+1=3\\C^Y_{21}&=C^Y_{12}=3 \\C^Y_{22}&=Cov(Y_2,Y_2)=Var(Y_2)=Var(X_1+X_2)=Var(X_1)+Var(X_2)+2Cov(X_1,X_2)=\\&=C^{X}_{11}+C^{X}_{22}+2C^{X}_{12}=2+2+2\cdot1=6\end{align*}$$ Hence $$C^Y=\begin{bmatrix} 2&3\\3&6 \end{bmatrix}$$

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  • $\begingroup$ What is wrong with using the definition I have? $\endgroup$ – dustin Oct 24 '14 at 20:32
  • $\begingroup$ I do not know, I have to check it. But it is not true that the matrix has to be diagonal. Check again your definition, perhaps it applies only for independent vectors or uncorrelated or something like that. $\endgroup$ – Jimmy R. Oct 24 '14 at 20:35

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