Question: Consider the normal random variables $X_1,\dots,X_n$, where $X_{i} = \theta X_{i-1} + \epsilon_{i-1}$ for $i=1,2,\dots,n$ and $X_0 = 0 $ and $\epsilon_i \stackrel{iid}{\sim} N(0,\sigma^2)$.
(a) Find the joint pdf of $\vec X$,
(b) show that the LRT for $\theta = 0$ versus $\theta \neq 0$ is based on the test statistic $- \left( \sum_{i=2}^n X_i X_{i-1} \right)^2 / \sum_{i=1}^{n-1} X_i^2$.
This is an old exam problem I'm having lots of trouble with.
Each $X_i$ has mean $0$, and $Cov(X_{t+k},X_t) = Cov(\theta X_{t+k-1} + \epsilon_{t+k-1}, X_t)= \cdots = \theta^{k} Cov(X_t,X_t)$.
From here I just use: \begin{align*} X_t = \theta^{t-1} \epsilon_1 + \theta^{t-2} \epsilon_2 + \cdots + \epsilon_t \\ Cov(X_t,X_t) = E[X_t^2] = \sigma^2 \sum_{k=0}^{t-1} \theta^{2k} = \sigma^2 \frac{\theta^{2t}-1}{\theta-1} \end{align*} Then, $X_1,\dots,X_n$ are jointly normal with mean $\mathbf 0$ and covariance matrix $\Sigma_{ij} = \theta^{|i-j|} \sigma^2 \dfrac{ \theta^{2 \min\{i,j\}}-1}{\theta-1}$. The joint pdf is then $\frac{1}{\sqrt{(2\pi)^n | \Sigma|}} e^{- \frac{1}{2} \mathbf x^T \Sigma^{-1} \mathbf x }$.
Now for part (b). I need to find $$ \dfrac{L(\Sigma; \theta = 0)}{ \sup \limits_{\theta \neq 0}L(\Sigma)} = \dfrac{\frac{1}{\sqrt{(2\pi)^n}} e^{- \frac{1}{2} \mathbf{x}^T \mathbf{x}}}{\sup \limits_{\theta \neq 0} L(\Sigma)} $$ but I have no idea how to tackle the denominator. I keep reminding myself that this is from an in-class exam, so the solution must be relatively uncomplicated . . . But no luck . . .