1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

Yes it is, you haven't missed anything in the differentiation, I am getting the same answer.

$\endgroup$
  • 1
    $\begingroup$ Certainly this is brief, but when the question is, "Is it correct?", this may legitimately be all there is to answer. I'm voting looks OK. $\endgroup$ – gung - Reinstate Monica Feb 28 '17 at 15:59

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.