No, the data are not heteroscedastic (by way of how you simulated them). Did you notice the 0 degrees of freedom of the test? That is a hint that something is going wrong here. The B-P test takes the squared residuals from the model and tests whether the predictors in the model (or any other predictors you specify) can account for substantial amounts of variability in these values. Since you only have the intercept in the model, it cannot account for any variability by definition.
Take a look at: http://en.wikipedia.org/wiki/Breusch-Pagan_test
Also, make sure you read help(bptest)
. That should help to clarify things.
One thing that is going wrong here is that the bptest()
function apparently does not test for this errant case and happens to throw out a tiny p-value. In fact, if you look carefully at the code underlying the bptest()
function, essentially this is happening:
format.pval(pchisq(0,0), digits=4)
which gives "< 2.2e-16"
. So, pchisq(0,0)
returns 0
and that is turned into "< 2.2e-16"
by format.pval()
. In a way, that is all correct, but it would probably help to test for zero dfs in bptest()
to avoid this sort of confusion.
EDIT
There is still lots of confusion concerning this question. Maybe it helps to really show what the B-P test actually does. Here is an example. First, let's simulate some data that are homoscedastic. Then we fit a regression model with two predictors. And then we carry out the B-P test with the bptest()
function.
library(lmtest)
n <- 100
x1i <- rnorm(n)
x2i <- rnorm(n)
yi <- rnorm(n)
mod <- lm(yi ~ x1i + x2i)
bptest(mod)
So, what is really happening? First, take the squared residuals based on the regression model. Then take $n \times R^2$ when regressing these squared residuals on the predictors that were included in the original model (note that the bptest()
function uses the same predictors as in the original model, but one can also use other predictors here if one suspects that the heteroscedasticity is a function of other variables). That is the test statistic for the B-P test. Under the null hypothesis of homoscedasticity, this test statistic follows a chi-square distribution with degrees of freedom equal to the number of predictors used in the test (not counting the intercept). So, let's see if we can get the same results:
e2 <- resid(mod)^2
bp <- summary(lm(e2 ~ x1i + x2i))$r.squared * n
bp
pchisq(bp, df=2, lower.tail=FALSE)
Yep, that works. By chance, the test above may turn out to be significant (which is a Type I error since the data simulated are homoscedastic), but in most cases it will be non-significant.
The Breusch-Pagan test regresses the square of the residual from your regression on your predictors. The square of the residual cannot have mean 0 and thus it requires an intercept in this regression in order to achieve unbiased results.
When your predictors don't include an intercept, an intercept is not included in the canned version of the BP test that you are running. As a result, you get biased coefficients and biased test statistics.
If you want to force your model to have a 0 intercept (generally not a good idea, but...), you can do a BP test by hand by getting the residuals from your model, squaring them, and regressing them on an intercept term and your predictors. You can do a Wald ($\chi^2$) test of the significance of the coefficients on your predictors to get the official BP test statistic, but the reported F
statistic and its p
-value that come standard in the output of the residual-squared ("auxiliary") regression are close approximations to the official/standard versions.
Best Answer
The problem isn't heteroskedasticity, that's why it's passing the test. The problem is that your model doesn't work well for (at least some of) your observations.
I've never seen anyone analyze stock prices without looking at their differences. Try a Dickey-Fuller test for a unit root---I bet that you can't reject that there is one, as @mpiktas alludes to in his comment.
If there isn't a unit root, perhaps there is a time trend or seasonality. You might try including a linear time trend or seasonal components.
Alternatively, you might try working with the log of the prices, which sometimes helps the fit.