ANLY 505 - Problem Set #2
Simulation of Linear Mixed Models
Anirudh Gurnani
2019-10-15
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?
library(Matrix)
library(lme4)
resp2 <- b0 + b1*elevation + b2*slope + standeff + ploteff
m1 <- lmer(resp2 ~ elevation + slope +(1|stand))
m1## Linear mixed model fit by REML ['lmerMod']
## Formula: resp2 ~ elevation + slope + (1 | stand)
## REML criterion at convergence: 81.9874
## Random effects:
## Groups Name Std.Dev.
## stand (Intercept) 1.099
## Residual 1.165
## Number of obs: 20, groups: stand, 5
## Fixed Effects:
## (Intercept) elevation slope
## -21.31463 0.02060 0.09511summary(m1)## Linear mixed model fit by REML ['lmerMod']
## Formula: resp2 ~ elevation + slope + (1 | stand)
##
## REML criterion at convergence: 82
##
## Scaled residuals:
## Min 1Q Median 3Q Max
## -1.65583 -0.62467 -0.01693 0.53669 1.41736
##
## Random effects:
## Groups Name Variance Std.Dev.
## stand (Intercept) 1.208 1.099
## Residual 1.358 1.165
## Number of obs: 20, groups: stand, 5
##
## Fixed effects:
## Estimate Std. Error t value
## (Intercept) -21.314628 6.602053 -3.228
## elevation 0.020600 0.004916 4.190
## slope 0.095105 0.016441 5.785
##
## Correlation of Fixed Effects:
## (Intr) elevtn
## elevation -0.991
## slope 0.049 -0.148Defined intercept(b0) is -1 which is totally different than estimated intercept -21.314628. Defined Elevation(b1) is .005 which is much smaller than estimated elevation 0.020600. Defined slope(b1) is 0.1 which is quite close to estimated slope 0.095105
- Create a function for your model and run 1000 simulations of that model.
set.seed(16)
m1_function = function(nstand = 5, nplot = 4, b0 = -1, b1 = 0.005, b2 = .1, sds = 2, sd = 1)
{
stand = rep(LETTERS[1:nstand], each = nplot)
standeff = rep(rnorm(nstand, 0, sds), 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
dframe = data.frame(resp2, elevation, slope, stand)
m1 = lmer(resp2 ~ elevation + slope + (1|stand), data = dframe)
}
set.seed(16)
m1_simul = replicate (1000, expr = m1_function())## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular- Extract the stand and residual variances from this simulation run. Print the first 6 rows of the data.
library(tidyverse)## ── Attaching packages ────────────────────────────────────────────────────────────── tidyverse 1.2.1 ──## ✔ ggplot2 3.1.1 ✔ purrr 0.3.2
## ✔ tibble 2.1.3 ✔ dplyr 0.8.1
## ✔ tidyr 0.8.3 ✔ stringr 1.4.0
## ✔ readr 1.3.1 ✔ forcats 0.4.0## ── Conflicts ───────────────────────────────────────────────────────────────── tidyverse_conflicts() ──
## ✖ tidyr::expand() masks Matrix::expand()
## ✖ dplyr::filter() masks stats::filter()
## ✖ dplyr::lag() masks stats::lag()library(purrr)
library(tidyr)
library(broom)
tidy(m1, 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("broom")
head(map_dfr(m1_simul, broom::tidy, .id = "model", effects = "ran_pars"))## # A tibble: 6 x 4
## model term group estimate
## <chr> <chr> <chr> <dbl>
## 1 1 sd_(Intercept).stand stand 1.10
## 2 1 sd_Observation.Residual Residual 1.17
## 3 2 sd_(Intercept).stand stand 2.36
## 4 2 sd_Observation.Residual Residual 0.975
## 5 3 sd_(Intercept).stand stand 1.62
## 6 3 sd_Observation.Residual Residual 1.05- 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?
m1_sims <- c(5, 20, 100) %>% set_names() %>% map(~replicate (1000, m1_function(nstand = .x)))## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingularlibrary(dplyr)
var_stand = m1_sims %>% modify_depth(2, ~tidy(.x, effects = "ran_pars", scales = "vcov")) %>% map_dfr(bind_rows, .id = "stand_number") %>% filter (group == "stand")
head(var_stand)## # A tibble: 6 x 4
## stand_number term group estimate
## <chr> <chr> <chr> <dbl>
## 1 5 var_(Intercept).stand stand 6.91
## 2 5 var_(Intercept).stand stand 1.92
## 3 5 var_(Intercept).stand stand 12.0
## 4 5 var_(Intercept).stand stand 6.84
## 5 5 var_(Intercept).stand stand 2.89
## 6 5 var_(Intercept).stand stand 2.52library(forcats)
var_stand = mutate(var_stand, stand_number = fct_inorder(stand_number))
add_prefix <- function(string) { paste(" #Stands= ", string, sep = " ")}
med_group = var_stand %>% group_by(stand_number) %>% summarise(mvar = median(estimate))
library(ggplot2)
ggplot(var_stand, aes(x = estimate)) + geom_density(fill = "blue", alpha = 0.25) + facet_wrap(~stand_number, labeller = as_labeller(add_prefix)) + geom_vline(aes(xintercept = 4, linetype = "True Variance"), size = 0.6) + geom_vline(data = med_group, aes(xintercept = mvar, linetype = "Median Variance"), size = 0.5) + 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 = "Density (Probability)")
When sample size increases, variance decreases. As the sample size increases from 5 to 20 its more evident but not so much when it increases from 20 to 100. Hence the threshold of the relationship between true variance and estimated is 20. So increase in sample will not have any effect on accuracy of variance estimate.
- Plot the coefficients of the estimates of elevation and slope. Hint: the x-axis should have 1000 values. Discuss the graphs.
coeff = m1_simul %>%
map_dfr(tidy, effects = "fixed") %>%
filter(term %in% c("elevation","slope")) %>%
group_by(term)
coeff## # A tibble: 2,000 x 4
## # Groups: term [2]
## term estimate std.error statistic
## <chr> <dbl> <dbl> <dbl>
## 1 elevation 0.0206 0.00492 4.19
## 2 slope 0.0951 0.0164 5.78
## 3 elevation -0.00546 0.00927 -0.589
## 4 slope 0.0868 0.0140 6.21
## 5 elevation 0.0152 0.00426 3.57
## 6 slope 0.121 0.0157 7.73
## 7 elevation 0.0156 0.0117 1.33
## 8 slope 0.124 0.0166 7.48
## 9 elevation 0.0130 0.00336 3.87
## 10 slope 0.0937 0.0100 9.32
## # … with 1,990 more rowslibrary(dplyr)
mean = coeff %>%
summarise_at("estimate", funs( mean))
mean## # A tibble: 2 x 2
## term estimate
## <chr> <dbl>
## 1 elevation 0.00492
## 2 slope 0.0997plot = ggplot(coeff, aes(x=rep(1:1000, each=2),y=estimate))+ geom_point(size=0.3) + facet_wrap(~term) + geom_hline(data=mean,aes(yintercept=mean$estimate,color = term), size = 0.9) + labs(x=" Numbers that replicated", y="Evaluated coefficients")
plot
The mean estimates of coefficient (elevation and slope) are 0.004 and 0.099 respectively which are close to the initial parameters of 0.005 and 0.1.
- Submit a link to this document in R Pubs to your Moodle. This assignment is worth 25 points.