Deriving ADF unit root test form for the time series with quadratic deterministic trend I have the following time series process $y_t $
$$\Delta y_t = \delta + \gamma t + \epsilon_t$$
where $e_t$ is white noise process with the variance of $\sigma^2$.
I guess that whereas $\Delta y_t$ is trend stationary process, $y_t$ is a random walk with a quadratic deterministic trend.
So, by using the process $y_t$ which is defined as a random walk with a quadratic deterministic trend, I want to apply the ADF unit root test. During the testing procedure, which form do we assign to $y_t$ under the null hypothesis?
What I did is as follows:
My solution is that! But my solution is totally wrong. I just want to exhibit my attempt, so I added them as a picture because this solution is too long to type it and it is wrong. (Sorry for not typing )
But, I have no idea other than this solution. Please share your ideas with me. Thank you a lot!


 A: Following this reference:
Given:  Time series process $y_t $
$$\Delta y_t = \delta + \gamma t + \epsilon_t$$
where $e_t$ is white noise process with the variance of $\sigma^2$.
Question:  For the ADF test, which form do we assign to $y_t$ under the null hypothesis?
ADF Test:
For a process of the form,
$$y_t = \beta_0 + \beta_1 t + \phi y_{t-1} + \sum_{j=1}^p \psi_j \Delta y_{t-j} + \epsilon_t$$
Test: $H_o: \phi = 1$
Answer:
For the given process, the higher order lag terms are not present:
Option 1: $y_t = \phi y_{t-1} + \delta + \gamma t + \epsilon_t, \ \ \ H_o: \phi = 1$
Option 2: $\Delta y_t = \pi y_{t-1} + \delta + \gamma t + \epsilon_t, \ \ \ H_o: \pi = 0$
Using the method of least squares with Option 2, fit this equation $\Delta y_t = \beta_0 + \beta_1 y_{t-1} + \beta_2 t + \epsilon_t$ and then test if $H_o: \beta_1 = 0$ using the normal t-test in regression results.
Example:
y <- numeric(100)
y[1] <- 3 + rnorm(1, 0, 2)
for (i in 2:100)
{
  y[i] <- y[i-1] + 3 + 0.5*(i-1) + rnorm(1, 0, 2)
}

test_data <- data.frame(
  dy = diff(y),
  t = 1:99,
  ym1 = y[1:99]
)

lm1 <- lm(dy ~ ym1 + t, data = test_data)
summary(lm1)
```

