# Differencing a time series

I am looking to find the ACF of a time series, but after it is differenced.

$y_t = a_1y_{t-1} + \epsilon_t , \mid a_1 \mid < 1$

I understand that to find the ACF this process needs to be stationary, but I am struggling to difference the equation due to the constant $a_1$

Any help would be greatly appreciated.

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This process is stationary since $\alpha$ is less than one in absolute value. – Dimitriy V. Masterov Apr 3 '14 at 15:57
Apologies I misconstrued my question, I have corrected now – Jobloggs Apr 3 '14 at 16:08

To difference the equation, note that $y_{t-1}=a_1y_{t-2}+\epsilon_{t-1}$ and deduct that from what you have. That is, deduct the LHS from your LHS and the RHS from your RHS. You get $y_t-y_{t-1}=a_1y_{t-1}+\epsilon_t-(a_1y_{t-2}+\epsilon_{t-1})=a_1(y_{t-1}-y_{t-2})+\epsilon_t-\epsilon_{t-1}$ which can be written as $$\Delta y_t=a_1\Delta y_{t-1}+\Delta \epsilon_t,$$ where $\Delta$ denotes the first difference. Is that the help you needed?

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That is exactly what I needed Karl, thank you for the quick response. – user43040 Apr 3 '14 at 18:20
Great! Please note that you have not signed in from the same account as when asking the question. Please do so and accept the answer. Good luck! – Student001 Apr 3 '14 at 18:21

if your time series non-stationary, then quite likely that $|a_1|>1$.

when $|a_1|<1$ the series are stationary.

$a_1$ is not a problem for differencing, but it's not very useful when $|a_1|<1$

to compute ACF you don't need the process be stationary, it's just the ACF would not make much sense, it'll be exponentially growing with a lag.

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Same as above, I have corrected now, apologies – Jobloggs Apr 3 '14 at 16:09