This is pure gold. Thank you for not skipping any steps or combining too many steps! There are a ton of videos and documents online describing how to derive and solve these equations, and the vast majority skip steps and then I'm lost. In order to understand every detail, I need to understand every step. I do, now! Thank you!
Outstanding explanation! I just had 2 homework problems that used this "trick" of equating the sum of Xi with the sum of Xbar - I would have been lost without this video! Thanks
@@BurkeyAcademy you showed how to get from the second to the first and I was able to follow. Taking the equality for granted, I can work backwards. But how do I go from the first to the second? I.e. from (Yi - ȳ)Xi to (Yi - ȳ)(Xi - x̄)? The purpose of (Yi - ȳ)(Xi - x̄) is to see that it's the equation of the co-variance but without knowing (Yi - ȳ)(Xi - x̄) = (Yi - ȳ)Xi, how do I find (Yi - ȳ)(Xi - x̄) from (Yi - ȳ)Xi? Am I making sense or is the answer in front of my eyes?
@@jas-hy3sy Like a lot of steps in derivations/proofs, it is hard to see how someone originally figured out that exact step-- personally I just have to convince myself that the step is true. So, as I tried to explain in the video: You have to just see that when you multiply both of these terms out: One: sum(yixi +ȳx̄-Yix̄-ȳXi) and Two: sum(yixi--ȳXi), that these are equal because the +ȳx̄ and -ȳXi cancel. Hopefully that helps- but if not, I can try again!
@@BurkeyAcademy Yes that part is clear. Thank you. And yes, the manipulation of the original to get to the second seems complex. Also, at 15:54 when you went back to add the summation signs, you changed the + sum(B1x̄Xi) to - sum(B1x̄Xi), but you ended up with the correct final answer.
Ok just did the calculation. The - sum(B1x̄Xi), should be + sum(B1x̄Xi), and before factoring the two terms, bring them to the right side, and then factor, to get what you got.
Hey, thanks, I spent a lot of time looking for an explanation for this (19:07), not even the most recognized econometrics books explain it (they take it for granted) =D
Can you watch 15:30-16:10 again please? I believe you may have accidentally switched the + E(B1*x-bar*Xi) to - E(B1*x-bar*Xi) when you are putting the summation signs (which I have represented with "E") into the equation. Would this change anything that follows?
Thanks for the catch. I added an annotation. The "+" sign is correct, and my writing the "-" down wasn't carried through to the next line. If you multiply out the collected term in the line below, you will get a plus sign back out of it.
At 17:41 I think you may have got them the wrong way around (shouldn't it be *sum of beta 1 multiplied by x bar minus xi (in brackets) multiplied by xi*? multiplying out gives the *sum of xixi minus the sum of x bar xi*...whereas in the previous expression it's the *sum of x bar xi minus the sum of xixi*? Or does this order not matter? Thanks - videos are v.helpful btw.
Thanks for the nice comment. As to the question: Since that term has a leading minus sign in front of it: -S[B1(xi-xb)xi], when you multiply it out you get -SB1xixi - - SB1xbxi. So the two negatives cancel. Does that help answer your question? [It is hard to discuss this using text and timestamps, isn't it? ☺ ]
It is really the same thing, except it gets very tedious unless you do it with matrix algebra. You could add a B2*X2i, take three derivatives, and solve three equations for three unknowns. Using Matrix algebra instead though, you find the solution is B-hat=(X'X)^(-1)(X'Y).
As I noted in the video description: Note: At 16:05 when I add the Sigma, I accidentally write down a "-" in front of it. This should be a +, but this error does not carry through to the rest of the video.
Yes, thanks for the catch- sorry for that! I have a note in the video description alerting people to this mistake- but aI know this is not a perfect fix. I hope one day my health and energy level will get better, so I don't make so many silly mistakes! Again. I thank you very much for catching my mistake. I'll go back and edit the original video and upload a better version.
Hi BurkeyAcademy! There were two B (Betas). How did you resolve it to One? Please explain... That's the only 1 thing remaining for me to get it fully. Thanks in Advance.
Simple case: a*b-a*c= a*(b-c) [multiply it back out to see that this is the same]. More complicated: If a*b*c*d-a*b*c*c = a*b*c*(d-c). Similarly, here I take sum(Bi*xi*xbar)-sum(Bi*xi*xi)=sum(Bi*xi*(xi-xbar), which I write as sum(Bi*(xi-xbar)*xi. Does that help?
+alejandro cuartas 1) There will not be a finite maximum, 2) Yes you could prove it, by checking the second order conditions, e.g. see www.sjsu.edu/faculty/watkins/2ndOrder.htm
+BurkeyAcademy Thank you very much, getting into lots of trouble when I start using second order condition to prove, will keep on trying with D-test.... Warm regards and thanks again, great vid!
I have assignments to complete. I want to get answers from you for that assignment. Can you help me and how can I send you that assignment to get answers☺️
Thanks for the great video. I know this is probably something rather insignificant but I'd like to understand the logic rather than memorize this step. When you do the chain rule and take both derivatives with respect to Bo and B1 why are the outsides of each equation multiplied by -1 and -Xi? I completely understand how the chain rule works but Im just not understanding how the derivative of the interior is -1 and -Xi... Please help quickly.. Thanks in advance!
Just simple chain rule... f(g(x))-- f(•) is squaring. So, derivative of that multiplies by the 2 and subtracts one from the exponent. Then take the derivative of the inside and multiply by it: The -B1xi in the first case and the -Bo in the second case. Does that help?
+BurkeyAcademy I have the same concern. As for -B1xi, I am approaching the problem as if I must use the product rule, namely (-B1)'(xi)+(-B1)(xi)'? I am a bit confused. At 5:45, I am not seeing how the last part of the chain for (-B1xi)' is simply -xi. Any help is appreciated.
Roughly speaking, the chain rule says that the derivative of f(g(b)) w.r.t. b is df/dg * dg/db. Here the f(•) is the square function, and g(•) is [y-bo-b1x]. So, df/dg = 2[y-bo-b1x] and the derivative of [y-bo-b1x] w.r.t. b1 is (-x). Are you suggesting we multiply the squared term out? In that case, with the product rule, you'd have -x•[y-bo-b1x]+ [y-bo-b1x]•-x=2[y-bo-b1x][-x], the same thing... If it still isn't clear, please try to let me know how I can help clarify!
Thank you for the clarification. I realized that I just needed to review partial derivation again to understand where I went wrong. Slowly making progress into the econometrics universe, over here. My textbook is a bit short of proofs so I am appreciative of you for taking the time to make the video and respond. Thanks a bunch.
No problem- It is pretty hard to either make a book or a class that gives all of the intuition, mathematics, and applied know-how that students need. Books are either way too applied, or way too mathematical. Let me know if you need help tracking anything down.
![Seungmin "그렇게, 천천히, 우리(As we are)" | [Stray Kids : SKZ-PLAYER]](http://i.ytimg.com/vi/kAzmhLHePqU/mqdefault.jpg)
This is pure gold. Thank you for not skipping any steps or combining too many steps! There are a ton of videos and documents online describing how to derive and solve these equations, and the vast majority skip steps and then I'm lost. In order to understand every detail, I need to understand every step. I do, now! Thank you!
I am so glad to have helped! I am the same way, and I just make the videos I wish I had when I was trying to learn this stuff! Good luck!
This is the best video explaining the derivation ever. Please teach my class
Thank you so much for this video, you derived the OLS and explained every step you made in a much better and understandable way than my lecturer!
Outstanding explanation! I just had 2 homework problems that used this "trick" of equating the sum of Xi with the sum of Xbar - I would have been lost without this video! Thanks
This is such a great explanation way better then my professor!!
THANK YOU!!!! IM ACTUALLY GONNA CRY CAUSE I GET IT
Of course you do! You can do this!
Really enjoyed your video. Learnt this today and I have exam tomorrow.
For the two star equations at 19:33 .. can you show how to get from the first one to the second one?
That's what I was doing from 19:33 - 24:00. What part wasn't clear?
@@BurkeyAcademy you showed how to get from the second to the first and I was able to follow. Taking the equality for granted, I can work backwards. But how do I go from the first to the second? I.e. from (Yi - ȳ)Xi to (Yi - ȳ)(Xi - x̄)? The purpose of (Yi - ȳ)(Xi - x̄) is to see that it's the equation of the co-variance but without knowing (Yi - ȳ)(Xi - x̄) = (Yi - ȳ)Xi, how do I find (Yi - ȳ)(Xi - x̄) from (Yi - ȳ)Xi? Am I making sense or is the answer in front of my eyes?
@@jas-hy3sy Like a lot of steps in derivations/proofs, it is hard to see how someone originally figured out that exact step-- personally I just have to convince myself that the step is true. So, as I tried to explain in the video: You have to just see that when you multiply both of these terms out: One: sum(yixi +ȳx̄-Yix̄-ȳXi) and Two: sum(yixi--ȳXi), that these are equal because the +ȳx̄ and -ȳXi cancel. Hopefully that helps- but if not, I can try again!
@@BurkeyAcademy Yes that part is clear. Thank you. And yes, the manipulation of the original to get to the second seems complex. Also, at 15:54 when you went back to add the summation signs, you changed the + sum(B1x̄Xi) to - sum(B1x̄Xi), but you ended up with the correct final answer.
Ok just did the calculation. The - sum(B1x̄Xi), should be + sum(B1x̄Xi), and before factoring the two terms, bring them to the right side, and then factor, to get what you got.
Hey, thanks, I spent a lot of time looking for an explanation for this (19:07), not even the most recognized econometrics books explain it (they take it for granted) =D
Your videos are extremely helpful.Thank you so much
Can you watch 15:30-16:10 again please? I believe you may have accidentally switched the + E(B1*x-bar*Xi) to - E(B1*x-bar*Xi) when you are putting the summation signs (which I have represented with "E") into the equation. Would this change anything that follows?
Thanks for the catch. I added an annotation. The "+" sign is correct, and my writing the "-" down wasn't carried through to the next line. If you multiply out the collected term in the line below, you will get a plus sign back out of it.
Well explained , perfectly helpful 🙏🏻
Thank you from Brazil!
At 17:41 I think you may have got them the wrong way around (shouldn't it be *sum of beta 1 multiplied by x bar minus xi (in brackets) multiplied by xi*? multiplying out gives the *sum of xixi minus the sum of x bar xi*...whereas in the previous expression it's the *sum of x bar xi minus the sum of xixi*? Or does this order not matter? Thanks - videos are v.helpful btw.
Thanks for the nice comment. As to the question: Since that term has a leading minus sign in front of it: -S[B1(xi-xb)xi], when you multiply it out you get -SB1xixi - - SB1xbxi. So the two negatives cancel. Does that help answer your question? [It is hard to discuss this using text and timestamps, isn't it? ☺ ]
BurkeyAcademy I think Ofir Hughes is right. However the mistake isn't evident because it just corrected for your +/- sign mistake earlier on.
perfectly explain! Thank you !
Hey! How would you derive the OLS estimates for multiple regression?
It is really the same thing, except it gets very tedious unless you do it with matrix algebra. You could add a B2*X2i, take three derivatives, and solve three equations for three unknowns. Using Matrix algebra instead though, you find the solution is B-hat=(X'X)^(-1)(X'Y).
Thank you so much i finally get it can't thank you enough
This was really helpful. Thank you.
can someone explain me how to get at min 17:26 sumatory of (xi - x_bar)? woudn't It be sum(x_i + x_bar)?
As I noted in the video description: Note: At 16:05 when I add the Sigma, I accidentally write down a "-" in front of it. This should be a +, but this error does not carry through to the rest of the video.
Amazing video thank you so much..
thanks for this video. it unloaded some of my burden
it's really helpful ,thanks a lot for sharing ,and may allah bless you
we explained. Thank you soo much
fantastic video
Thank you! 😃
thank you
at 16:08 while distributing the summation, you have changed the (+) to (-), shouldnt that be (+)
Yes, thanks for the catch- sorry for that! I have a note in the video description alerting people to this mistake- but aI know this is not a perfect fix. I hope one day my health and energy level will get better, so I don't make so many silly mistakes! Again. I thank you very much for catching my mistake. I'll go back and edit the original video and upload a better version.
Really helpful, thank you!
Well explained
Thank you
This has been helpful, Pls can you do a video explaining the derivation of OLS estimators in Multiple linear regression WITHOUT using matrix algebra
Hi BurkeyAcademy! There were two B (Betas). How did you resolve it to One? Please explain... That's the only 1 thing remaining for me to get it fully. Thanks in Advance.
Can you give me an approximate time (minutes & seconds) in the video where I did this? That way I can make sure I am focusing on the right part.
At 17:26
Simple case: a*b-a*c= a*(b-c) [multiply it back out to see that this is the same]. More complicated: If a*b*c*d-a*b*c*c = a*b*c*(d-c). Similarly, here I take sum(Bi*xi*xbar)-sum(Bi*xi*xi)=sum(Bi*xi*(xi-xbar), which I write as sum(Bi*(xi-xbar)*xi. Does that help?
how can we be sure that the results are the min and not the max, is it possible to prove?
+alejandro cuartas 1) There will not be a finite maximum, 2) Yes you could prove it, by checking the second order conditions, e.g. see www.sjsu.edu/faculty/watkins/2ndOrder.htm
+BurkeyAcademy Thank you very much, getting into lots of trouble when I start using second order condition to prove, will keep on trying with D-test.... Warm regards and thanks again, great vid!
I have assignments to complete. I want to get answers from you for that assignment. Can you help me and how can I send you that assignment to get answers☺️
Sure, I can help you cheat- for the low, low price of $100,000 US. ☺
this is great, thanks!!!
is this true if we know that the intercept is equal to 0?
Of course, but there is also an easier way in that case, which I have a video on.
this is some black magic
Thank you kind sir 🙏
Thanks alot
thank you so much!
Thanks for the great video.
I know this is probably something rather insignificant but I'd like to understand the logic rather than memorize this step.
When you do the chain rule and take both derivatives with respect to Bo and B1 why are the outsides of each equation multiplied by -1 and -Xi?
I completely understand how the chain rule works but Im just not understanding how the derivative of the interior is -1 and -Xi...
Please help quickly.. Thanks in advance!
Just simple chain rule... f(g(x))-- f(•) is squaring. So, derivative of that multiplies by the 2 and subtracts one from the exponent. Then take the derivative of the inside and multiply by it: The -B1xi in the first case and the -Bo in the second case. Does that help?
+BurkeyAcademy I have the same concern. As for -B1xi, I am approaching the problem as if I must use the product rule, namely (-B1)'(xi)+(-B1)(xi)'? I am a bit confused. At 5:45, I am not seeing how the last part of the chain for (-B1xi)' is simply -xi. Any help is appreciated.
Roughly speaking, the chain rule says that the derivative of f(g(b)) w.r.t. b is df/dg * dg/db. Here the f(•) is the square function, and g(•) is [y-bo-b1x]. So, df/dg = 2[y-bo-b1x] and the derivative of [y-bo-b1x] w.r.t. b1 is (-x). Are you suggesting we multiply the squared term out? In that case, with the product rule, you'd have -x•[y-bo-b1x]+ [y-bo-b1x]•-x=2[y-bo-b1x][-x], the same thing... If it still isn't clear, please try to let me know how I can help clarify!
Thank you for the clarification.
I realized that I just needed to review partial derivation again to understand where I went wrong.
Slowly making progress into the econometrics universe, over here. My textbook is a bit short of proofs so I am appreciative of you for taking the time to make the video and respond. Thanks a bunch.
No problem- It is pretty hard to either make a book or a class that gives all of the intuition, mathematics, and applied know-how that students need. Books are either way too applied, or way too mathematical. Let me know if you need help tracking anything down.
you are great
Does it bother anyone that he sort of sounds like Robert Downey Jr?
I like it😂
Hetvi Shah oh my god😂 so true
nicely explained.. but just throw in a little more speed without missing any intermediate steps.. :)
It's help full sir, but i don't see the PDF of OLS
How I get it Sir
Thank you for such kind of Explanation👌👍
As I said at As I said at 6:00, go to my website and click under FILES. Or, to make it easier, I included a link in the video description.
@@BurkeyAcademy Many thanks Sir
I get it 👍👍👍
Your voice sounds a lot like Robert Downey Jr.
One more vote for Iron Man... I think that makes it 20 for him, versus only 4 for Tom Hanks! ☺
10q it is easiest way!!
Horrible