This is THE BEST channel for stats ever!!! It touches every tiny teeny detail of the assumptions that explains each step without any confusion! Much Appreciation! God Bless!
Wish i could buy stocks, this channel deserves a lot more attention! Haven't found anything on RUclips this thorough with the math, which I really appreciate.
I think Var(b_0) can be more easily calculated. b_0 can be written as \sum **( 1/n - \bar X \frac{X - \bar X}{S_{xx}} )** Y_i. The term within asterisks is non-stochastic, so let's just call it m_i. Using the formula for variance of sum of random variables, Var(b_0) can be written as \sum {m_i}^2 Var(Y_i) + 2 \sum_{j>i} m_i m_j Cov(Y_i, Y_j). Var(Y_i) = \sigma^2, Cov(Y_i, Y_j) can be easily proved to be zero since there is no auto correlation and it's all smooth sailing from there. Gets rid of all the filler steps. Hope you understand TeX.
Great Video, everything is explained very well! I found a faster an easier way to derive the variance of b0 by using the centred model, like you did in many of your other videos. Using the centred model doesn't require us to calculate the covariance and we clearly see that the var(b0)=var(Ybar)+(xbar)^2*Var(b1), hence proving at the same time that the covariance is 0.
I don't understand why b1 is treated as non random in proof of var(b1) where var(yi)=sigma^2 (meaning var(B1)=0), but it is treated as random variable in proof of var(b0). Or is it because b1=sxy/sxx so it is random and B1 is a constant non random population value? also, ybar is supposed to be constant, it is not a variable right?
I was wondering if you would clarify assumption 2: Is it simply that that the Covariance of of Error Term i and j are 0 as they are not equal to each other. Example If error term i is 2 then error term j cannot be 2 for them to have a 0 covariance? Is it as simple as as that? What happens if they have the same covariance? thank you
I don't think covariance relates to any absolute values. It shows the relationship. For example, here you can see uncorrelated residuals, d2mvzyuse3lwjc.cloudfront.net/doc/en/UserGuide/images/Graphic_Residual_Analysis/Graphic_Residual_Analysis-4.jpg?v=10731
X is non random, though X can take different values. Sxx would be zero if X was always the same constant. Though there can be (and usually is) non random variability in X. For example, an experimenter may pre determine different values for X, for which they measure the random Y outcome
This is THE BEST channel for stats ever!!! It touches every tiny teeny detail of the assumptions that explains each step without any confusion! Much Appreciation! God Bless!
Thanks so much for this feedback and encouragement!
Wish i could buy stocks, this channel deserves a lot more attention! Haven't found anything on RUclips this thorough with the math, which I really appreciate.
I think Var(b_0) can be more easily calculated.
b_0 can be written as \sum **( 1/n - \bar X \frac{X - \bar X}{S_{xx}} )** Y_i.
The term within asterisks is non-stochastic, so let's just call it m_i.
Using the formula for variance of sum of random variables,
Var(b_0) can be written as \sum {m_i}^2 Var(Y_i) + 2 \sum_{j>i} m_i m_j Cov(Y_i, Y_j).
Var(Y_i) = \sigma^2, Cov(Y_i, Y_j) can be easily proved to be zero since there is no auto correlation and it's all smooth sailing from there. Gets rid of all the filler steps.
Hope you understand TeX.
Excellent. Great and clear explanation. Best I've found. Thanks.
Great Video, everything is explained very well! I found a faster an easier way to derive the variance of b0 by using the centred model, like you did in many of your other videos. Using the centred model doesn't require us to calculate the covariance and we clearly see that the var(b0)=var(Ybar)+(xbar)^2*Var(b1), hence proving at the same time that the covariance is 0.
Yup. That works too! Thanks for sharing! :-)
Exactly what I was looking for. Subscribed.
This very helpful and well explained. Thnk you.
how did you get Sigma (x_i-x_bar)y_i please?
I don't understand why b1 is treated as non random in proof of var(b1) where var(yi)=sigma^2 (meaning var(B1)=0), but it is treated as random variable in proof of var(b0). Or is it because b1=sxy/sxx so it is random and B1 is a constant non random population value?
also, ybar is supposed to be constant, it is not a variable right?
Should I assume that Sxy = sum{[Xi-Xbar][Yi-Ybar]} is equal to sum(Xi-Xbar)Yi? Is this always the case at all times or there are exceptions?
yeah: sum{[Xi-Xbar][Yi-Ybar]} = sum{XiYi - XiYbar - YiXbar + XbarYbar} =
= sum{XiYi} - Ybar*sum{Xi} - Xbar*sum{Yi} + nXbarYbar =
= sum{XiYi} - (sum{Yi}/n)*sum{Xi} - Xbar*sum{Yi} + n*Xbar*(sum{Yi}/n) =
= sum{XiYi} - sum{Yi}*(sum{Xi}/n) - Xbar*sum{Yi} + Xbar*sum{Yi} =
= sum{XiYi} - sum{Yi}*(sum{Xi}/n) = sum{XiYi - YiXbar} = sum{Yi(Xi - Xbar)}
It is always equal and equal to: sum{Xi(Yi - Ybar)}
I was wondering if you would clarify assumption 2: Is it simply that that the Covariance of of Error Term i and j are 0 as they are not equal to each other. Example If error term i is 2 then error term j cannot be 2 for them to have a 0 covariance? Is it as simple as as that? What happens if they have the same covariance? thank you
I don't think covariance relates to any absolute values. It shows the relationship. For example, here you can see uncorrelated residuals, d2mvzyuse3lwjc.cloudfront.net/doc/en/UserGuide/images/Graphic_Residual_Analysis/Graphic_Residual_Analysis-4.jpg?v=10731
If X is non-random, shouldnt S_xx be 0?
X is non random, though X can take different values. Sxx would be zero if X was always the same constant. Though there can be (and usually is) non random variability in X. For example, an experimenter may pre determine different values for X, for which they measure the random Y outcome
Isn't Y-Bar a constant and therefore have no covariance with b1?
No, as y_i is random.
Weldone well explained 🙏🏽🙌🏽
you are genius