ANLY 505 - Problem Set #2

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):

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?

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

## 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 -0.148

From the result, the estimated beta2 is close to initial parameters. The estimated beta0 as well as beta1 is far from initial parameter.

Create a function for your model and run 1000 simulations of that model.

myfunction=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
  data1=data.frame(resp2,elevation,slope,stand)
  lmer(resp2~1+elevation+slope+(1|stand),data=data1)
}
myfunction()

## Linear mixed model fit by REML ['lmerMod']
## Formula: resp2 ~ 1 + elevation + slope + (1 | stand)
##    Data: data1
## 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

simulation=replicate(n=1000,expr=myfunction())

## 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| = 0.00350333
## (tol = 0.002, component 1)

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.00455142
## (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| = 0.0036354
## (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| = 0.00439557
## (tol = 0.002, component 1)

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.00239512
## (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

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

variances=simulation %>% map_dfr(tidy,effects="ran_pars",scales="vcov")
variances[1:6,]

## # A tibble: 6 x 3
##   term                     group    estimate
##   <chr>                    <chr>       <dbl>
## 1 var_(Intercept).stand    stand       2.61 
## 2 var_Observation.Residual Residual    1.11 
## 3 var_(Intercept).stand    stand       9.73 
## 4 var_Observation.Residual Residual    1.36 
## 5 var_(Intercept).stand    stand       0.827
## 6 var_Observation.Residual Residual    0.914

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?

plan(multiprocess)
stand_sim=c(5, 50, 100) %>% 
  set_names(c("size_5","size_50","size_100")) %>% 
  future_map(function(.size)replicate(n=1000,expr=myfunction(nstand=.size)))

## 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| = 0.00290078
## (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| = 0.00306721
## (tol = 0.002, component 1)

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.0062793
## (tol = 0.002, component 1)

## 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| = 0.00504501
## (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
## 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

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.00211455
## (tol = 0.002, component 1)

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.00206189
## (tol = 0.002, component 1)

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.00205707
## (tol = 0.002, component 1)

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.00209298
## (tol = 0.002, component 1)

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.00201666
## (tol = 0.002, component 1)

## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl =
## control$checkConv, : Model failed to converge with max|grad| = 0.00210085
## (tol = 0.002, component 1)

stand_var=stand_sim %>%
     modify_depth(2,~tidy(.x,effects="ran_pars",scales="vcov")) %>%
     map_dfr(bind_rows,.id="stand_num") %>%
     filter(group=="stand")
stand_var

## # A tibble: 3,000 x 4
##    stand_num term                  group estimate
##    <chr>     <chr>                 <chr>    <dbl>
##  1 size_5    var_(Intercept).stand stand    6.53 
##  2 size_5    var_(Intercept).stand stand    0.657
##  3 size_5    var_(Intercept).stand stand    3.44 
##  4 size_5    var_(Intercept).stand stand    5.34 
##  5 size_5    var_(Intercept).stand stand    4.65 
##  6 size_5    var_(Intercept).stand stand    0.234
##  7 size_5    var_(Intercept).stand stand    1.21 
##  8 size_5    var_(Intercept).stand stand    2.55 
##  9 size_5    var_(Intercept).stand stand    0.660
## 10 size_5    var_(Intercept).stand stand    0.993
## # ... with 2,990 more rows

ggplot(stand_var,aes(x=estimate))+
     geom_density(fill="red",alpha=.25)+
     facet_wrap(~stand_num)+
     geom_vline(xintercept=4)

From the result, variance decreases when sample size increases (from 5 to 50 to 100).

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

coef=simulation %>%
  map_dfr(tidy,effects="fixed") %>%
  filter(term %in% c("elevation","slope"))
coef

## # A tibble: 2,000 x 4
##    term      estimate std.error statistic
##    <chr>        <dbl>     <dbl>     <dbl>
##  1 elevation   0.0152   0.00426     3.57 
##  2 slope       0.121    0.0157      7.73 
##  3 elevation   0.0156   0.0117      1.33 
##  4 slope       0.124    0.0166      7.48 
##  5 elevation   0.0130   0.00336     3.87 
##  6 slope       0.0937   0.0100      9.32 
##  7 elevation   0.0135   0.0157      0.856
##  8 slope       0.0767   0.0152      5.05 
##  9 elevation  -0.0235   0.0132     -1.78 
## 10 slope       0.127    0.0106     12.0  
## # ... with 1,990 more rows

coef %>% 
  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()

The plot shows the estimated coefficients are around 0.0047 (red line) and 0.0998 (green line). These are close to the initial parameters of 0.005 and 0.1 respectively. Each vertical line indicates one estimated coefficients beta1 and beta2 in each plot. These are around the line but are scattered, meaning the model might not fit well.

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

ANLY 505 - Problem Set #2

Simulation of Linear Mixed Models

Li Li

2019-08-26