Hello,Chris. I have recently started to learn a linear regression in my third year at university. My professor is trying to mathematically explain many concepts you deliver in this video clip Thanks for your help, I can get an deep insight how the robust regression is working and effective in controlling outliers in a large data set. Thanks.
Something doesn't quite follow here; @18:50, you mention bounded influence regression as a possible countermeasure against points that are highly influential; so influential in fact that they pull the regression line towards themselves. However, we may run into a case where such a point has a very low residual (due to its pulling of the line), in that case, the point would not be a candidate for elimination by our least trimmed squares rule. Am I missing something here?
Ok now, after a few minutes of reflection I may have an answer to that: is it because when you say residuals you are talking about standardized residuals, such that a point such as the one I proposed above, would still have a high standardized residual value due to its high leverage?
Hello,Chris. I have recently started to learn a linear regression in my third year at university. My professor is trying to mathematically explain many concepts you deliver in this video clip Thanks for your help, I can get an deep insight how the robust regression is working and effective in controlling outliers in a large data set. Thanks.
Excelent presentation!
Congrats!
Thank you very much for this video, sir! It was very helpful and easy to understand!
Thank you for posting the video, the lecture is really illuminating!
Thank you for the explanation !
comment: least median squares is similar to the least median absolute value, X is a random variable we would have Median(X^2) = (Median(|X|))^2.
Thanks
Isn't the absolute value function continuous but not differentiable at the origin? The video @3:06 says that its has a discontinuity at the origin.
You are absolutely right. I meant to say that the derivative has a discontinuity at the origin.
How can I decide K in Huber estimetor
K involves a trade-off between robustness and efficiency. Most people just use the value that Huber recommended, as given in the video.
Something doesn't quite follow here; @18:50, you mention bounded influence regression as a possible countermeasure against points that are highly influential; so influential in fact that they pull the regression line towards themselves. However, we may run into a case where such a point has a very low residual (due to its pulling of the line), in that case, the point would not be a candidate for elimination by our least trimmed squares rule. Am I missing something here?
Ok now, after a few minutes of reflection I may have an answer to that: is it because when you say residuals you are talking about standardized residuals, such that a point such as the one I proposed above, would still have a high standardized residual value due to its high leverage?
If one point is far different from the others, then it can't have a very low residual so long as there are a sufficient number of data points.
Thank you
please take a real data and show as how to evaluate the given data by using robust method