# 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!

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?

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: $$H_o: \phi = 1$$

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)
$$$$
`