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There is a Panel with \(M\) individuals and each individual has \(Q\) records/rows. A function generate an individual specific outcome given the \(Q\) individual specific inputs, along with shared parameters/values stored as variables that contain common values for each of the \(M\) individuals.
For example, suppose we have a dataframe of individual wage information from different countries (the number of countries is \(M\)). Each row is an individual from one country, giving us \(Q \cdot M\) observations of wages.
We want to generate country specific gini based on the individual wage data for each country in the dataframe. Additionally, perhaps the gini formula requires not just individual wages but some additional parameters or shared dataframes as inputs. We will use the ff_dist_gini_vector_pos.html function from REconTools.
First, we simulate a dataframe with \(M\) countries, and up to \(Q\) people in each country. The countries share the same mean income, but have different standard deviations.
# Parameter Setups
it_M <- 10
it_Q_max <- 100
fl_rnorm_mu <- 1
ar_rnorm_sd <- seq(0.01, 0.2, length.out=it_M)
set.seed('789')
ar_it_q <- sample.int(it_Q_max, it_M, replace=TRUE)
# N by Q varying parameters
mt_data <- cbind(ar_it_q, ar_rnorm_sd)
tb_M <- as_tibble(mt_data) %>% rowid_to_column(var = "ID") %>%
rename(sd = ar_rnorm_sd,
Q = ar_it_q) %>%
mutate(mean = fl_rnorm_mu) %>%
select(ID, Q,
mean, sd)
# Show table
kable(tb_M, caption = paste0("M=", it_M,
" countries (ID is country ID), observation per country (Q)",
", mean and s.d. of wages each country")) %>%
kable_styling_fc()| ID | Q | mean | sd |
|---|---|---|---|
| 1 | 45 | 1 | 0.0100000 |
| 2 | 12 | 1 | 0.0311111 |
| 3 | 42 | 1 | 0.0522222 |
| 4 | 26 | 1 | 0.0733333 |
| 5 | 99 | 1 | 0.0944444 |
| 6 | 37 | 1 | 0.1155556 |
| 7 | 100 | 1 | 0.1366667 |
| 8 | 43 | 1 | 0.1577778 |
| 9 | 67 | 1 | 0.1788889 |
| 10 | 70 | 1 | 0.2000000 |
Second, we now expand the dataframe so that each country has not just one row, but \(Q_i\) of observations (\(i\) is country), or randomly drawn income based on the country-specific income distribution. Note that there are three ways of referring to variable names with dot, which are all shown below:
# A. Normal Draw Expansion, Explicitly Name
set.seed('123')
tb_income_norm_dot_dollar <- tb_M %>% group_by(ID) %>%
do(income = rnorm(.$Q, mean=.$mean, sd=.$sd)) %>%
unnest(c(income)) %>%
left_join(tb_M, by="ID")
# Normal Draw Expansion again, dot dollar differently with string variable name
set.seed('123')
tb_income_norm_dollar_dot <- tb_M %>% group_by(ID) %>%
do(income = rnorm(`$`(., 'Q'), mean = `$`(., 'mean'), sd = `$`(., 'sd'))) %>%
unnest(c(income)) %>%
left_join(tb_M, by="ID")
# Normal Draw Expansion again, dot double bracket
set.seed('123')
svr_mean <- 'mean'
svr_sd <- 'sd'
svr_Q <- 'Q'
tb_income_norm_dot_bracket_db <- tb_M %>% group_by(ID) %>%
do(income = rnorm(.[[svr_Q]], mean = .[[svr_mean]], sd = .[[svr_sd]])) %>%
unnest(c(income)) %>%
left_join(tb_M, by="ID")Third, we print the first set of rows of the dataframe, and also summarize income by country groups.
# Show dataframe dimension
print(dim(tb_income_norm_dot_bracket_db))## [1] 541 5# Show first 20 rows
kable(head(tb_income_norm_dot_bracket_db, 20),
caption = "ID = country ID, wage draws"
) %>% kable_styling_fc()| ID | income | Q | mean | sd |
|---|---|---|---|---|
| 1 | 0.9943952 | 45 | 1 | 0.01 |
| 1 | 0.9976982 | 45 | 1 | 0.01 |
| 1 | 1.0155871 | 45 | 1 | 0.01 |
| 1 | 1.0007051 | 45 | 1 | 0.01 |
| 1 | 1.0012929 | 45 | 1 | 0.01 |
| 1 | 1.0171506 | 45 | 1 | 0.01 |
| 1 | 1.0046092 | 45 | 1 | 0.01 |
| 1 | 0.9873494 | 45 | 1 | 0.01 |
| 1 | 0.9931315 | 45 | 1 | 0.01 |
| 1 | 0.9955434 | 45 | 1 | 0.01 |
| 1 | 1.0122408 | 45 | 1 | 0.01 |
| 1 | 1.0035981 | 45 | 1 | 0.01 |
| 1 | 1.0040077 | 45 | 1 | 0.01 |
| 1 | 1.0011068 | 45 | 1 | 0.01 |
| 1 | 0.9944416 | 45 | 1 | 0.01 |
| 1 | 1.0178691 | 45 | 1 | 0.01 |
| 1 | 1.0049785 | 45 | 1 | 0.01 |
| 1 | 0.9803338 | 45 | 1 | 0.01 |
| 1 | 1.0070136 | 45 | 1 | 0.01 |
| 1 | 0.9952721 | 45 | 1 | 0.01 |
# Display country-specific summaries
REconTools::ff_summ_bygroup(tb_income_norm_dot_bracket_db, c("ID"), "income")$df_table_grp_statsFourth, there is only one input for the gini function ar_pos. Note that the gini are not very large even with large SD, because these are normal distributions. By Construction, most peple are in the middle. So with almost zero standard deviation, we have perfect equality, as standard deviation increases, inequality increases, but still pretty equal overall, there is no fat upper tail.
# Gini by Group
tb_gini_norm <- tb_income_norm_dot_bracket_db %>% group_by(ID) %>%
do(inc_gini_norm = REconTools::ff_dist_gini_vector_pos(.$income)) %>%
unnest(c(inc_gini_norm)) %>%
left_join(tb_M, by="ID")
# display
kable(tb_gini_norm,
caption = paste0(
"Country-specific wage GINI based on income draws",
", ID=country-ID, Q=sample-size-per-country",
", mean=true-income-mean, sd=true-income-sd"
)) %>%
kable_styling_fc()| ID | inc_gini_norm | Q | mean | sd |
|---|---|---|---|---|
| 1 | 0.0052111 | 45 | 1 | 0.0100000 |
| 2 | 0.0137174 | 12 | 1 | 0.0311111 |
| 3 | 0.0245939 | 42 | 1 | 0.0522222 |
| 4 | 0.0303468 | 26 | 1 | 0.0733333 |
| 5 | 0.0527628 | 99 | 1 | 0.0944444 |
| 6 | 0.0544053 | 37 | 1 | 0.1155556 |
| 7 | 0.0786986 | 100 | 1 | 0.1366667 |
| 8 | 0.0818873 | 43 | 1 | 0.1577778 |
| 9 | 0.1014639 | 67 | 1 | 0.1788889 |
| 10 | 0.0903825 | 70 | 1 | 0.2000000 |