ANLY 505 - Problem Set #2
Simulation of Linear Mixed Models
Linfang li
2019-04-23
The statistical model:
\(y_t = \beta_0 + \beta_1 * (Elevation_s)_t + \beta_2 * Slope_t + (b_s)_t + \epsilon_t\)
Where:
- \(\beta_0\) is the mean response when both Elevation and Slope are 0
- \(\beta_1\) is the change in mean response for a 1-unit change in elevation. Elevation is measured at the stand level, so all plots in a stand share a single value in elevation.
- \(\beta_2\) is the change in mean response for a 1-unit change in slope. Slope is measured at the plot level, so every plot potentially has a unique value of slope.
Let’s define the parameters:
- the intercept \((\beta_0)\) will be -1
- the coefficient for elevation \((\beta_1)\) will be set to 0.005
- the coefficient for slope \((\beta_2)\) will be set to 0.1
nstand = 5
nplot = 4
b0 = -1
b1 = .005
b2 = .1
sds = 2
sd = 1Simulate other variables:
set.seed(16)
stand = rep(LETTERS[1:nstand], each = nplot)
standeff = rep( rnorm(nstand, 0, sds), each = nplot)
ploteff = rnorm(nstand*nplot, 0, sd)Simulate elevation and slope:
elevation = rep( runif(nstand, 1000, 1500), each = nplot)
slope = runif(nstand*nplot, 2, 75)Simulate response variable:
resp2 = b0 + b1*elevation + b2*slope + standeff + ploteff Your tasks (complete each task in its’ own code chunk, make sure to use echo=TRUE so I can see your code):
- Fit a linear mixed model with the response variable as a function of elevation and slope with stand as a random effect. Are the estimated parameters similar to the intial parameters as we defined them?
b0 = -1b1 = 0.005b2 = 0.1
b1 = 0.005, which is much smaller than the initial parameter as we defined; b2 = 0.1, which is close to the initial parameter as we defined;
library(lme4)## Loading required package: Matrixfit1 = lmer(resp2 ~ 1 + elevation + slope+ (1|stand))
cat("b0 = ", b0, sep = "")## b0 = -1cat("b1 = ", b1, sep = "")## b1 = 0.005cat("b2 = ", b2, sep = "")## b2 = 0.1- Create a function for your model and run 1000 simulations of that model.
twolevel_fun<-function(nstand = 5, nplot = 4, b0 = -1, b1 = 0.005, b2 = 0.1, sds = 2, sd = 1) {
standeff = rep( rnorm(nstand, 0, sds), each = nplot)
stand = rep(LETTERS[1:nstand], each = nplot)
ploteff = rnorm(nstand*nplot, 0, sd)
elevation = rep( runif(nstand, 1000, 1500), each = nplot)
slope = runif(nstand*nplot, 2, 75)
resp2 = b0 + b1*elevation + b2*slope + standeff + ploteff
dat = data.frame(resp2, elevation, slope, stand)
lmer(resp2 ~ 1 + elevation + slope+ (1|stand))
}
sims <-replicate(1000, twolevel_fun() )
sims[[1000]]## Linear mixed model fit by REML ['lmerMod']
## Formula: resp2 ~ 1 + elevation + slope + (1 | stand)
## REML criterion at convergence: 83.9657
## Random effects:
## Groups Name Std.Dev.
## stand (Intercept) 2.628
## Residual 1.053
## Number of obs: 20, groups: stand, 5
## Fixed Effects:
## (Intercept) elevation slope
## 3.0727412 0.0001542 0.1027728- Extract the stand and residual variances from this simulation run. Print the first 6 rows of the data.
library(broom)
tidy(fit1)## Warning in bind_rows_(x, .id): binding factor and character vector,
## coercing into character vector## Warning in bind_rows_(x, .id): binding character and factor vector,
## coercing into character vector## # A tibble: 5 x 5
## term estimate std.error statistic group
## <chr> <dbl> <dbl> <dbl> <chr>
## 1 (Intercept) -21.3 6.60 -3.23 fixed
## 2 elevation 0.0206 0.00492 4.19 fixed
## 3 slope 0.0951 0.0164 5.78 fixed
## 4 sd_(Intercept).stand 1.10 NA NA stand
## 5 sd_Observation.Residual 1.17 NA NA Residualtidy(fit1, effects = "fixed")## # A tibble: 3 x 4
## term estimate std.error statistic
## <chr> <dbl> <dbl> <dbl>
## 1 (Intercept) -21.3 6.60 -3.23
## 2 elevation 0.0206 0.00492 4.19
## 3 slope 0.0951 0.0164 5.78tidy(fit1, effects = "ran_pars", scales = "vcov")## # A tibble: 2 x 3
## term group estimate
## <chr> <chr> <dbl>
## 1 var_(Intercept).stand stand 1.21
## 2 var_Observation.Residual Residual 1.36library(purrr) # v. 0.2.4
suppressPackageStartupMessages( library(dplyr) ) # v. 0.7.4
library(ggplot2) # v. 2.2.1
stand_sims = c(5, 20, 100) %>%
set_names() %>%
map(~replicate(1000, twolevel_fun(nstand = .x) ) )
stand_vars = stand_sims %>%
modify_depth(2, ~tidy(.x, effects = "ran_pars", scales = "vcov") ) %>%
map_dfr(bind_rows, .id = "stand_num") %>%
filter(group == "stand")
head(stand_vars)## # A tibble: 6 x 4
## stand_num term group estimate
## <chr> <chr> <chr> <dbl>
## 1 5 var_(Intercept).stand stand 1.92
## 2 5 var_(Intercept).stand stand 12.0
## 3 5 var_(Intercept).stand stand 6.84
## 4 5 var_(Intercept).stand stand 2.89
## 5 5 var_(Intercept).stand stand 2.52
## 6 5 var_(Intercept).stand stand 0.966- Choose three different sample sizes (your choice) and run 1000 model simulations with each sample size. Create 3 visualizations that compare distributions of the variances for each of the 3 sample sizes. Make sure that the axes are labelled correctly. What do these graphs say about the relationship between sample size and variance?
As we can see the real variance of stand is 4. The estimates are more accurate when the sample is increasing.
library(purrr) # v. 0.2.4
suppressPackageStartupMessages( library(dplyr) ) # v. 0.7.4
library(ggplot2) # v. 2.2.1
stand_sims = c(5, 20, 100) %>%
set_names(c("size_5", "size_20", "size_100")) %>%
map(~replicate(1000, twolevel_fun(nstand = .x) ) )
stand_vars = stand_sims %>%
modify_depth(2, ~tidy(.x, effects = "ran_pars", scales = "vcov") ) %>%
map_dfr(bind_rows, .id = "stand_num") %>%
filter(group == "stand")
head(stand_vars)## # A tibble: 6 x 4
## stand_num term group estimate
## <chr> <chr> <chr> <dbl>
## 1 size_5 var_(Intercept).stand stand 7.09
## 2 size_5 var_(Intercept).stand stand 2.53
## 3 size_5 var_(Intercept).stand stand 10.3
## 4 size_5 var_(Intercept).stand stand 1.62
## 5 size_5 var_(Intercept).stand stand 9.98
## 6 size_5 var_(Intercept).stand stand 3.19ggplot(stand_vars, aes(x = estimate) ) +
geom_density(fill = "blue", alpha = .25) +
facet_wrap(~stand_num) +
geom_vline(xintercept = 4)
stand_vars = mutate(stand_vars, stand_num = forcats::fct_inorder(stand_num) )
groupmed = stand_vars %>%
group_by(stand_num) %>%
summarise(mvar = median(estimate) )
ggplot(stand_vars, aes(x = estimate) ) +
geom_density(fill = "blue", alpha = .25) +
facet_wrap(~stand_num) +
geom_vline(aes(xintercept = 4, linetype = "True variance"), size = .5 ) +
geom_vline(data = groupmed, aes(xintercept = mvar, linetype = "Median variance"),
size = .5) +
theme_bw() +
scale_linetype_manual(name = "", values = c(2, 1) ) +
theme(legend.position = "bottom",
legend.key.width = unit(.1, "cm") ) +
labs(x = "Estimated Variance", y = NULL)
stand_vars %>%
group_by(stand_num) %>%
summarise(mean(estimate < 4) )## # A tibble: 3 x 2
## stand_num `mean(estimate < 4)`
## <fct> <dbl>
## 1 size_5 0.604
## 2 size_20 0.559
## 3 size_100 0.54- Plot the coefficients of the estimates of elevation and slope. Hint: the x-axis should have 1000 values. Discuss the graphs.
The estimates fluctuate aroung the real value.
library(furrr)## Loading required package: future## Warning: package 'future' was built under R version 3.5.2simul_estimates <- sims %>%
future_map(tidy, effects = "fixed") %>%
bind_rows()
simul_estimates## # A tibble: 3,000 x 4
## term estimate std.error statistic
## <chr> <dbl> <dbl> <dbl>
## 1 (Intercept) 10.6 10.3 1.03
## 2 elevation -0.00546 0.00927 -0.589
## 3 slope 0.0868 0.0140 6.21
## 4 (Intercept) -15.9 5.63 -2.83
## 5 elevation 0.0152 0.00426 3.57
## 6 slope 0.121 0.0157 7.73
## 7 (Intercept) -13.9 13.3 -1.05
## 8 elevation 0.0156 0.0117 1.33
## 9 slope 0.124 0.0166 7.48
## 10 (Intercept) -12.3 4.45 -2.75
## # ... with 2,990 more rowssimul_estimates %>%
dplyr::filter(term %in% c("elevation", "slope")) %>%
group_by(term) %>%
mutate(x = 1:1000) %>%
ungroup() %>%
mutate(real_value = ifelse(term == "elevation", 0.005, 0.1)) %>%
ggplot(aes(x = x, y = estimate)) +
geom_line() +
facet_wrap(~ term) +
geom_hline(aes(yintercept = real_value, color = term), linetype = 2, size = 0.8) 
- Submit a link to this document in R Pubs to your Moodle.