Hi, Just wondering what the formula for calculating the standard error of slope and intercept coefficients in a regression is? Many thanks
Var (B1 hat) = s^2/ sigma (Xi-X bar)^2 Var (B0 hat) = s^2 * sigma Xi^2/(n*sigma (Xi-X bar)^2) let me know if u wanan know how this comes about. it actually comes from some simple algebraic manipulation and making use of the no serial correlaiton assumption.
Thanks- is the s^2 in the numerator the variance of the independent variable?
yes you got that right it is in the numerator.
HydrogenRainbow Wrote: ------------------------------------------------------- > Var (B1 hat) = s^2/ sigma (Xi-X bar)^2 > Var (B0 hat) = s^2 * sigma Xi^2/(n*sigma (Xi-X > bar)^2) > let me know if u wanan know how this comes about. > it actually comes from some simple algebraic > manipulation and making use of the no serial > correlaiton assumption. Correct me if I’m wrong, but that is the formula for the Variance of the regression coefficients and the standard error of the regression coefficients would be the square route of the these formulas. Additionally, is the sigma in the formulas lower case sigma as in “standard deviation” or is it upper case sigma as in “sum of”?
sigma = summation. std error (B1 hat) = sqrt of var(B1 hat) std error (B0 hat) = sqrt of var (B0 hat)
HydrogenRainbow (what is that anyway?) did a fine job answeringthe question, but just so nobody freaks - these are not formuulas that you should know for the exam and are not even especially useful formulas in the world (you wouldn’t use them in code for example).
Whether or not the information will be on the test is less important than being able to apply the logic of the information to another area of our future analyses. Thanks for he help Hydrogen.
No problem. But Jerrod, the standard error of both B1 hat and B0 hat are always given in the statistical outputs and so cranking out the figures isnt of great importance like what JDV has said. 
HydrogenRainbow Wrote: ------------------------------------------------------- > Var (B1 hat) = s^2/ sigma (Xi-X bar)^2 > Var (B0 hat) = s^2 * sigma Xi^2/(n*sigma (Xi-X > bar)^2) > let me know if u wanan know how this comes about. > it actually comes from some simple algebraic > manipulation I would like to know how did you derive these equations…
B1 hat= sigma (Xi - Xbar)(Yi-Y bar) / sigma (Xi-Xbar)^2 ------ (1) Y bar = B0 + B1 hat X bar + e bar------(2) Yi = B0 + B1Xi + ei Substituting (2) and (3) into (1) we get (i skipped the intermediate steps. i believe you should be able to manage basic algebraic manipulations) B1 hat = B1 + sigma (Xi - X bar)ei/ sigma (Xi - X bar)^2 - sigma (Xi-X bar)e bar / sigma (Xi-X bar)^2 The last term = 0 because you can write it as e bar * sigma (Xi-X bar) / sigma (Xi-X bar)^2 and sigma (Xi-X bar ) = 0 Therefore we get B1 hat = B1 + sigma (Xi - X bar)ei/ sigma (Xi - X bar)^2 Taking variance conditional on Xi on both sides Var (B1 Hat | Xi) = var(B1 + sigma (Xi - X bar)ei/ sigma (Xi - X bar)^2 | Xi) Since it is conditional on Xi, this means that sigma (Xi - X bar)^2 can be treated as a constant and so we get Var (B1 Hat | Xi) = var(B1 + sigma (Xi - X bar)ei/ sigma (Xi - X bar)^2 | Xi) =[sigma (Xi - X bar)^2]^-2 var (sigma (Xi - X bar)ei) -----(1) var (sigma (Xi - X bar)ei) = ?? To solve for the question marks we rely on the basic definition of Var (x) = E(X - E(X))^2. E(sigma (Xi - X bar)ei) = sigma (Xi - X bar ) E (ei) = 0 since E(ei)=0 [take note i am actually making use of conditional expectations here, i.e. i am taking expectations conditional on Xi so sigma (Xi- Xbar) is actually a constant). Therefore, var (sigma (Xi - X bar)ei) = E(sigma (Xi - X bar)ei)^2 = E[(sigma (Xi - X bar))^2*(ei)^2)] + cross product terms. = sigma (Xi - X bar )^2 E(ei^2)+E (cross pdt terms) and tjhe last term = 0 because of the assumption that ei~iid [ei and ej are independent and so cov (ei,ej)=0] Therefore var (sigma (Xi - X bar)ei) =sigma (Xi - X bar )^2 *E(ei^2)=sigma (Xi - X bar )^2 *sigma ^2 [from homoskedasticity assumption] hence, var (b1 hat) = sigma (Xi - X bar )^2 *sigma ^2*[sigma (Xi - X bar)^2]^-2 = sigma^2/sigma (Xi - X bar )^2 (PLUgging the above into (1))
Thanks HydrogenRainbow for all the effort. I am better off knowing those formulas than deriving it on-the-fly.
- Yeah I don’t deny it is such a b**ch deriving the formula…BUt the good thing is that the CFA exam won’t be testing us on the formula for the std error of the slope parameter
HydrogenRainbow…Have u written your thesis on this? Because I lost track of it halfway through the reading. Brilliant, must add though.
No - this is no thesis for this cos this is considered “elementary” I believe. The derivation can be found in the better introductory econometrics textbooks, e.g. Gujarati or Wooldridge [which I used on my own for my class cos the original text sucks]. U may have lost track of it because it is kinda hard to type the mathematical symbols here. =/
I understand that…was kidding about the thesis thing. But you guys do a great job and take a lot of pains to get doubts solved.
I tried to understand that myself initially looking at the screen. Gave up. Then tool a printout and converted the symobls b hat and b bars on the paper. Gave up. Then took it to other colleague of mine and asked for his help. He gave up. We both went to the 3rd teammate and asked his expertise. He gave up. Finally we hit the gym in vengeance and I replied back of a job well done but a tangential thesis worth a Noble cause.
Hmm. Not sure if JDV can do a better job in presenting the proof…