40 Assignment 2 (Guide)

1. A mathematical model is a simplification and description of some real-world phenomenon. The model is usually expressed and communicated using a mathematical equation. The equation is deterministic and does not allow for the real-world phenomenon to have any randomness in it.

2. A statistical model is a simplification and description of some real-world phenomenon. The model is usually expressed and communicated using a mathematical equation and a probability density/mass function (e.g., the normal distribution). The combination of the mathematical equation and probability density/mass function enables to model to be stochastic and account for randomness in a real-world phenomenon that the model describes.

3. Everyone did great answering this question!!!

4. A <- matrix(c(1,2,3,2,5,6,3,6,9),3,3,byrow=TRUE)
   B <- diag(1,3)
   C <- t(A)

   A+C

   ##      [,1] [,2] [,3]
   ## [1,]    2    4    6
   ## [2,]    4   10   12
   ## [3,]    6   12   18

   A-C

   ##      [,1] [,2] [,3]
   ## [1,]    0    0    0
   ## [2,]    0    0    0
   ## [3,]    0    0    0

   A%*%B

   ##      [,1] [,2] [,3]
   ## [1,]    1    2    3
   ## [2,]    2    5    6
   ## [3,]    3    6    9

   solve(A+B)%*%(C+B)

   ##               [,1]          [,2]          [,3]
   ## [1,]  1.000000e+00 -8.881784e-16 -1.554312e-15
   ## [2,]  1.665335e-16  1.000000e+00  1.110223e-16
   ## [3,] -5.551115e-17  1.110223e-16  1.000000e+00

5. Which of the models below are a linear statistical model? Please justify your answer using words.

   1. Multiple answers are correct here. This is a linear model because the only non-linear part is where the predictor variable (i.e., $\mathbf{x}$) is squared. Any linear or non-linear transformation of the predictor variable does not change if a model is linear or non-linear. Another correct answer is that this is not a linear model because the error term (i.e., $\boldsymbol{\varepsilon}$) is missing. If the error term was not missing, then it would be a linear model.
   2. This is a linear model because the only non-linear part is where Euler’s number (i.e., $e$) is raised to the $\mathbf{x}$ power. Any linear or non-linear transformation of the predictor variable does not change if a model is linear or non-linear.
   3. This is not a linear model because Euler’s number is raised to the parameter $\beta_{1}$ power (i.e., the parameter shows up in the model inside of a non-linear function).
   4. Multiple answers are correct here. This is not a linear model because the parameter shows up in the model inside of a non-linear function. Another correct answer is that you can easily simplify the model and show that it is the same as $\mathbf{y}=\beta_{0}+\beta_{1}\mathbf{x}+\boldsymbol{\varepsilon}.$

6. x <- seq(0, 3, by = 0.01)
   beta <- c(-0.34, 1.04)
   E.y_a <- beta[1] + beta[2] * x^2
   E.y_b <- beta[2] * exp(x)
   E.y_c <- beta[1] + exp(beta[2] * x)
   E.y_d <- beta[1] + beta[2] * x
   col <- c("black", "deepskyblue", "purple", "gold")
   
   plot(x, E.y_a, ylim = c(range(E.y_a, E.y_b, E.y_c, E.y_d)), typ = "l", col = col[1],
       xlab = expression(italic(x)), ylab = "Expected value", lwd = 2)
   points(x, E.y_b, typ = "l", col = col[2], lwd = 2)
   points(x, E.y_c, typ = "l", col = col[3], lwd = 2)
   points(x, E.y_d, typ = "l", col = col[4], lwd = 2)
   legend(x = 0, y = 22, legend = c("2a", "2b", "2c", "2d"), bty = "n", lty = 1, col = col)