Consider the following multiplicative model:

$$ y_t = \alpha y_{t-1} \epsilon_t ~~~~~,~\text{ln}(\epsilon_t) \sim NID(0,\sigma^2)$$

In order to find the logarithmic transformation of the parameter $\alpha$, I did the following passages:

$$ ln(y_t) = ln(\alpha) + ln(y_{t-1}) + ln(\epsilon_t) $$ $$S(\hat{\alpha})= \sum[(ln(y_t)-ln(\alpha)-ln(y_{t-1}))^2]$$

Now I take the derivative:

$$ 0 = -2\sum[(ln(y_t)-ln(\alpha)-ln(y_{t-1})]\dfrac{1}{\alpha} $$ $$ (T-1) ln(\hat{\alpha})=\sum ln(y_t) - \sum ln(y_{t-1}) $$ $$ ln(\hat{\alpha})=\dfrac{1}{T-1} \sum ln(\dfrac{y_t}{y_{t-1}}) $$

Is it correct? Thanks in advance.

Consider the following multiplicative model:

$$ y_t = \alpha y_{t-1} \epsilon_t,\qquad \text{ln}(\epsilon_t) \sim NID(0,\sigma^2)$$

In order to find the logarithmic transformation of the parameter $\alpha$, I did the following passages:

\begin{align} \ln(y_t) &= \ln(\alpha) + \ln(y_{t-1}) + \ln(\epsilon_t) \\[5pt] S(\hat{\alpha}) &= \sum\big[(\ln(y_t)-\ln(\alpha)-\ln(y_{t-1}))^2\big] \end{align}

Now I take the derivative:

\begin{align} 0 &= -2\sum\big[(\ln(y_t)-\ln(\alpha)-\ln(y_{t-1})\big]\dfrac{1}{\alpha} \\[5pt] (T-1) \ln(\hat{\alpha}) &= \sum \ln(y_t) - \sum \ln(y_{t-1}) \\[10pt] \ln(\hat{\alpha}) &= \dfrac{1}{T-1} \sum \ln\bigg(\dfrac{y_t}{y_{t-1}}\bigg) \end{align}

Is it correct?