The Breusch Pagan test for heteroscedasticity

This video explains the intuition and motivation behind the Breusch-Pagan test for heteroscedasticity. Check out ben-lambert.co... for course materials, and information regarding updates on each of the courses. Quite excitingly (for me at least), I am about to publish a whole series of new videos on Bayesian statistics on youtube. See here for information: ben-lambert.co... Accompanying this series, there will be a book: www.amazon.co....

[@paulschneider1267]

Impressed by your ability to say heteroscedasticity

[@11runv]

Thank you so much. I sit in class for two hours and don't understand these things and then watch these quick videos and it just clicks.

[@dennisgavrilenko]

Ben, you are an absolute legend. Thanks so much for these videos!!

[@SpartacanUsuals]

Hi, thanks for your comment. This is an interesting point. I think that you are correct that if the residuals squared were constant then this would mean the R Squared for the regression would be high - falsely leading one to conclude that there was heteroskedasticity along one of the independent variables. However, actually having constant residuals squared is itself indicative of other problems - namely serial correlation

[@SpartacanUsuals]

Constant residuals squared can only come about if the residuals are equal in magnitude. This can only happen if your model has serious serial correlation. Hence this highlights the importance of visually inspecting your model and residuals - since this form of serial correlation should be quite obvious

[@television80]

Thank you for the time making this video, Saludos!!!

[@ambroseezzat2703]

Thank you Dr. Lambert! Very clear explanation with great insights! Thanks again

[@SpartacanUsuals]

Practically, this is a situation which I haven't come across before, and I wouldn't worry about its effects too much. The intuition that a model with heteroskedasticity errors will see a large R squared on the auxiliary regression is good for the majority of situations. Let me know if you have any more questions and I will do my best to answer them. Best, Ben

[@BrunoCardelli]

Hi nice video!, one question... Why LM=N.R^2 ~ chi.squer??

[@dilshaniranawaka6043]

I read somewhere that BPG has a chi-squared distribution. If that so, can you conduct an F test for a Chi-square distribution? Also thank you for the videos..... quick demonstration on all the hot topics helps us ace our exams. Thanks

[@danx2932]

Thanks for the videos. They are very helpful. I have a question with regards to the difference between "homoscedastic error" and "zero conditional mean of error". "homoscedastic error" refers to var(u_i) does not change along with xi. "zero conditional mean of error" mean s E(u_i|x_i) = 0, meaning knowing x doesn't help predict u. If understanding above is correct, isnt Breusch-Pagan testing "zero conditional mean of error", instead of "homoscedastic error" ?

[@reddevil2744]

Hi, I am from Germany. I have a question to the LM Statistic at 8:30. You said that the null hypothesis Ho: is a x² (chi -square distiribuiton) with p degress of freedom. How do I know how much p is in this case? Is p equal to the number of variables of the auxiliary regression u²^ = delta(0) + delta(1)X1 + delta (2)X2 + delta(3)X3 In this case we have 3 independent variables that means p = 3?

[@thcarm]

it would be great if you enable caption on your videos. as a brazilian student, it is a bit hard to understante british accent.

[@JustMe-pt7bc]

homoscedasticity essentially means constant error variance. Does it mean that that variance is zero or can be any constant sigma squared?

[@dragonEX123456]

+CH D variance is not necessarily zero, just that variance is a constant

[@avavaviv1]

It cannot be zero - that brakes one of the Gauss-Markov assumptions

[@ahmadnomanalnoor5986]

in computing LM , do we take the R square, the adjusted R square or the multiple R square?

[@JJ-hq1eu]

THANK YOU SO MUCH FOR YOUR VIDEOS

[@Sarghilani]

Thank you very much for your great video,....I got what I didn't understand in the class.

[@sandrobednorz3065]

Is this method valid only for OLS regressions? Could you also use it for Kalman Filter residuals?

[@zk7309-k2l]

very well-made video, 谢谢你

[@ktaepan2331]

This is really helpful. Thank you!

[@invarietateconcordia3627]

such a great explanation!

[@Darieee]

Awesome video, thanks a lot ! You're using \ for divisions, why is that :) ??

[@VinceJordan3]

Thanks Ben, that makes a lot of sense. Very helpful!

[@VinceJordan3]

the LM-statistic is what confuses me. The R-sq might be high, but not necessarily because you can "explain your variation in the residuals with your auxillary variables". Say your residual^2 are perfectly constant, then you run this regression, the betas are all zero but the r-sq is 1.0 because the constant in the regression is so good at explaining the variance in the residuals. So, a high LM results from a regression where the residual^2's are perfectly constant (homoskedastic)?

[@shawnmartin9080]

nice video I still just always get confused when to reject the null

[@SpartacanUsuals]

Hi, thanks for your message. We reject the null only if our statistic is greater than the critical value for that particular distribution under H0. Hope that helps, Best, Ben

[@pranjalsupratim612]

What is Auxiliary Regression in the video ? Can someone clarify that part ? Thanks :)

[@StallionBear7947]

Thanks for your wonderful videos. They are succinct and get straight to the point. Regarding the LM test, it seems that it must be designed for a small number of data points (small N), correct? In the world of Big Data, N samples where N in is in the millions or billions will, when multiplied by R-squared value, surely exceed the Chi-Squared critical value, implying Heteroskedasticity is always present in large data sets. Surely that can't be correct?

[@NhatLinhNguyen82]

+Karl Gierach Remember the R-squared is of auxiliary model, where its value should be low if variation of any of x explain none of the variation of in u squared (homoscedastic) . N actually helps to combat the difference in sample size. In a big sample size R-squared SHOULD be more accurate, thus any R-squared should be penalized more and in a low sample size, the random sampling alone could produce a bigger R-squared, therefore lower N, makes this error less severe. I hope i explained the intuition.

[@final0915]

+Karl Gierach That's why there is a chi distribution table for you to look at

[@hebe-x4z]

You are sooooooo brilliant ！

[@dominicj7977]

Many statistical softwares like R and python gives LM statistic as Breusch Pagan test.

[@dominicj7977]

@@johnnyjonas564 Yea later I figured it out. It is all the same test but measuring different things

[@dominicj7977]

@@johnnyjonas564 Both give similar results. Higher the Fstatistic, higher will be the value of R2 and higher the LM statistic. Also for the F test, we have to worry about an additional parameter for the degree of freedom of the auxiliary model. But for LM, we only have to worry about one degree of freedom

[@nichananwanchai9910]

thx you a lot

[@Blkpll]

How is the White test different from the Breusch Pagan test?

[@jenniferale3333]

The white test accounts for quadratic and cubic explanatory variables

[@segundomz5685]

Could you prove the lm because it doesn't make sense. If you look closely the U^2 terms follow a chi square distribution with one degree of freedom. So the right hand side as well. I dont get why you need to square the explained values and dive them for the square of u^2. I mean you are squaring values that follow a chi square distrib. Besides, the f test follows the same doubt

[@rex-wn2td]

U r great

[@VinceJordan3]

btw, your videos are awesome too

[@reddevil2744]

If you want and you have time you can go to www.gutefrage.net/frage/oekonometrie-statistik here I have an eview about the results of an auxiliary regression about u²^ with four independet variables. u²^ = delta(0) + delta(1)FE + delta(2)ED + delta(3)EX + delta(4)EX² I don't know how good your German is. The question here is: "What conclusions do you receive from the results and how do you proceed?" It would be very nice from you if you could help me here. Kind regards from Germany