quick question, looking at the equation visible on the screen at 8:05 minutes into the video. If you argue that the sum of xi-x(bar) is equal to 0 the denominator of both fractions would be the xi multiple of 0 which should be 0 as well. Essentially then the equation would be ß0*0/0+ß1*0/0. Is that legal maths?
@@jfrench Never mind, i figured it out.. simple arithmetic lol If sum of xi - xbar is equal to (x1 - xbar) + (x2 - xbar) + (x3 + xbar), and the sum of it is equal to zero, then for an example: (10-5) + (12-5) + (-7-5) = 5 + 7 -12 = 0 Then, sum of (xi - xbar) * xi, is equal to: ((10-5) * 10) + ((12-5) * 12) + ((-7-5) * (-7)) = 50 + 84 + 84 = 218
At 6:55 of the video, can you please clarify why (xi-xbar) is a constant? Unless you are referring to the sum of (xi-xbar), each individual (xi-xbar) varies with i. Edit: I eventually figured it out. Turns out each individual (xi-xbar) is not constant, and also you skipped a step at 7:01. You should have shown how exactly you distributed the summation at the numerator rather than jumping straight to bringing out B0. Although it may seem minor, these two factors led me to be stuck on the problem for about 2 hours. The good thing is I now have a greater understanding of the concept due to working on it myself after being unable to follow your explanation. With that being said, I am a big fan of your videos and appreciate the effort! Slowly going through all the steps (except for this one) is why I found your videos helpful in the first place. Thank you and have a great day :)
fran green Sure. Follow this on paper and it will make a lot more sense. First of all, pause at 6:55 and only look at Σ ((xi - xbar)(B0+B1xi)) and ignore 1/Σ((xi - xbar)xi). We can expand the brackets and make it Σ((xi - xbar)(B0) + (xi - xbar)(B1xi)). This is the same as Σ(xi - xbar)(B0) + Σ(xi - xbar)(B1xi). So what we have done is split the thing into 2 terms. we can divide each term by the denominator individually. Now what we find is that Σ((xi - xbar)(B0) is a constant, which is what the narrator meant at 6:55. And since the denominator is constant as well, a constant divided by a constant is still a constant. Hope that helps, let me know if there is anything I didn’t make clear 😊
Thanks for the feedback Shun. I've tried to clarify in the comments! Unfortunately, RUclips doesn't allow the creator to replace a video with a new one without creating a whole new video. But hopefully the comments provide enough clarification.
at 4:24 you arrived at 1/nƩ(β₀+β₁y)= n(β₀)/n +β₁Ʃy/n = β₀+ β₁ȳ however i was expecting 1/nƩ(β₀+β₁y)= (β₀)/n +β₁Ʃy/n = (β₀)/n +β₁ȳ. can you explain that step
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extremely simple, and lucid! Thanks
Amazing ❤
Thanks fr ur simplified version
Great help🙌
quick question, looking at the equation visible on the screen at 8:05 minutes into the video. If you argue that the sum of xi-x(bar) is equal to 0 the denominator of both fractions would be the xi multiple of 0 which should be 0 as well. Essentially then the equation would be ß0*0/0+ß1*0/0. Is that legal maths?
The sum of xi - xbar = 0. However, the sum of (xi - xbar) * xi is NOT equal to 0 (in general). That is why that doesn't result in 0/0.
How is this possible? Am I missing something?
If the [sum of xi - xbar] = 0
then [sum of xi - xbar ] * xi
= 0 * xi
= 0
@@jfrench Never mind, i figured it out.. simple arithmetic lol
If sum of xi - xbar is equal to (x1 - xbar) + (x2 - xbar) + (x3 + xbar), and the sum of it is equal to zero, then for an example:
(10-5) + (12-5) + (-7-5) = 5 + 7 -12 = 0
Then, sum of (xi - xbar) * xi, is equal to:
((10-5) * 10) + ((12-5) * 12) + ((-7-5) * (-7)) = 50 + 84 + 84 = 218
At 6:55 of the video, can you please clarify why (xi-xbar) is a constant? Unless you are referring to the sum of (xi-xbar), each individual (xi-xbar) varies with i.
Edit: I eventually figured it out. Turns out each individual (xi-xbar) is not constant, and also you skipped a step at 7:01. You should have shown how exactly you distributed the summation at the numerator rather than jumping straight to bringing out B0. Although it may seem minor, these two factors led me to be stuck on the problem for about 2 hours. The good thing is I now have a greater understanding of the concept due to working on it myself after being unable to follow your explanation.
With that being said, I am a big fan of your videos and appreciate the effort! Slowly going through all the steps (except for this one) is why I found your videos helpful in the first place. Thank you and have a great day :)
Can you explain to me why they are constant please?
fran green Sure. Follow this on paper and it will make a lot more sense. First of all, pause at 6:55 and only look at Σ ((xi - xbar)(B0+B1xi)) and ignore 1/Σ((xi - xbar)xi). We can expand the brackets and make it Σ((xi - xbar)(B0) + (xi - xbar)(B1xi)). This is the same as Σ(xi - xbar)(B0) + Σ(xi - xbar)(B1xi). So what we have done is split the thing into 2 terms. we can divide each term by the denominator individually. Now what we find is that Σ((xi - xbar)(B0) is a constant, which is what the narrator meant at 6:55. And since the denominator is constant as well, a constant divided by a constant is still a constant. Hope that helps, let me know if there is anything I didn’t make clear 😊
Shun Yat Cheung thank you!!
Thanks for the feedback Shun. I've tried to clarify in the comments! Unfortunately, RUclips doesn't allow the creator to replace a video with a new one without creating a whole new video. But hopefully the comments provide enough clarification.
at 4:24 you arrived at 1/nƩ(β₀+β₁y)= n(β₀)/n +β₁Ʃy/n = β₀+ β₁ȳ however i was expecting 1/nƩ(β₀+β₁y)= (β₀)/n +β₁Ʃy/n = (β₀)/n +β₁ȳ. can you explain that step
(1/n)*Ʃ(β₀) = (1/n)*(nβ₀) = β₀
@@nevmiku any idea how i can show that B1 is NOT unbiased
@@maiden5427 prove E(B1) =/ B1