Practice Questions 1

Let’s say our utility function is:

\[v_j = \beta_1 x_j^{\mathrm{price}} + \beta_2 x_j^{\mathrm{cacao}} + \beta_3 \delta_j^{\mathrm{hershey}} + \beta_4 \delta_j^{\mathrm{lindt}}\]

And we estimate the following coefficients:

Parameter Coefficient
\(\beta_1\) -0.1
\(\beta_2\) 0.1
\(\beta_3\) -2.0
\(\beta_4\) -0.1

What are the expected probabilities of choosing each of these bars using a logit model?

Attribute Bar 1 Bar 2 Bar 3
Price $1.20 $1.50 $3.00
% Cacao 10% 60% 80%
Brand Hershey Lindt Ghirardelli 
# Create the model coefficient variables
b1 <- -0.1
b2 <- 0.1
b3 <- -2.0
b4 <- -0.1

# Compute the observed utility for each chocolate bar
v1 <- b1*1.2 + b2*0.1 + b3
v2 <- b1*1.5 + b2*0.6 + b4
v3 <- b1*3 + b2*0.8

# Compute the probabilities using the logit fraction
denom <- exp(v1) + exp(v2) + exp(v3)
p1 <- exp(v1) / denom
p2 <- exp(v2) / denom
p3 <- exp(v3) / denom
p1
#> [1] 0.06925051
p2
#> [1] 0.4723548
p3
#> [1] 0.4583946

Practice Questions 2

Observations - Height of students (inches):

#>  [1] 65 69 66 67 68 72 68 69 63 70
  1. Let’s say we know that the height of students, \(\tilde{x}\), in a classroom follows a normal distribution. A professor obtains the above height measurements students in her classroom. What is the log-likelihood that \(\tilde{x} \sim \mathcal{N} (68, 4)\)? In other words, compute \(\ln \mathcal{L} (\mu = 68, \sigma = 4)\).
# Load the data
x <- c(65, 69, 66, 67, 68, 72, 68, 69, 63, 70)

# Compute the value of f(x) for each x
f_x <- dnorm(x, 68, 4)

# The likelihood is just the product of the probabilities in f_x
prod(f_x)
#> [1] 1.447528e-11
# But this is a really tiny number, so computing the log-likelihood is helpful
log(prod(f_x))
#> [1] -24.95858
# Of course, the way we typically compute the log-likelihood is by summing up
# the log of the values in f_x
sum(log(f_x))
#> [1] -24.95858
  1. Compute the log-likelihood function using the same standard deviation \((\sigma = 4)\) but with the following different values for the mean, \(\mu: 66, 67, 68, 69, 70\). How do the results compare? Which value for \(\mu\) produces the highest log-likelihood?
library(tidyverse)

# Create a vectors of values for the mean
means <- c(66, 67, 68, 69, 70)

# Compute the likelihood using different values for the mean:
L1 <- sum(log(dnorm(x, means[1], 4)))
L2 <- sum(log(dnorm(x, means[2], 4)))
L3 <- sum(log(dnorm(x, means[3], 4)))
L4 <- sum(log(dnorm(x, means[4], 4)))
L5 <- sum(log(dnorm(x, means[5], 4)))
logLiks <- c(L1, L2, L3, L4, L5)

# Create a data frame of the results
df <- data.frame(means, logLiks)
df
#>   means   logLiks
#> 1    66 -25.83358
#> 2    67 -25.08358
#> 3    68 -24.95858
#> 4    69 -25.45858
#> 5    70 -26.58358
# Filter out the row with the maximum likelihood value:
df %>%
  filter(logLiks == max(logLiks))
#>   means   logLiks
#> 1    68 -24.95858
# Plot the result:
df %>%
  ggplot(aes(x = means, y = logLiks)) +
  geom_line() +
  geom_point() +
  theme_bw() +
  labs(x = "Mean Value", y = "Log-likelihood Values")

Practice Questions 3

Suppose we estimate the following utility model describing preferences for cars:

\[ u_j = \alpha p_j + \beta_1 x_j^{mpg} + \beta_2 x_j^{elec} + \varepsilon_j \]

The estimated model produces the following results:

Parameter Coefficient
\(\alpha\) -0.7
\(\beta_1\) 0.1
\(\beta_2\) -0.4

Hessian:

\[ \begin{bmatrix} -6000 & 50 & 60 \\ 50 & -700 & 50 \\ 60 & 50 & -300 \end{bmatrix} \]

Compute a 95% confidence interval around the coefficients using:

  1. Standard errors
  2. Simulated draws
# 1. Get coefficients
beta <- c(price = -0.7, mpg = 0.1, elec = -4.0)

# 2. Get covariance matrix
hessian <- matrix(c(
    -6000,   50,   60,
       50, -700,   50,
       60,   50, -300),
    ncol = 3, byrow = TRUE)

covariance <- -1*solve(hessian)
  1. Solution using standard errors
se <- sqrt(diag(covariance))
lower <- beta - 2*se
upper <- beta + 2*se

coef_ci <- data.frame(
    mean = beta, 
    lower = lower, 
    upper = upper
)

coef_ci
#>       mean       lower     upper
#> price -0.7 -0.72585699 -0.674143
#> mpg    0.1  0.02392002  0.176080
#> elec  -4.0 -4.11629585 -3.883704
  1. Solution using simulated draws
draws <- as.data.frame(MASS::mvrnorm(10^4, beta, covariance))
coef_ci <- maddTools::ci(draws, ci = 0.95)

coef_ci
#>             mean       lower     upper
#> price -0.7001425 -0.72497169 -0.675090
#> mpg    0.1004593  0.02665909  0.174476
#> elec  -3.9998885 -4.11361564 -3.883845