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 = 1

Simulate 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):

1. 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: Matrix

fit1 = lmer(resp2 ~ 1 + elevation + slope + (1|stand))
fit1

## Linear mixed model fit by REML ['lmerMod']
## Formula: resp2 ~ 1 + 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      0.09511

Ans: Based on the results, we can see that estimated β2 (0.09511) is close to the initial parameter 0.1. However, estimated β0 (-21.31463) and estimated β1 (0.02060) are not close to the initial parameter setting, which are -1 and 0.005 respectively.

2. 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 ['lmerMod']
## 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     0.086839

times <- 1000
set.seed(16)
simsmix <- rerun(times, mix_fun())

## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.00350336
## (tol = 0.002, component 1)

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.0045514
## (tol = 0.002, component 1)

## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.00363544
## (tol = 0.002, component 1)

## boundary (singular) fit: see ?isSingular

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.00439561
## (tol = 0.002, component 1)

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.00239513
## (tol = 0.002, component 1)

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.00219495
## (tol = 0.002, component 1)

## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular
## boundary (singular) fit: see ?isSingular

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.00322495
## (tol = 0.002, component 1)

## boundary (singular) fit: see ?isSingular

3. Extract the stand and residual variances from this simulation run. Print the first 6 rows of the data.

library(broom)
library(kableExtra)


variances <- simsmix %>% map_dfr(tidy, effects = "ran_pars", scales = "vcov")
variances %>% print(n = 6)

## # A tibble: 2,000 x 3
##   term                     group    estimate
##   <chr>                    <chr>       <dbl>
## 1 var_(Intercept).stand    stand       1.21 
## 2 var_Observation.Residual Residual    1.36 
## 3 var_(Intercept).stand    stand       5.56 
## 4 var_Observation.Residual Residual    0.951
## 5 var_(Intercept).stand    stand       2.61 
## 6 var_Observation.Residual Residual    1.11 
## # … with 1,994 more rows

4. 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)

## 
## Attaching package: 'dplyr'

## The following object is masked from 'package:kableExtra':
## 
##     group_rows

## The following objects are masked from 'package:stats':
## 
##     filter, lag

## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union

stand_sims = c(10, 50, 200) %>%
  set_names() %>%
  map(~replicate(1000, mix_fun(nstand = .x)))

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.00848934
## (tol = 0.002, component 1)

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.00617958
## (tol = 0.002, component 1)

## boundary (singular) fit: see ?isSingular

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.0154332
## (tol = 0.002, component 1)

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.00201409
## (tol = 0.002, component 1)

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.0135423
## (tol = 0.002, component 1)

stand_vars = stand_sims %>%
  modify_depth(2, ~tidy(.x, effects = "ran_pars", scales = "vcov")) %>%
  map_dfr(bind_rows, .id = "num") %>%
  filter(group == "stand")

groupmed <- stand_vars %>%
  group_by(num) %>%
  summarise(medvar=median(estimate))

f1 <- ggplot(stand_vars, aes(x=estimate))+
  geom_density(fill="orange", alpha=0.5)+
  facet_wrap(~num, labeller=as_labeller(function(x) paste("Number of stands:", x, sep=" ")))+
  geom_vline(aes(xintercept=sds^2, linetype="True Variance"), size=0.4)+
  geom_vline(data=groupmed, aes(xintercept=medvar, linetype="Median Variance"), size=0.4)+
  scale_linetype_manual(name="", values=c(2,1))+
  labs(x="Estimated Variance", y="Density")+
  theme_bw()
f1

Ans: Based on the plots, we can see that the variace decreases as sample size increases.

5. Plot the coefficients of the estimates of elevation and slope. Hint: the x-axis should have 1000 values. Discuss the graphs.

 coef <- simsmix %>%
  map_dfr(tidy, effects="fixed") %>%
  filter(term %in% c("elevation", "slope"))
coef$sequence <- rep(1:times, each=2)

coefmed <- coef %>%
  group_by(term) %>%
  summarise(med=median(estimate))
coefmed 

## # A tibble: 2 x 2
##   term          med
##   <chr>       <dbl>
## 1 elevation 0.00472
## 2 slope     0.0998

f2 <- 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="Sequence number", y="Estimated coefficients")+
  theme_bw()
f2

Ans : Based on the plot, the median of elevation is around 0.0047 and the median of slope is 0.0998. The values are close to initial parameter setting (0.005, 0.1). The estimates are rather scattered. Visually speaking, variance of slope estimates is larger (more sparse) than variance of elevation estimates.

6. Submit a link to this document in R Pubs to your Moodle. This assignment is worth 25 points.