ANLY 505 - Simulated Data and Linear Mixed Models
Week 2
Xiao Zhou
2019-11-11
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(lme4)## Loading required package: Matrixfit1 = lmer(resp2 ~ elevation + slope + (1|stand))
fit1## 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(fit1)## 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 -0.148Estimated β0 is -21.3146, which is far from the initial parameter -1. Estimated β1 is 0.0206, which is not close to initial parameter 0.005. Estimated β2 is 0.0951, which is much close to the initial parameter 0.1
- Create a function for your model and run 1000 simulations of that model.
library(purrr)
set.seed(18)
mixed_fun = function(nstand = 5, nplot = 4, b0 = -1, b1 = 0.005, b2 = 0.1, 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)
fit1 = lmer(resp2~elevation+slope+(1|stand), data = dframe)
}
mixed_fun()
simsthousand = rerun(1000, mixed_fun())## 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- 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 3.2.0 ✔ readr 1.3.1
## ✔ tibble 2.1.3 ✔ dplyr 0.8.3
## ✔ tidyr 1.0.0 ✔ stringr 1.4.0
## ✔ ggplot2 3.2.0 ✔ forcats 0.4.0## ── Conflicts ───────────────────────────────────────────────────────────────────────────────── tidyverse_conflicts() ──
## ✖ tidyr::expand() masks Matrix::expand()
## ✖ dplyr::filter() masks stats::filter()
## ✖ dplyr::lag() masks stats::lag()
## ✖ tidyr::pack() masks Matrix::pack()
## ✖ tidyr::unpack() masks Matrix::unpack()library(broom)
library(purrr)
variances = simsthousand %>% map_dfr(tidy, effects = "ran_pars", scales = "vcov")
head(variances,6)## # A tibble: 6 x 3
## term group estimate
## <chr> <chr> <dbl>
## 1 var_(Intercept).stand stand 2.02
## 2 var_Observation.Residual Residual 1.03
## 3 var_(Intercept).stand stand 1.03
## 4 var_Observation.Residual Residual 0.812
## 5 var_(Intercept).stand stand 5.76
## 6 var_Observation.Residual Residual 1.22- 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?
library(ggplot2)
library(dplyr)
stand_sims = c(10, 20, 100) %>%
set_names() %>%
map(~replicate(1000, mixed_fun(nstand = .x)))
stand_vars = stand_sims %>%
modify_depth(2, ~tidy(.x, effects = "ran_pars", scales = "vcov")) %>%
map_dfr(bind_rows, .id = "id") %>%
filter(group == "stand")
ggplot(stand_vars, aes(x = estimate)) +
geom_density(fill = "orange", alpha = "0.25") +
facet_wrap(~id) +
geom_vline(xintercept = 4) +theme(legend.position = "bottom", legend.key.width = unit(.1, "cm")) +
labs(x = "Estimated Variance", y = "Density")
As shown in the graphs, when sample size increases, variance will decrease.
- Plot the coefficients of the estimates of elevation and slope. Hint: the x-axis should have 1000 values. Discuss the graphs.
coef = simsthousand %>%
map_dfr(tidy, effects="fixed") %>%
filter(term %in% c("elevation", "slope"))
coef$sequence = rep(1:1000, each=2)
coefmed = coef %>%
group_by(term) %>%
summarise(med=median(estimate))
plot2 = ggplot(coef, aes(sequence, estimate))+
geom_point(size=0.2)+
facet_wrap(~term)+
geom_hline(data=coefmed, aes(yintercept=med), size=0.4)+
labs(x="Replicated numbers", y="Estimate of coefficients")+
theme_bw()
plot2
As shown in the graphs, the median estimated coefficients values are 0.0047 and 0.0998, which are much close to initial parameters 0.005 and 0.1. But the estimated β1 and β2 are scattered, some β1 value are negative. The model does not fit well.
- Submit a link to this document in R Pubs to your Moodle. This assignment is worth 25 points.