What is the OLS estimate for $\Delta{y_t} = py_{t-1} +\epsilon_t$

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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.

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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?