I'm not adding anything new to what Jochen and Robin have said, but I'll try stating it in another way in case that helps clarify anything

R2 measures the proportion of variance in a dataset that is described by a model.

To understand what this is needs an explanation of variance. In the context of calculating R2, variance is the deviation of the observations from the mean. Subtracting a uniform number (mean or otherwise) from data has no effect on the pattern of deviation about the mean - it just shifts the mean.

Since your manipulation of the data had no impact on the pattern of deviation from the mean, this means the deviations remaining after the manipulation are identical to those before and divide out to 1. Since you have made no difference to the variance you get an R2 of 0.

'This represents a poor fit, when it is not'  Subtracting a uniform value from a dataset is a poor (to be precise, zero) fit of variance. Is it some other aspect of the dataset you wish to fit? If you can advise us as to what information you are trying to get from your procedure we may be able to recommend a more appropriate statistical test.

Hi,

What you observed is not a poor fit, it's just that fitting a model with intercept only prevent you from calculating a r-square, as the total sum of squares will be the same as the residual sum of squares (http://en.wikipedia.org/wiki/Coefficient_of_determination). So, trying to fit a linear model without intercept in a stationary or non-stationary cases is just like computing their mean and variance. If you want to compare them, there are plenty of tests for that, if you just want to describe them just use their mean and variance. 

So, if you want a goodness of fit, you need to fit a linear model of degree 1 (if it's a time series just do it over time or even on a index of 1 to the length of your series if it's regular) and you will get your R2, which shouldn't be 0 unless there is no trend in your data.

What do you want?

I think you try to use R² for something what R² is not meant for...

R² is the faction of explained variance ("explained" by the model by reducing the SSE [sum of squared errors] = your "sum(residuals squared)").

The residuals are always defined to some model. No model, no residuals.

The simplest possible model is one without any predictor. However, this simples possible model, by definition, has an intercept term what will model the expected value of the data. (the maximum likelihood solution for the expected value is the sample mean), and thus the errors are the differences to the expected value, and the expected sum of the squared errors is ... yes: the variance.

Hence, the SSE of a "no-predictor-model" is estimated as n*s²(x). The SSE of a model with (effective) predictors will be smaller than this and is sum(res²).

R² = (n*s²(x) - sum(res²)) / n*s²(x) = 1 - sum(res²)/n*s²(x)

R² =1 - average(res²)/s²(x)

The average squared residuals is the variance of the data in the "no-predictor model". This is how "variance" is defined. So, by definition, the "no-predictor model" does not reduce the variance relative to the "no-predctor model". Souds stupid, but so is the taks to "compare 'a linear model of order 0' with the 'no-predictor model' (i.e. with itself)" to see how this would reduce the SSE.

You can also argue via the deviance, because the deviance is the SSE.

I'm not adding anything new to what Jochen and Robin have said, but I'll try stating it in another way in case that helps clarify anything

R2 measures the proportion of variance in a dataset that is described by a model.

To understand what this is needs an explanation of variance. In the context of calculating R2, variance is the deviation of the observations from the mean. Subtracting a uniform number (mean or otherwise) from data has no effect on the pattern of deviation about the mean - it just shifts the mean.

Since your manipulation of the data had no impact on the pattern of deviation from the mean, this means the deviations remaining after the manipulation are identical to those before and divide out to 1. Since you have made no difference to the variance you get an R2 of 0.

'This represents a poor fit, when it is not'  Subtracting a uniform value from a dataset is a poor (to be precise, zero) fit of variance. Is it some other aspect of the dataset you wish to fit? If you can advise us as to what information you are trying to get from your procedure we may be able to recommend a more appropriate statistical test.

The data must be stationary before ARIMA fitting, by first testing it for unit root or visualizing stationarity. If unit root cannot be rejected, you have to transform the time series through first difference, log, square, log difference, and seasonal difference. After model specification, you move on to estimation with a proper comprehensive diagnostic checking on residuals for normality, autocorrelation, and heteroskedasticity (correctable by G-ARCH or its varients). Finally, you perform forecasting.

Dear all,

Thanks for your comments.

I am trying to check if a straight line can fit the time series - if the time series has a constant mean, or there is a trend, or any pattern for that matter. If there is a trend, then R2 is poor. However, specifically, if there is an oscillatory behaviour, then the straight line gives a good fit, but clearly this is not a good fit. This is where I get R2=0. Basically, I am using the wrong test, I believe, since there is no model that I am fitting, except for intercept. So, I need a test to check for the latter case.

I hope that clarifies the objective.

Laya, there is no (statistical) test for that.

A statistical test is a procedure that tells you how likely you can expect the observed (or more "extreme") data given a particular model.

But you seem to want develop an idea what a reasonable model may be. This requires a good understanding of the science behind the data, and it requires well-defined aims. The most helpful tool is to plot the data and let your eye decide if there is a pattern. The supposedly recognized pattern should then be aligned with the science behind the data. If this all seems to be reasonable, this may be then taken as an hypothesis about the relationship (expressed as some model).

In the next step you may identify the relevant parameter(s) of the model [again given the scientific context] and specify a hypothesis about these parameters. Then, finally, you may collect new data and formally test this hypothesis using a statistical hypothesis test.

You should follow the Box-Jenkins approach. A unit root test can solve your problem not to mention more robust, reliable, and powerful. If the time series is non stationary with a unit root, it will often have a seasonal trend or be a noisy random walk. It is inappropriate to use r-squared fitness test from regression analysis (what's your regressor?) to check for stationarity. 

To build on Jochen's excellent answer and tie it back to R2. This is how R2 works, but the steps are much more intuitive than they appear when typed in black and white and most modelling and curve fitting software does all of these steps automatically for you. 

1) you must determine a reasonable hypothesis (or a limited set of reasonable hypotheses) for what mathematical pattern you data follows. From your additional explanation it sounds like you are expecting a cyclical pattern. Do you understand the underlying science enough to judge what kind of oscillating equation would fit, e.g. sine/cosine? Is it suitable for fast fourier transform? What about a wavelet decomposition?

2) use your experimental data to fit the parameters of that mathematical pattern - this gives you a model

3) you calculate the values that this model predicts for each of your observations

4) you determine how much these predictions deviate from the true value (this is the standard error of the estimate, SEE)

5) you determine how much this deviation is relative to the deviation from the mean (the sum of squared error, SSE) and subtract this from 1. This tells you how much of the deviation in your observations is not captured by your model

The danger when comparing models with different degrees of complexity (degrees of freedom) is that more complex models will always fit data better (even noise) so if you are comparing models with different levels of complexity you should use adjusted R2, which corrects for the degree of complexity.

Ideally once you have fitted a model you should, as Jochen recommends, collect new data to validate that the model transfers to new data. This step is critical for ensuring that you model is not over-fitted and capturing variation due to noise.

Hi Beattie,

The original poster (OP) has a time-series and he or she wants to check for a trend essentially if the data is stationary or not. Using r-squared for checking on stationarity would be inappropriate. That's the issue here not the way of how to interpret r-squared. In fact, r-squared is used to check for model prediction power. And for a non-stationary time series, r-squared is good to be low because most have already been explained by the built-in residual pattern. Standard error in my opinion is much better way to test for forecasting accuracy. I 100% support Dr. Robert F. Nau's position on r-squared topic: http://people.duke.edu/~rnau/rsquared.htm

Richard

I am joining this thread a bit late, and I have looked at the question as posed by Laya:  :I am trying to check if a straight line can fit the time series - if the time series has a constant mean, or there is a trend, or any pattern for that matter. If there is a trend, then R2 is poor. However, specifically, if there is an oscillatory behaviour, then the straight line gives a good fit, but clearly this is not a good fit. This is where I get R2=0. Basically, I am using the wrong test, I believe, since there is no model that I am fitting, except for intercept. So, I need a test to check for the latter case."

I am not sure I understand the problem Laya is trying to solve here, and I am assuming R-square above is referring to the square of Cor(Yt, t), but a time series does not have to have a linear trend, it could be a more complicated function of t (e.g., cubic polynomial in t) in which case looking at r-squared won't work. You have two approaches: (i) use a sctochastic trend model, as suggested by Juehui and first or second-order differences to detrend the data, or (ii) plot Yt vs t, then decide on the polynomial trend, fit a regression model to Yt as a function of this trend function + any other predictors, look at auto-correlation in residuals from the model, then fit an ARIMA model with trend and predictors. 

R2 = 0 is worrying you?  Why?  It should be negative as well (see http://en.wikipedia.org/wiki/Coefficient_of_determination for more explanation).  If your goal is to decide whether the time series is best modeled as noisy constant value then the proper tool seems to be the ordinary chi squared test.  Of course, you should make chi squared as small as possible first by adjusting the presumed constant level. R2 was designed for other purposes.  Well, you may have troubles to find the limiting values for chi squared test when the number of your observations exceeds 30.  But then you can calculate the chi squared divided by number of degrees of freedom (= number of data points -1 in your case).  The value between 1/2 and 2 means your hypothesis is not bad, the data are most likely indeed constant up to the noise level.  Contrary, the values of (normalized) chi squared exceeding 2 indicate that this model is most likely incorrect.  Too small values are also suspected: most likely your estimate of individual data variances are oversized.

Linear Regression is useful when the underlying model is (approximately) linear. When the underlying model is oscillating, it is clearly not linear, and therefore linear models are not useful. R2=0 implies that the linear model is not better than the model using the mean, namely a confirmation that indeed it is not appropriate. You need another approach.

As I see it, there are two options:

1. Use a linear model but with kernel methods, whereby each variable is transformed using a kernel, resulting in non-oscillating data is somewhat linear; you should be able to see such 'linearity' visually on a chart. Your kernel could be, for example, sin(x_i) or cos(x_i) or a polynomial combination of various trigonometry functions.

2. Use other models, e.g. non-linear-regression against a series of trigonometry functions; this is would be similar to a least square fit for a Fourier series.

In either case, you need to justify why a specific model type is indeed a reasonable representation of the data.