I am quite new to the R and the ARIMA model, and I have a question on the ARIMA model that I obtained in R.
I will use the US unemployment rate as an example, the data range is from Jan, 1948 to Feb, 2015, total of 806 observations. After looking at the AICc, I decided to use ARIMA(2,1,2) model. (BTW I am using Arima() function from "forecast" package in R)
The output is the following:
Series: log.unemp
ARIMA(2,1,2)
ar1 1.6406
ar2 -0.7499
ma1 -1.5943
ma2 0.7893
sigma^2 estimated as 0.001307: log likelihood=1530.14
AIC=‐3050.27 AICc=‐3050.2 BIC=‐3026.82
The code is
fit.best <- Arima(log.unemp, c(2, 1, 2), include.constant=FALSE)
print(fit.best)
Then I want to measure the forecast performance of this model. That is, to calculate things like RMSE, Theil's U, etc. But I do not know how to do that. The reason is that it seems that I do not know how to derive the forecast equation from this output to calculate the fitted values.
So could anyone help me on this? How should I derive the forecast equation from this output? Also, after obtaining the equation, how can I do the forecast in Excel to calculate the fitted values from the first data point (there are some numbers that are not available when you are calculating the fitted value for t=1)?
Thanks!
you can use summmary(fit.best) to view RMSE.
Or if you want to caluate by yourself you can derive residuals and fitted values like this:
fitted=log.unemp-fit.best$residuals
about the equation see this
You can use forecast package
fit.best <- Arima(log.unemp, c(2, 1, 2), include.constant=FALSE)
my_forecast <- forecast(fit.best, h=10)
my_forecast #will show the next 10 periods
# or use some detailed data like
plot(my_forecast$residuals)
Use arima model as below code:
arimafit = arima(log.unemp, order=c(2,1,2))
Forecast using the below code:
arima_future = forecast(arimafit, h=3)
where forecast is the function to forecast for next whatever months you want.
h=3 means it will forecast for next 3 months.
If you want to check RMSE on the test data you can use DMwr package:
metrics = as.data.frame(DMwR::regr.eval(<test_data_vector>, arima_future$point_forecast))
test_data_vector - is the test data vector which you can create while dividing your main dataset into train and test dataset.
arima_future$point_forecast - Is the vector point forecast vector you will get in step2.
Related
I have a problem with my code. I want to forecast stock returns with an ARIMA model in R but I can not get my data stationary. Besides transforming the stock prices into returns, I also tried the diff function for differencing my time-series. I always assumed that data becomes stationary by using one of the 2 methods. However, when I run an augmented dickey fuller test (adf.test in R) my p-value shows me that the data remains non-stationary. What am I doing wrong?
enter image description here
Thanks in advance.
You must perform time series decomposition into data, seasonal, trend and residuals:
library('forecast')
library('tseries')
data$moving_average=ma(data$original, order=7)
moving_average = ts(na.omit(data$moving_average), frequency=30)
decomposition = stl(moving_average, s.window="periodic")
stationary <- seasadj(decomposition)
plot(decomposition)
You will get:
I have a question about this time series analysis, with mean monthly air temperature (Deg. F) Nottingham Castle 1920-1939:
https://datamarket.com/data/set/22li/mean-monthly-air-temperature-deg-f-nottingham-castle-1920-1939#!ds=22li&display=line
When I ran
auto.arima(x.t,trace=True)
it gave me "ARIMA(5,0,1) with non-zero mean" and "AIC=1198.42" as the lowest AIC. However, when I manually input the arima model, I came across a model with even lower aic.
arima(x = x.t, order = c(3, 1, 3))
aic = 1136.95.
When I run the function auto.arima(x.t,trace = TRUE,d=1), It gave me ARIMA(2,1,2) with AIC of 1221.413. While ARIMA(3,1,3) with drift gives 1209.947 and ARIMA(3,1,3) gives 1207.859.
I am really confused. I thought auto.arima should automatically suggest you the number of differencing. Why is auto.arima AIC different than the arima AIC while they have the same model?
You're fitting two different ARIMA models. Obviously an ARIMA(5,0,1) model is not the same as an ARIMA(3,1,3) model. In the former, you model p=5 time lags with no differencing, whereas in the latter you consider p=3 time lags with d=1 degree of differencing. Additionally, your model's MA components are also different: q=1 vs. q=3.
Different models will obviously give you different quality metrics (i.e. different AICs).
My goal: I want to understand a time series, a strongly auto-regressive one (ACF and PACF output told me that) and make a forecast.
So what I did was I first transformed my data into a ts, then decomposed the time series, checked its stationarity (the series wasn't stationary). Then I conducted a log transformation and found an Arima model that fits the data best - I checked the accuracy with accuracy(x) - I selected the model with the accuracy output closest to 0.
Was this the correct procedure? I'm new to statistics and R and would appreciate some criticism if that wasn't correct.
When building the Arima model I used the following code:
ARIMA <- Arima(log(mydata2), order=c(2,1,2), list(order=c(0,1,1), period=12))
The result I received was a log function and the data from the past (the data I used to build the model) wasn't displayed in the diagram. So then to transform the log into the original scale I used the following code:
ARIMA_FORECAST <- forecast(ARIMA, h=24, lambda=0)
Is that correct? I found it somewhere on the web and don't really understand it.
Now my main question: How can I plot the original data and the ARIMA_FORECAST in one diagram? I mean displaying it the way the forecasts are displayed if no log transformation is undertaken - the forecast should be displayed as the extension of the data from the past, confidence intervals should be there too.
The simplest approach is to set the Box-Cox transformation parameter $\lambda=0$ within the modelling function, rather than take explicit logarithms (see https://otexts.org/fpp2/transformations.html). Then the transformation will be automatically reversed when the forecasts are produced. This is simpler than the approach described by #markus. For example:
library(forecast)
# estimate an ARIMA model to log data
ARIMA <- auto.arima(AirPassengers, lambda=0)
# make a forecast
ARIMA_forecast <- forecast(ARIMA)
# Plot forecasts and data
plot(ARIMA_forecast)
Or if you prefer ggplot graphics:
library(ggplot2)
autoplot(ARIMA_forecast)
The package forecast provides the functions autolayer and geom_forecast that might help you to draw the desired plot. Here is an example using the AirPassengers data. I use the function auto.arima to estimate the model.
library(ggplot2)
library(forecast)
# log-transform data
dat <- log(AirPassengers)
# estimate an ARIMA model
ARIMA <- auto.arima(dat)
# make a forecast
ARIMA_forecast <- forecast(ARIMA, h = 24, lambda = 0)
Since your data is of class ts you can use the autoplot function from ggplot2 to plot your original data and add the forecast with the autolayer function from forecast.
autoplot(AirPassengers) + forecast::autolayer(ARIMA_forecast)
The result is shown below.
I'm trying to use an ARIMA model in R to forecast data. A slice of my time series looks like this:
This is just a slice of time for you get a sense of it. I have daily data from 2010 to 2015.
I want to forecast this into the future. I'm using the forecast library, and my code looks like this:
dt = msts(data$val, seasonal.periods=c(7, 30))
fit = auto.arima(dt)
plot(forecast(fit, 300))
This results in:
This model isn't good or interesting. My seasonal.periods were defined by me because I expect to see weekly and monthly seasonality, but the result looks the same with no seasonal periods defined.
Am I missing something? Very quickly the forecast predictions change very, very little from point to point.
Edit:
To further show what I'm talking about, here's a concrete example. Let's say I have the following fake dataset:
x = 1:500
y = 0.5*c(NA, head(x, -1)) - 0.4*c(NA, NA, head(x, -2)) + rnorm(500, 0, 5)
This is an AR(2) model with coefficients 0.5 and 0.4. Plotting this time series yields:
So I create an ARIMA model of this and plot the forecast results:
plot(forecast(auto.arima(y), 300))
And the results are:
Why can't the ARIMA function learn this obvious model? I don't get any better results if I use the arima function and force it to try an AR(2) model.
auto.arima does not handle multiple seasonal periods. Use tbats for that.
dt = msts(data$val, seasonal.periods=c(7, 30))
fit = tbats(dt)
plot(forecast(fit, 300))
auto.arima will just use the largest seasonal period and try to do the best it can with that.
Hello I use the package forecast in order to do times-series prevision. I would like to know how to un-log a series on the final forecast plot. With the forecast package I don't know how to un-log my series. Here is an example:
library(forecast)
data <- AirPassengers
data <- log(data) #with this AirPassengers data not nessesary to LOG but with my private data it is...because of some high picks...
ARIMA <- arima(data, order = c(1, 0, 1), list(order = c(12,0, 12), period = 1)) #Just a fake ARIMA in this case...
plot(forecast(ARIMA, h=24)) #but my question is how to get a forecast plot according to the none log AirPassenger data
So the image is logged. I want to have the same ARIMA modell but witht the none loged data.
It is not necessary to use the hack proposed by #ndoogan. forecast.Arima has built-in facilities for undoing transformations. The following code will do what is required:
fc <- forecast(ARIMA, h=24, lambda=0)
Better still, build the transformation into the model itself:
ARIMA <- Arima(data, order=c(1,0,1), list(order=c(1,0,1),period=12)), lambda=0)
fc <- forecast(ARIMA, h=24)
Note that you need to use the Arima function from the forecast package to do this, not the arima function from the stats package.
#Hemmo is correct that this back-transformation will not give the mean of the forecast distribution, and so is not the optimal MSE forecast. However, it will give the median of the forecast distribution, and so will give the optimal MAE forecast.
Finally, the fake model used by #Swiss12000 makes little sense as the seasonal part has frequency 1, and so is confounded with the non-seasonal part. I think you probably meant the model I've used in the code above.
The problem with #ndoogan's answer is that logarithm is not a linear transformation. Which means that E[exp(y)] != exp(E[y]). Jensen's inequality gives actually that E[exp(y)] >= exp(E[y]). Here's a simple demonstration:
set.seed(1)
x<-rnorm(1000)
mean(exp(x))
[1] 1.685356
exp(mean(x))
[1] 0.9884194
Here's a case concerning the prediction:
# Simulate AR(1) process
set.seed(1)
y<-10+arima.sim(model=list(ar=0.9),n=100)
# Fit on logarithmic scale
fit<-arima(log(y),c(1,0,0))
#Simulate one step ahead
set.seed(123)
y_101_log <- fit$coef[2]*(1-fit$coef[1]) +
fit$coef[1]*log(y[100]) + rnorm(n=1000,sd=sqrt(fit$sigma2))
y_101<-exp(y_101_log) #transform to natural scale
exp(mean(y_101_log)) # This is exp(E(log(y_101)))
[1] 5.86717 # Same as exp(predict(fit,n.ahead=1)$pred)
# differs bit because simulation
mean(y_101) # This is E(exp(log(y_101)))=E(y_101)
[1] 5.904633
# 95% Prediction intervals:
#Naive way:
pred<-predict(fit,n.ahead=1)
c(exp(pred$pred-1.96*pred$se),exp(pred$pred+1.96*pred$se))
pred$pred pred$pred
4.762880 7.268523
# Correct ones:
quantile(y_101,probs=c(0.025,0.975))
2.5% 97.5%
4.772363 7.329826
This also provides a solution to your problem in general sense:
Fit your model
Simulate multiple samples from that model (for example one step ahead predictions as above)
For each simulated sample, make the inverse transformation to get the values in original scale
From these simulated samples you can compute the expected value as a ordinary mean, or if you need confidence intervals, compute empirical quantiles.
This is a bit of a hack, but it seems to do what you want. Based on your fitted model ARIMA:
fc<-forecast(ARIMA,h=24)
fc$mean<-exp(fc$mean)
fc$upper<-exp(fc$upper)
fc$lower<-exp(fc$lower)
fc$x<-exp(fc$x)
Now plot it
plot(fc)