ANLY 505 - Simulated Data and Linear Mixed Models
Week 2
Hua Qiu
2019-11-09
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? From the result, the b1 =0.005 and it is much smaller than the initial Parameter.The b2=0.1 and it is close to the initial parameter.The estimated intercept -21.31 is complete different from real value -1.
library(lme4)## Loading required package: MatrixQ1 <- lmer(resp2 ~ 1 + elevation + slope + (1|stand))
summary(Q1)## Linear mixed model fit by REML ['lmerMod']
## Formula: resp2 ~ 1 + 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 -0.148library(lmerTest)##
## Attaching package: 'lmerTest'## The following object is masked from 'package:lme4':
##
## lmer## The following object is masked from 'package:stats':
##
## stepcat("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.
library(purrr)
mix_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
dat = data.frame(resp2, elevation, slope, stand)
lmer(resp2 ~ 1 + elevation + slope + (1|stand), data = dat)
}
mix_fun()## Linear mixed model fit by REML ['lmerModLmerTest']
## Formula: resp2 ~ 1 + elevation + slope + (1 | stand)
## Data: dat
## REML criterion at convergence: 80.9781
## Random effects:
## Groups Name Std.Dev.
## stand (Intercept) 2.3573
## Residual 0.9754
## Number of obs: 20, groups: stand, 5
## Fixed Effects:
## (Intercept) elevation slope
## 10.584601 -0.005464 0.086839simsmix = rerun(1000, mix_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- 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 0.8.3 √ stringr 1.4.0
## √ ggplot2 3.2.0 √ forcats 0.4.0## -- Conflicts ----------------------------------------------------------------------------------------------------------------------------------- tidyverse_conflicts() --
## x tidyr::expand() masks Matrix::expand()
## x dplyr::filter() masks stats::filter()
## x dplyr::lag() masks stats::lag()library(broom)
library(broom.mixed)## Registered S3 methods overwritten by 'broom.mixed':
## method from
## augment.lme broom
## augment.merMod broom
## glance.lme broom
## glance.merMod broom
## glance.stanreg broom
## tidy.brmsfit broom
## tidy.gamlss broom
## tidy.lme broom
## tidy.merMod broom
## tidy.rjags broom
## tidy.stanfit broom
## tidy.stanreg broom##
## Attaching package: 'broom.mixed'## The following object is masked from 'package:broom':
##
## tidyMCMCvariances <- simsmix %>% map_dfr(tidy, effects = "ran_pars", scales = "vcov")
variances %>% print(n = 6)## # A tibble: 2,000 x 4
## effect group term estimate
## <chr> <chr> <chr> <dbl>
## 1 ran_pars stand var__(Intercept) 2.61
## 2 ran_pars Residual var__Observation 1.11
## 3 ran_pars stand var__(Intercept) 9.73
## 4 ran_pars Residual var__Observation 1.36
## 5 ran_pars stand var__(Intercept) 0.827
## 6 ran_pars Residual var__Observation 0.914
## # ... with 1,994 more rows- 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? The real variance of stand is 4. The estimate becomes more accurate as the sample size increases as We can see from the plot.The variance decrease as the sample size increases
library(ggplot2)
library(dplyr)
stand_sims = c(10, 50, 200) %>%
set_names() %>%
map(~replicate(1000, mix_fun(nstand = .x)))## boundary (singular) fit: see ?isSingularstand_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)
- Plot the coefficients of the estimates of elevation and slope. Hint: the x-axis should have 1000 values. Discuss the graphs. From the graph ,we can tell Defined Elevation(0.005) and slope(0.1) and mean of Elevation is exact same.
library(furrr)## Loading required package: futuresimsest <- simsmix %>%
future_map(tidy, effects = "fixed") %>%
bind_rows()
simsest %>%
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 = 4, size = 0.5) +
theme_bw()
- Submit a link to this document in R Pubs to your Moodle. This assignment is worth 25 points.