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I have an equation $\Delta{y_t} = py_{t-1} +\epsilon_t$. I want to know what is the p value estimated using OLS. I have drawn the below calculations based on my bounded knowledge on OLS estimation with simple formulas.

Calculations:

$\Delta{y_t} = py_{t-1} +\epsilon_t$ => $y_t-y_{t-1} = py_{t-1} +\epsilon_t $

$\therefore y_t = y_{t-1}(p+1) +\epsilon_t$

We know,

$S = \sum[y_t - (y_{t-1}(p+1) + \epsilon_t)]^2$ [based on OLS]

Take differentiation w.r.t to p unknown variable

$\frac{\partial s }{\partial p} = \frac {\partial }{\partial p}(\sum[{y_t} - (y_{t-1}(p+1)+\epsilon_t)]^2$

$= 2 * \sum[{y_t} - [y_{t-1}(p+1) + \epsilon_t]]*[0 - \frac {\partial y_{t-1} (p+1)} {\partial p} + \frac{\partial \epsilon_t}{\partial p}]$

$=2*\sum[y_t - [y_{t-1}(p+1) + \epsilon_t] * y_{t-1}$

Equating to 0

$0 = \sum y_t y_{t-1} - \sum(y_{t-1})^2(p+1)- \sum \epsilon_ty_{t-1}$

$(p+1) \sum (y_{t-1})^2 +\epsilon_t \sum y_{t-1} = \sum y_t y_{t-1}$

$p+1 = \frac{\sum y_t y_{t-1} - \epsilon_t \sum y_{t-1}}{\sum (y_{t-1})^2}$

$\therefore$

$p = \frac{\sum y_t y_{t-1} - \epsilon_t \sum y_{t-1}} {\sum (y_{t-1})^2} - 1$

I am obtaining the p value as above. Can someone say whether the calculations done are in the right manner or not? Thanx for any guidance.

Devi
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  • Is $\epsilon$ white noise and thus not part of the regression? Also are the summations over $t$ in which case if $\epsilon$ is meant to be included it should stay inside the summation. – user121049 Oct 31 '17 at 08:43
  • It is part of regression. So the equation be $p = \frac{\sum y_t y_{t-1} - \sum \epsilon_ty_{t-1}}{\sum (y_{t-1})^2} - 1$. So is my logic correct ? – Devi Oct 31 '17 at 10:08
  • ε should not be part of the regression. As it is written, your S would identically be zero in the first step, no? Anyway, why are your trying to derive the basic OLS result to find the coefficients? Any package (including Excel) can do OLS for you. You just need to find the regression to apply to your problem. – Mathemagical Nov 01 '17 at 08:37
  • @Mathemagical I want to implement and see the results :) – Devi Nov 01 '17 at 08:54
  • I have differentiated S with respect to p and equated to 0 ..not S = 0. Can u tell what is the change I have to add in the above derivation ? – Devi Nov 01 '17 at 11:09

1 Answers1

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Ditching the ε terms in the estimator (you can read about the derivation at many places on the web).

For the equation

$ y_t = y_{t-1}(p+1) +\epsilon_t$

The OLS estimator is $p = \frac{\sum y_t y_{t-1} }{\sum (y_{t-1})^2} - 1$

Is this what you are implementing for your Dickey Fuller problem?

Mathemagical
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  • no..I was trying with $p = \sum yx/\sum x^2$...I have messed with the wrong equation....I gave a try with above equation and new result is −1.31944953 which after seasonal->first diff->first diff I have obtained. – Devi Nov 04 '17 at 06:53
  • Pls see the edit above – Devi Nov 04 '17 at 07:03