(a) The OLS estimator for Î² minimizes the Sum of Squared Residuals: n Î²Ë†=argmin ô°…ô°„(y âˆ’Î²x)2ô°† Î²ii i=1 Take the first-order condition to show that (b) Show that Ë† ô°ƒni=1 xiyi Î²=ô°ƒn x2. i=1 i Ë† ô°ƒni=1 xiÎµi Î²=Î²+ ô°ƒni=1x2i What is E[Î²Ë† | Î²] and Var(Î²Ë† | Î²)? Use this to show that conditional on Î² Î²Ë† has the following distribution: Ë† ô° Ïƒ2 ô°‚ Î²|Î²âˆ¼NÎ²ô°ƒn x2. i=1 i 1 (c) Suppose we believe that Î² is distributed normally with mean 0 and variance Ïƒ2 ; that is Î» Î² âˆ¼ N(0 Ïƒ2 ). Additionally assume that Î² is independent of Îµi. Compute the mean and Î» variance of Î²Ë†. That is what is E[Î²Ë†] and Var(Î²Ë†)?(Hint you might find useful: E[w1] = E[E[w1 | w2]] and Var(w1) = E[Var(w1 | w2)] + Var(E[w1 | w2]) for any random variables w1 and w2.) Question 2 Let us consider the linear regression model yi = Î²0 + Î²1xi + ui (i = 1 … n) which satisfies Assumptions MLR.1 through MLR.5 (see Slide 7 in â€œLinear_regression_reviewâ€ under â€œModulesâ€ on Canvas)1. The xis (i = 1 … n) and Î²0 and Î²1 are nonrandom. The randomness comes from uis (i = 1 … n) where var (ui) = Ïƒ2. Let Î²Ë†0 and Î²Ë†1 be the usual OLS estimators (which are unbiased for ï£« y1 ï£¶ ï£«1ï£¶ ï£¬ y2 ï£· ï£¬1ï£·Î²0 and Î²1 respectively) obtained from running a regression of ï£¬ . ï£· on ï£¬ . ï£· (the intercept column) and ï£¬ . ï£· ï£¬ï£ y n âˆ’ 1 ï£·ï£¸ on ï£· ï£¬ . ï£· only ï£«x1 ï£¬ x2 ï£¬ . ï£¬ . ï£¶ ï£· ï£·. Suppose you also run a regression of ï£· ï£·ï£¸ ï£« y1 ï£¶ ï£«x1 ï£¶ ï£¬ x2 ï£· ï£¬ï£ x n âˆ’ 1 xn yn xn a) Give the expression of Î² Ìƒ1 as a function of yis and xis (i = 1 … n). (excluding the intercept column) to obtain another estimator Î² Ìƒ1 of Î²1. ô° Ìƒô°‚ Ìƒ

b) Derive E Î²1 in terms of Î²0 Î²1 and xis. Show that Î²1 is unbiased for Î²1 when Î²0 = 0. If Î²0 Ì¸= 0 when will Î² Ìƒ1 be unbiased for Î²1? c) Derive Varô°Î² Ìƒ ô°‚ the variance of Î² Ìƒ in terms of Ïƒ2 and x s (i = 1…n). 11i 1The model is a simple special case of the general multiple regression model in â€œLinear_regression_reviewâ€. Solving this question does not require knowledge about matrix operations. ï£¬ï£·ï£¬ï£· ï£¬ï£ y n âˆ’ 1 ï£·ï£¸ ï£¬ï£ 1 ï£·ï£¸ yn 1 ï£¬ y2 ï£· ï£¬ . ï£· ï£¬ .ï£¬ï£ x n âˆ’ 1 ï£·ï£¸ 2 d) Show that Varô°Î² Ìƒ ô°‚ is no greater than Varô°Î²Ë† ô°‚; that is Var ô°Î² Ìƒ ô°‚ â‰¤ Var ô°Î²Ë† ô°‚. When do 1111 you have Var ô°Î² Ìƒ ô°‚ = Var ô°Î²Ë† ô°‚? (Hint you might find useful: use ô°ƒn x2 â‰¥ ô°ƒn (x âˆ’ x Ì„)2 where 11 i=1ii=1i x Ì„ = n1 ô°ƒni=1 xi.)e) Choosing between Î²Ë†1 and Î² Ìƒ1 leads to a tradeoff between the bias and variance. Comment on this tradeoff.

Question 3 Let vË† be an estimator of the truth v. Show that E (vË† âˆ’ v)2 = Var (vË†) + [Bias (vË†)]2 where Bias (vË†) = E (vË†) âˆ’ v. (Hint: The randomness comes from vË† only and v is nonrandom). Applied questions (with the use of R) For this question you will be asked to use tools from R for coding. Installation

Requirements: answer the question | .doc file

Question 1 Supposethatwehaveamodelyi =Î²xi+Îµi (i=1…n)wherey= n1 ô°ƒni=1yi =0x=ô°ƒni=1xi =0 and Îµi is distributed normally with mean 0 and variance Ïƒ2; that is Îµi âˆ¼ N(0Ïƒ2).

(d) Since everything is normally distributed it turns out that E[Î² | Î²Ë†] = E[Î²] + Cov(Î² Î²Ë†) Â· (Î²Ë† âˆ’ E[Î²Ë†]). Var(Î²Ë†) Let Î²Ë†RR = E[Î² | Î²Ë†]. Compute Cov(Î²Î²Ë†) and use the value of E[Î²] along with the values of E[Î²Ë†] Cov(Î²Î²Ë†) and Var(Î²Ë†) you have computed to show that Ë†RR Ë† ô°ƒni=1 x2i Ë† Î² = E[Î² | Î²] = ô°ƒni=1 x2i + Î» Â· Î² (Hint: Cov(w1 w2) = E[(w1 âˆ’ E[w1])(w2 âˆ’ E[w2])] and E[w1w2] = E[w1E[w2 | w1]] for any random variables w1 and w2)

(e) Does Î²Ë†RR increase or decrease as Î» increases? How does this relate to Î² being distributed N(0 Ïƒ2 )? Î» To install R please see

https://www.r-project.org/.

Once you install R please install also R Studio https://rstudio.com/products/rstudio/ download/.

You will need to use R Studio to solve the problem set. Question 1 Supposethatwehaveamodelyi =Î²xi+Îµi (i=1…n)wherey= n1 ô°ƒni=1yi =0x=ô°ƒni=1xi =0 and Îµi is distributed normally with mean 0 and variance Ïƒ2; that is Îµi âˆ¼ N(0Ïƒ2).

(a) The OLS estimator for Î² minimizes the Sum of Squared Residuals: nÎ²Ë†=argmin ô°…ô°„(y âˆ’Î²x)2ô°† Î²iii=1 Take the first-order condition to show that

(d) Since everything is normally distributed it turns out that E[Î² | Î²Ë†] = E[Î²] + Cov(Î² Î²Ë†) Â· (Î²Ë† âˆ’ E[Î²Ë†]). Var(Î²Ë†) Let Î²Ë†RR = E[Î² | Î²Ë†]. Compute Cov(Î²Î²Ë†) and use the value of E[Î²] along with the values of E[Î²Ë†] Cov(Î²Î²Ë†) and Var(Î²Ë†) you have computed to show that Ë†RR Ë† ô°ƒni=1 x2i Ë† Î² = E[Î² | Î²] = ô°ƒni=1 x2i + Î» Â· Î² (Hint: Cov(w1 w2) = E[(w1 âˆ’ E[w1])(w2 âˆ’ E[w2])] and E[w1w2] = E[w1E[w2 | w1]] for any random variables w1 and w2)

(e) Does Î²Ë†RR increase or decrease as Î» increases? How does this relate to Î² being distributed N(0 Ïƒ2 )? Î»

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