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I am trying to make sales prediction from time series data.

Original plot

After performing a log transformation to the original data and differencing it by 1, I got a stationary dataset. So I plotted ACF and PACF graphs.

ACF graph: acf

PACF graph: pacf

Does this mean I should use ARIMA(0,1,0) to perform time series analysis? Also, if I got the value of p,d and q, do I use the original data to perform ARIMA model? For example,

ARIMA(original_data, order=(0,1,0))?
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ARIMA(original_data, order=(0,1,0))?

The d term should be chosen after you analyze the autocorrelation in your data and before you difference the data. You can also conduct the stationarity test to determine whether you need to take differences. And if you log-transform the raw data in the first place, then after calculating your forecast, you should make a back transform using $e$.

The p term reflects autocovariance structure, but I don't see significant values in your plots at lags 1 to 5 (typical range to search in...), so p = 0 most likely.

If you want to automate this routine, you can check out auto.arima function that is available in a library for some software, like R or Python.

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    $\begingroup$ And if you log-transform the raw data in the first place, then after calculating your forecast, you should make a back transform using $e$. Right. And since exponentiation is not a linear operation, beware that a point forecast of the mean in logs is not the point forecast of the mean in levels. Also, the shortest possible forecast interval with x% coverage in logs is not the shortest possible forecast interval with x% coverage when transformed back to levels by taking the endpoints and exponentiating them. $\endgroup$
    – Richard Hardy
    Nov 1, 2019 at 12:44
  • $\begingroup$ @RichardHardy, thank you. Are you making a reference to forecast interval boundaries? I understand, but is this possible to estimate them correctly in this case? $\endgroup$
    – Alexey Burnakov
    Nov 1, 2019 at 12:52
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    $\begingroup$ The transformed forecast interval still has correct coverage but is no longer the shortest possible. If a particular distribution of errors (innovations, shocks) is assumed, then a shorter interval could be derived, I think. I am not sure a shorter interval could be derived without such an assumption. $\endgroup$
    – Richard Hardy
    Nov 1, 2019 at 12:54

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