Solved – perform chow test on time series

chow-testlinear modelstructural-change

I am currently writing on my master thesis, therefore i have to find out if a specific moment had an impact on a time series.

By just looking at the plot I would say that after the prize reached his maximum at observation 831, there was a shift in the trend… in other words i would say there is a structual break. I wanted to confirm my assumption by performing a chow test

on the first look i was very happy with the result.

p-value < 2.2e-16

which means that there is a structual break at observation 832

But then i replaced 832 with several different observation numbers and i noticed that the p-value declines if I run the chow test with lower observation numbers.

when i run the test on observation 1 there was a p-value of 0,23

pvalue at observation 200 = 0,19

pvalue at overvation 300 was 0,01

I can perform sctest on every single observation after 300 and it will always show a p-value < 0,01.

what am i doing wrong? is my formula Prize~Trend correct? (I am not that used to R or time series analysis in general)

I am very glad and grateful for your help

Regards

Best Answer

(1) The Chow test is for a change in the coefficients of a regression model at a known time. If you don't know at which point in time the (hypothesized) structural change occurs, don't use a Chow test. A natural alternative is to use Andrew's sup$F$ test which formalizes your approach (conducting the Chow test for all possible timings) but appropriately adjusts the corresponding $p$-values. It rejects if the maximum of the $F$ (or Chow) statistics becomes to large. See vignette("strucchange-intro", package = "strucchange") for worked examples and more references. Also, citation("strucchange") gives you more pointers.

(2) The model Prize ~ Trend is surely not appropriate for your data. This would suggest that it is stationary around a deterministic linear trend. Even if you allow for structural breaks and relax it to a piecewise linear trend, you won't find a good model for the Prize time series. Probably it would make more sense to model the returns rather than the levels of this time series. But more context would be required for a better recommendation. I suggest you talk to your advisor and ask for more guidance and suitable references.

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Solved – How to identify structural change using a Chow test on Eviews

I am assuming that you are treating each country separately, and are attempting to determine if there is a break-point in the level of a series. Here are three (EDIT: four) main points that I hope will help:

The Chow test assumes that there is a known break-point in the series. If this point is not know, the Chow test is not appropriate (there are alternatives, although inference will be difficult in such a small sample).
The degrees of freedom in the F-test will be the same for each test of break-point. That is, it will always be F(2,47). The F-statistic calculated (7.438332 in your example) should be different at each tested point. However, given that you have a relatively small sample, such a test may suggest that there is a structural break at every point in the series.
Have you considered alternatives to the full structural break? For example, including a dummy variable for 1991 that could pick up an exogenous shock (such as a policy implementation that impacted GDP growth only in that period, but the economy returned to trend after). Alternatively, you could consider a broken trend model, if you think that the trend growth in GDP has shifted but not the intercept.
EDIT: Following from another user's point (mpiktas) that GDP may have a unit root. You should probably be looking at GDP as a natural logarithm (as we often see GDP moving with an exponential trend, due to the nature of population growth, etc.). Inference from a trend model on the log of GDP should be fine (log-GDP is probably trend-stationary - although you should do some testing - which implies that once accounting for the trend the residual series is stationary).

From your example: $$ y_t = \beta_0 + \beta_1 t + \epsilon_t \qquad (1)$$ The basic form of the Chow test is:

Construct a dummy variable $D_t$ that is $=0$ before the break and $=1$ after the break.
Run a regression: $$ y_t = \beta_0 + \beta_1 t + \gamma_0 D_t + \gamma_1 t D_t + \nu_t \qquad (2) $$
Test the sum of squared residuals from (1) against (2) where: $$ H_0 : \gamma_0 = \gamma_1 = 0 $$ $$ H_1 \text{: At least one coefficient not equal to zero} $$ And, $ F = \frac{SSR_{(1)} - SSR_{(2)}}{SSR_{(1)}} \frac{N-k}{q} $ Where $q$ is the number of restrictions (the number of equals signs in the null hypothesis $H_0$ above, and $k$ is the number of parameters in the restricted model (after applying the null hypothesis, so just $\beta_0$ and $\beta_1$).

Hope this helps.

Time-Series – How to Detect and Quantify a Structural Break in Time-Series (R): Comprehensive Guide

Questions

Q0: The time series looks rather right-skewed and the level shift is accompanied by a scale shift. Hence, I would analyze the time series in logs rather than levels, i.e., with multiplicative rather than additive errors. In logs, it seems that an AR(1) model works quite well in each segment. See e.g. acf() and pacf() before and after the break.

pacf(log(window(myts1, end = c(2018, 136))))
pacf(log(window(myts1, start = c(2018, 137))))

Q1: For a time series without breaks in the mean, you can simply use the squared (or absolute) residuals and run a test for level shifts again. Alternatively, you can run tests and breakpoint estimation based on a maximum likelihood model where the error variance is another model parameter in addition to the regression coefficients. This is Zeileis et al. (2010, doi:10.1016/j.csda.2009.12.005). The corresponding score-based CUSUM tests are available in strucchange as well but the breakpoint estimation is in fxregime. Finally, in the absence of regressors when looking only for changes in mean and variance the changepoint R package also provides dedicated functions.

Having said that, it seems that a least-squares approach (treating the variance as a nuisance parameter) is sufficient for the time series you posted. See below.

Q2: Yes. I would simply fit separate models to each segment and analyze these "as usual" Bai & Perron (2003, Journal of Applied Econometrics) also argue that this is justified asymptotically due to the faster convergence of the breakpoint estimates (with rate $n$ rather than $\sqrt{n}$).

Q3: I'm not fully sure what you are looking for here. If you want to run the tests sequentially to monitor incoming data, then you should adopt a formal monitoring approach. This is also discussed in Zeileis et al. (2010).

Analysis code snippets:

Combine log series with its lags for subsequent regression.

d <- ts.intersect(y = log(myts1), y1 = lag(log(myts1), -1))

Testing with supF and score-based CUSUM tests:

fs <- Fstats(y ~ y1, data = d)
plot(fs)
lines(breakpoints(fs))

sc <- efp(y ~ y1, data = d, type = "Score-CUSUM")
plot(sc, functional = NULL)

This highlights that both intercept and autocorrelation coefficient change significantly at the time point visible in the original time series. There is also some fluctuation in the variance but this is not significant at 5% level.

A BIC-based dating also clearly finds this one breakpoint:

bp <- breakpoints(y ~ y1, data = d)
coef(bp)
##                       (Intercept)        y1
## 2016(123) - 2018(136)    3.926381 0.3858473
## 2018(137) - 2019(1)      3.778685 0.2845176

Clearly, the mean drops but also the autocorrelation slightly. The fitted model in logs is then:

plot(log(myts1), col = "lightgray", lwd = 2)
lines(fitted(bp))
lines(confint(bp))

Re-fitting the model to each segments can then be done via:

summary(lm(y ~ y1, data = window(d, end = c(2018, 136))))
## Call:
## lm(formula = y ~ y1, data = window(d, end = c(2018, 136)))
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -0.73569 -0.18457 -0.04354  0.12042  1.89052 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  3.92638    0.21656   18.13   <2e-16 ***
## y1           0.38585    0.03383   11.40   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.2999 on 742 degrees of freedom
## Multiple R-squared:  0.1491, Adjusted R-squared:  0.148 
## F-statistic: 130.1 on 1 and 742 DF,  p-value: < 2.2e-16

summary(lm(y ~ y1, data = window(d, start = c(2018, 137))))
## Call:
## lm(formula = y ~ y1, data = window(d, start = c(2018, 137)))
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -0.43663 -0.13953 -0.03408  0.09028  0.99777 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  3.61558    0.33468   10.80  < 2e-16 ***
## y1           0.31567    0.06327    4.99  1.2e-06 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.2195 on 227 degrees of freedom
## Multiple R-squared:  0.09883,    Adjusted R-squared:  0.09486 
## F-statistic:  24.9 on 1 and 227 DF,  p-value: 1.204e-06

Best Answer

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