- streetlegend
**Posts:**12**Joined:**

Hi guys,I have a question about ARIMA notation - ie. ARIMA (1,0,1)From what I understand, the notation explains the (autoregressive processes, unit roots, moving average processes).So take for instance Y(t) is an AR(2) process with 1 unit root and 1 moving average process.Taking the difference, deltaY(t) is an AR(1) process with 0 unit roots and 1 moving average processes.I understand that the notation for deltaY(t) = ARIMA(1,0,1)However, reading in notes, the Y(t) notation = ARIMA(1,1,1)I am wondering why the Y(t) notation doesn't = ARIMA(2,1,1)?Thanks.

what you are asking is not very clear (to me)but you almost have everything as you write:Quotethe notation explains the (autoregressive processes, unit roots, moving average processes)let the ARIMA(p,d,q) be.then :p = order of the autoregressive partd = order of differencing leading to a stationary ARMA processq = order of the moving average part

Last edited by tagoma on December 4th, 2010, 11:00 pm, edited 1 time in total.

- StructCred
**Posts:**301**Joined:**

Because ARIMA(0,1,0) is an AR(1) process with coefficient of 1?

ARIMA(0,1,0) is an I(1) process, that is a random walk.

- streetlegend
**Posts:**12**Joined:**

QuoteOriginally posted by: edouardwhat you are asking is not very clear (to me)My question is why an AR(2) process is noted as ARIMA(1,x,x) as opposed to ARIMA(2,x,x)

Quotewhy an AR(2) process is noted as ARIMA(1,x,x) where did you read that ? what is your source ?

Last edited by tagoma on December 13th, 2010, 11:00 pm, edited 1 time in total.