Question

Statistics 108, Homework4 Due: November 2nd, 2018, In Class (turn in paper form) You need to show the steps to get the full credits. This homework is to transit from simple linear regression to multiple lincar regression. Total: 90 points. 1. (40 points) Understanding the coefficient of determination, R2 (a) Visualization. For each of the seven datasets under Canvas/Files/Datasets (datal.txt data7.txt), plot the scatter plot, add the fitted lincar regression line, and compute the R2. What do you find? (Plot each dataset on the same y-scale to compare across the datasets.) (b) True/False. For each of the following statement, say whether t is true or false and explain why. (i) A large R2 always means that the fitted linear regression line is a good fit of the data. () A sall R2 always means that the predictor and the response are not related. ii) If all observations Y fall on on straight line and the line is not horizontal, then 2. (40 points) Simple linear regression in matrir form. Here, we analyze the data in Homework (a) What are the dimensions of the response vector Y and the design matrix X. Write down R2 = 1. 1, problem 2, again, but using the matrix form the first five rows of them. (b) Calculate the following two quantities: XX, XTY (c) Caleulate the least squares estimators by Compare the results here with those from that in Homework 1. Are they the same? (d) Give an estimate of the variance of β. Based on this, what are the standard error of %, the standard error of β1, and the estimate of Cov(A, β)? Compare these with Homework 2, problem 2, what do you find? 3. (10 points) Rigorous derivations. Suppose and ad- be 0. Let ad-be -ea Show that Therefore, by definition B is the inverse of A, ie., B = A-1

Answer 1

For 1) and 2) data is required.

3) Given $A=\begin{bmatrix} a &b \\ c& d \end{bmatrix}$ and suppose $ad-bc\neq 0$ .

Let $B=\frac{1}{ad-bc}\begin{bmatrix} d &-b \\ -c& a \end{bmatrix}$ . Using the product rule of matrix multiplication,

$BA=\frac{1}{ad-bc}\begin{bmatrix} d &-b \\ -c& a \end{bmatrix}\begin{bmatrix} a &b \\ c& d \end{bmatrix}\\ BA=\frac{1}{ad-bc}\begin{bmatrix} da-bc &db-bd \\ -ca+ac& -bc+ad \end{bmatrix}\\ BA=\frac{1}{ad-bc}\begin{bmatrix} da-bc &0 \\0& -bc+ad \end{bmatrix}\\ BA=\begin{bmatrix} 1 &0 \\0& 1 \end{bmatrix}\\ BA=I_2$

Also,

$AB=\frac{1}{ad-bc}\begin{bmatrix} a &b \\ c& d \end{bmatrix}\begin{bmatrix} d &-b \\ -c& a \end{bmatrix}\\ AB=\frac{1}{ad-bc}\begin{bmatrix} da-bc &ab-ab \\ -cd+dc& -bc+ad \end{bmatrix}\\ AB=\frac{1}{ad-bc}\begin{bmatrix} da-bc &0 \\0& -bc+ad \end{bmatrix}\\ AB=\begin{bmatrix} 1 &0 \\0& 1 \end{bmatrix}\\ AB=I_2$

Thus . $I_2$ is the identity matrix of order 2. Therefore, by definition is the inverse of . $B=A^{-1}$ .

The proof is complete.

We are required to solve only one question. Please post the remaining questions as another post.