on the seminal paper of Robert Engle about Arch Process (Engle, R. F. (1982). Autoregressive conditional heteroscedasticity with estimates of the variance of United Kingdom inflation. Econometrica: Journal of the Econometric Society, 987-1007.) the first order conditions for maximum likelihood estimation for an ARCH(1) process is derived. Consider the next statements:
\begin{equation} y_t = \epsilon_t\cdot h_t^{\frac{1}{2}} \\ h_t = \alpha_0 + \alpha_1 \cdot y_{t-1}^2 \end{equation}
Where $E{[}\epsilon_t{]}=0$ and $Var{[}\epsilon_t{]}=1$. Now, the log-likelihood estimator of $\alpha$ named $l_t$ is:
\begin{equation} l_t = -\frac{1}{2} log\, h_t - \frac{1}{2} \frac{y_t^2}{h_t} \end{equation}
where $log\,x$ is the natural logarithm of $x$. Well, the paper state that the first order conditions for the likelihood estimation are:
\begin{equation} \frac{\partial l_t}{\partial \alpha} = \frac{1}{2h_t} \frac{\partial h_t}{\partial \alpha} \left(\frac{y_t^2}{h_t} - 1\right) \end{equation}
along with the Hessian. My question is how get the expression for $ \frac{\partial l_t}{\partial \alpha}$ from $l_t$? I tried to deduce the expression but I get a different result. In the calculation of the partial derivative of $y_t^2$ with respect to $\alpha$, is $y_t$ treated as independent of $\alpha$? and if the answer is yes why is that? (considering that $y_t = \epsilon_t \cdot h_t^{\frac{1}{2}}$ where $h_t$ itself depends on $\alpha$).
In advance, thanks for your time.
Well, checking my calculations after a long time, there's the answer. First, the model used is:
\begin{align} y_t | \psi_{t-1} &\sim \mathcal{N}(0, h_t)\\ h_t &= \alpha_0 + \alpha_1 y_{t-1}^2 \\ h_t &= h(y_{t-1}, y_{t-2}, \ldots,y_{t-p}, \alpha) \end{align}
where $p$ is the order of the process. As you can see $y_t$ is not equal to $\epsilon_t \cdot h^{1/2}$. So:
\begin{align} \frac{\partial{h_t}}{\partial\alpha} &= - \frac{1}{2} \frac{\partial{log(h_t)}}{\partial\alpha} - \frac{1}{2} \frac{\partial{\frac{y^2_t}{h_t}}}{\partial\alpha}\\ \frac{\partial{h_t}}{\partial\alpha} &= - \frac{1}{2 h_t} \frac{\partial{h_t}}{\partial\alpha} - \frac{1}{2} \frac{-y^2_t}{h^2_t} \frac{\partial h_t}{\partial \alpha}\\ \frac{\partial{h_t}}{\partial\alpha} &= - \frac{1}{2 h_t} \frac{\partial{h_t}}{\partial\alpha} \left(1 - \frac{y^2_t}{h^2_t}\right) \\ \frac{\partial{h_t}}{\partial\alpha} &= \frac{1}{2 h_t} \frac{\partial{h_t}}{\partial\alpha} \left(\frac{y^2_t}{h^2_t} - 1\right) \end{align}
which is the expression needed.