Auto regressive process, maximum likelihood estimator A first-order autoregressive process, $X_0,\dots,X_n$, is given through the following conditional distributions:
$X_i | X_{i-1},\dots,X_0 \sim \mathcal{N}(\alpha X_{i-1},1)$,
for $i = 1,2,\dots,n$ and $X_0 \sim \mathcal{N}(0,1)$.
I know that the log-likelihood function $\ell{(\alpha)}$ is of the form:
$\ell(\alpha) = - \frac{1}{2} \sum (x_i - \alpha x_{i-1})^2 + c$. But I don't know how to show that.
I found for $\hat{\alpha}_{ML}$ the following solution: $\hat{\alpha}_{ML} = \frac{s}{t}, \mathrm{where} \; s = \sum x_1 x_{i-1} \mathrm{and} \; t = \sum (x_{i-1})^2$. Is this right?
Then I have to show that this is the global maximum. If I take the second derivative I get a constant. Is this the sign that I got the global maximum, because the first derivative is linear wrt to $\alpha$? Right?
 A: Using the chain rule, the joint density here can be decomposed as (denoting $\mathbf X$ the collection of the $n+1$ random variables)
$$f_{\mathbf X}(x_n,x_{n-1},...,x_0) = f(x_n\mid x_{n-1},...,x_0)\cdot f(x_{n-1}\mid x_{n-2},...,x_0)\cdot f(x_{n-2}\mid x_{n-3},...,x_0) \cdot...\cdot f(x_0)$$
$$=\left(\prod_{i=1}^{n}\frac {1}{\sqrt{2\pi}}\exp\left\{-\frac {(x_i-\alpha x_{i-1})^2}{2}\right\}\right)\frac {1}{\sqrt{2\pi}}\exp\left\{-\frac {x_0^2}{2}\right\}$$
Viewed as a likelihood function of $\alpha$, and taking its natural logarithm, we have
$$\ln L(\alpha \mid \mathbf X) = -\frac 12\sum_{i=1}^n (x_i-\alpha x_{i-1})^2 +c$$
...where in $c$ is also included the density of $x_0$ (but $x_0$ affects estimation of $\alpha$ through its presence in the conditional density related to $X_1$).
Then 
$$\frac {\partial \ln L(\alpha \mid \mathbf X)}{\partial \alpha} = \frac {\partial }{\partial \alpha} \left(-\frac 12\sum_{i=1}^n (x_i-\alpha x_{i-1})^2\right)$$
$$=-\frac 12\frac {\partial }{\partial \alpha} \left(\sum_{i=1}^n (x_i^2-2\alpha x_ix_{i-1}+\alpha^2x_{i-1}^2)\right)  $$
$$=-\frac 12\frac {\partial }{\partial \alpha} \left(\sum_{i=1}^n x_i^2-2\alpha \sum_{i=1}^nx_ix_{i-1}+\alpha^2\sum_{i=1}^nx_{i-1}^2)\right) $$
$$=\sum_{i=1}^n x_ix_{i-1} -\alpha\sum_{i=1}^nx_{i-1}^2$$
Setting 
$$\frac {\partial \ln L(\alpha \mid \mathbf X)}{\partial \alpha} =0\Rightarrow \hat \alpha_{ML} = \frac {\sum_{i=1}^n x_ix_{i-1}}{\sum_{i=1}^nx_{i-1}^2}$$
while $$\frac {\partial^2 \ln L(\alpha \mid \mathbf X)}{\partial \alpha^2} = -\sum_{i=1}^nx_{i-1}^2 <0$$
which guarantees a global and unique maximum, since it is negative irrespective of $\alpha$.
