1
$\begingroup$

The following is an assignment, I just need some help understanding what I've done so far.

Suppose $X_1 \sim N(0, \sigma^2 / (1-\rho^2))$ and $U_2, \dots, U_n \sim N(0, \sigma^2)$, independent between them and with $X_1$. We define:

$$ X_k = U_k + \rho X_{k-1},$$

for $k=2,\dots,n$.

  1. Show using induction that $(X_{k-1}, X_k)$ follow a bivariate normal. I used the equation:

$$(X_{k-1}, X_k) = \begin{bmatrix} 1 & 0\\ \rho & 1 \end{bmatrix} \begin{bmatrix} X_{k-1}\\ U_k \end{bmatrix}.$$ For the base case I used $k=2$: $$(X_1, X_2) = \begin{bmatrix} X_1\\ U_{2}+\rho X_1 \end{bmatrix},$$ I have already proved that $U_{k+1}+\rho X_k$ follows $N(0, \sigma^2 / (1-\rho^2))$ using induction.

For the induction step I assume it holds for $k$ and we want to prove it holds also for $k+1$: $$(X_k, X_{k+1}) = \begin{bmatrix} X_k\\ U_{k+1}+\rho X_k \end{bmatrix},$$ which, as said before, I have already proved.

  1. Compute the covariance matrix of $(X_{k-1}, X_k)$ For the main diagonal I guess we can simply take the variance of the normal distributions: $\sigma^2 / (1-\rho^2)$. For the covariance I used the following equality:

$$2Cov(X_{k-1}, X_k) = Var(X_{k-1} + X_k) - Var(X_{k-1}) - Var(X_k).$$

This gives me that $Cov(X_{k-1}, X_k) = 0$.

  1. Compute $Cov(X_{k-j}, X_k)$ for $k=2,\dots,n$ and $j=1,\dots,k-1$ using the model for $X_k$.

Basically I computed just $E[X_{k-j}X_k]$. Suppose I compute $Cov(X_{k-1}, X_k)$:

$$ E[(U_k + \rho X_{k-1})(U_{k-1} + \rho X_{k-2})] = E[U_k]E[U_{k-1}] + \rho E[U_k]E[X_{k-2}] + \rho E[U_{k-1}X_{k-1}] + \rho^2 E[X_{k-1}X_{k-2}] = \rho^3E[X_{k-1}X_{k-2}] = \dots.$$

Is this sufficient?

  1. Compute the covariance matrix of $(X_1, \dots, X_n)$. For this I need a suggestion. Thanks.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.