Practice Questions 1

Suppose we estimate a model and get the following results:

\[ \hat{\beta} = \begin{bmatrix} -0.4 \\ 0.5 \end{bmatrix} \]

\[ \nabla_{\beta}^2 \ln(\mathcal{L}) = \begin{bmatrix} -6000 & 60 \\ 60 & -700 \end{bmatrix} \]

Use the hessian to compute the standard errors for \(\hat{\beta}\)

hessian <- matrix(c(-6000, 60, 60, -700), ncol = 2, byrow = TRUE)
covariance <- -1*solve(hessian)
se <- sqrt(diag(covariance))
sigma1 <- se[1]
sigma2 <- se[2]
sigma1

#> [1] 0.01291548

sigma2

#> [1] 0.03781266

Use the standard errors to compute a 95% confidence interval around \(\hat{\beta}\)

beta1 <- -0.4
beta2 <- 0.5
ci1 <- c(beta1 - 2*sigma1, beta1 + 2*sigma1)
ci2 <- c(beta2 - 2*sigma2, beta2 + 2*sigma2)
ci1

#> [1] -0.425831 -0.374169

ci2

#> [1] 0.4243747 0.5756253

Practice Questions 2

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:

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

Generate 10,000 draws of the model coefficients using the estimated coefficients and hessian. Use the mvrnorm() function from the MASS library.

beta <- c(-0.7, 0.1, -4.0)

hessian <- matrix(c(
    -6000,   50,   60,
       50, -700,   50,
       60,   50, -300),
    ncol = 3, byrow = TRUE)

covariance <- -1*solve(hessian)

draws <- MASS::mvrnorm(10^5, beta, covariance)
head(draws)

#>            [,1]       [,2]      [,3]
#> [1,] -0.6883767 0.07455516 -3.981057
#> [2,] -0.6849568 0.05760901 -4.105302
#> [3,] -0.7153545 0.16959248 -3.967045
#> [4,] -0.6972684 0.13993164 -3.860456
#> [5,] -0.6963814 0.09933565 -3.977531
#> [6,] -0.6853383 0.14693863 -3.895686

Use the draws to compute the mean and 95% confidence intervals of each parameter estimate.

# Mean and CI for par 1
mean(draws[, 1])

#> [1] -0.6999935

quantile(draws[, 1], c(0.025, 0.975))

#>       2.5%      97.5% 
#> -0.7254501 -0.6746434

# Compare to computed CI
sigma <- sqrt(diag(covariance))
c(beta[1] - 2*sigma[1], beta[1] + 2*sigma[1])

#> [1] -0.725857 -0.674143

# Mean and CI for par 2 & 3
mean(draws[, 2])

#> [1] 0.1000915

quantile(draws[, 2], c(0.025, 0.975))

#>       2.5%      97.5% 
#> 0.02538884 0.17540911

mean(draws[, 3])

#> [1] -3.999676

quantile(draws[, 3], c(0.025, 0.975))

#>      2.5%     97.5% 
#> -4.114231 -3.885681

# Solution using the class package ci() function:
maddTools::ci(draws, ci = 0.95)

#>         mean       lower      upper
#> 1 -0.6999935 -0.72545007 -0.6746434
#> 2  0.1000915  0.02538884  0.1754091
#> 3 -3.9996755 -4.11423052 -3.8856808

In Class Question 1: Long <–> Wide

Write code to read in the following two files:

pv_cells.csv: Data on solar photovoltaic cell production by country
milk_production.csv: Data on milk production by state

milk_production <- read_csv(here('data', 'milk_production.csv'))
pv_cells <- read_csv(here('data', 'pv_cells.csv'))

Now modify the format of each:

If the data are in “wide” format, convert it to “long” with pivot_longer()
If the data are in “long” format, convert it to “wide” with pivot_wider()

# milk_production is in long format - convert to wide
milk_wide <- milk_production %>%
    pivot_wider(
        names_from = state, 
        values_from = milk_produced)
head(milk_wide)

#> # A tibble: 6 × 51
#>    year   Maine `New Hampshire` Vermont Massachusetts `Rhode Island` Connecticut
#>   <dbl>   <dbl>           <dbl>   <dbl>         <dbl>          <dbl>       <dbl>
#> 1  1970  6.19e8       356000000  1.97e9     658000000       75000000   661000000
#> 2  1971  6.29e8       359000000  2.02e9     658000000       70000000   660000000
#> 3  1972  6.38e8       353000000  2.04e9     629000000       69000000   643000000
#> 4  1973  6.14e8       337000000  1.95e9     595000000       64000000   614000000
#> 5  1974  6.11e8       329000000  1.94e9     593000000       63000000   613000000
#> 6  1975  6.29e8       336000000  2.01e9     601000000       63000000   608000000
#> # … with 44 more variables: New York <dbl>, New Jersey <dbl>,
#> #   Pennsylvania <dbl>, Delaware <dbl>, Maryland <dbl>, Michigan <dbl>,
#> #   Wisconsin <dbl>, Minnesota <dbl>, Ohio <dbl>, Indiana <dbl>,
#> #   Illinois <dbl>, Iowa <dbl>, Missouri <dbl>, North Dakota <dbl>,
#> #   South Dakota <dbl>, Nebraska <dbl>, Kansas <dbl>, Virginia <dbl>,
#> #   West Virginia <dbl>, North Carolina <dbl>, Kentucky <dbl>, Tennessee <dbl>,
#> #   South Carolina <dbl>, Georgia <dbl>, Florida <dbl>, Alabama <dbl>, …

# pv_cells is in wide format - convert to long
pv_cells_long <- pv_cells %>%
    pivot_longer(
        cols = China:World,
        names_to = "country", 
        values_to = "numCells")

# Could also do this
pv_cells_long <- pv_cells %>%
    pivot_longer(
        cols = -Year,
        names_to = "country", 
        values_to = "numCells")

head(pv_cells_long)

#> # A tibble: 6 × 3
#>    Year country     numCells
#>   <dbl> <chr>          <dbl>
#> 1  1995 China           NA  
#> 2  1995 Taiwan          NA  
#> 3  1995 Japan           16.4
#> 4  1995 Malaysia        NA  
#> 5  1995 Germany         NA  
#> 6  1995 South Korea     NA

Practice Questions

October 27, 2021

Practice Questions 1

Practice Questions 2

In Class Question 1: Long <–> Wide