WAMBS Blavaan Tutorial

In this tutorial you follow the steps of the When-to-Worry-and-How-to-Avoid-the-Misuse-of-Bayesian-Statistics – checklist (the WAMBS-checklist)

Preparation

This tutorial expects:

  • Any installed version of JAGS
  • Installation of R packages rjags, lavaan and blavaan.This tutorial was made using Blavaan version 0.3-2 and Lavaan version 0.6-1
  • Basic knowledge of hypothesis testing
  • Basic knowledge of correlation and regression
  • Basic knowledge of Bayesian inference
  • Basic knowledge of coding in R
Check the WAMBS checklist here

WAMBS checklist

When to worry, and how to Avoid the Misuse of Bayesian Statistics

To be checked before estimating the model

  1. Do you understand the priors?

To be checked after estimation but before inspecting model results

  1. Does the trace-plot exhibit convergence?
  2. Does convergence remain after doubling the number of iterations?
  3. Does the posterior distribution histogram have enough information?
  4. Do the chains exhibit a strong degree of autocorrelation?
  5. Do the posterior distributions make substantive sense?

Understanding the exact influence of the priors

  1. Do different specification of the multivariate variance priors influence the results?
  2. Is there a notable effect of the prior when compared with non-informative priors?
  3. Are the results stable from a sensitivity analysis?
  4. Is the Bayesian way of interpreting and reporting model results used?

 

Example Data

The data we be use for this exercise is based on a study about predicting PhD-delays (Van de Schoot, Yerkes, Mouw and Sonneveld 2013). The data can be downloaded here.Among many other questions, the researchers asked the Ph.D. recipients how long it took them to finish their Ph.D. thesis (n=333). It appeared that Ph.D. recipients took an average of 59.8 months (five years and four months) to complete their Ph.D. trajectory. The variable B3_difference_extra measures the difference between planned and actual project time in months (mean=9.96, minimum=-31, maximum=91, sd=14.43). For more information on the sample, instruments, methodology and research context we refer the interested reader to the paper.

For the current exercise we are interested in the question whether age (M = 30.7, SD = 4.48, min-max = 26-69) of the Ph.D. recipients is related to a delay in their project.

The relation between completion time and age is expected to be non-linear. This might be due to that at a certain point in your life (i.e., mid thirties), family life takes up more of your time than when you are in your twenties or when you are older.

So, in our model the gap (B3_difference_extra) is the dependent variable and age (E22_Age) and age\(^2\)(E22_Age_Squared ) are the predictors. The data can be found in the file phd-delays.csv .

Question: Write down the null and alternative hypotheses that represent this question. Which hypothesis do you deem more likely?
Answer

\(H_0:\) Age\(^2\) is not related to a delay in the PhD projects.

\(H_1:\) Age\(^2\) is related to a delay in the PhD projects.

 

Preparation – Importing and Exploring Data

You can find the data in the file phd-delays.csv , which contains all variables that you need for this analysis. Although it is a .csv-file, you can directly load it into R using the following syntax:

library(rjags) # if you dont have this, or any other package you need, installed yet, please use the install.packages("rjags") command.
library(blavaan)
data <- read.csv2(file="phd-delays.csv")
colnames(data)<- c("diff", "child", "sex","age","age2")
library(psych) #to get some extended summary statistics
library(tidyverse) # needed for data manipulation.

Once you loaded in your data, it is advisable to check whether your data import worked well. Therefore, first have a look at the summary statistics of your data. you can do this by using the describe() function.

Question: Have all your data been loaded in correctly? That is, do all data points substantively make sense? If you are unsure, go back to the .csv-file to inspect the raw data.
Answer
describe(data)
##       vars   n    mean     sd median trimmed    mad min  max range  skew
## diff     1 333    9.97  14.43      5    6.91   7.41 -31   91   122  2.21
## child    2 333    0.18   0.38      0    0.10   0.00   0    1     1  1.66
## sex      3 333    0.52   0.50      1    0.52   0.00   0    1     1 -0.08
## age      4 333   31.68   6.86     30   30.39   2.97  26   80    54  4.45
## age2     5 333 1050.22 656.39    900  928.29 171.98 676 6400  5724  6.03
##       kurtosis    se
## diff      5.92  0.79
## child     0.75  0.02
## sex      -2.00  0.03
## age      24.99  0.38
## age2     42.21 35.97

The descriptive statistics make sense:

diff: Mean (9.97), SE (0.79)

age: Mean (31.68), SE (0.38)

age2: Mean (1050.22), SE (35.97)

 

Step 1: Do you understand the priors?

1.Do you understand the priors?

Before actually looking at the data we first need to think about the prior distributions and hyperparameters for our model. For the current model, there are four priors:

  • the intercept
  • the two regression parameters (\(\beta_1\) for the relation with AGE and \(\beta_2\) for the relation with AGE2)
  • the residual variance (\(\in\))

We first need to determine which distribution to use for the priors. Let’s use for the

  • intercept a normal prior with \(\mathcal{N}(\mu_0, \sigma^{2}_{0})\), where \(\mu_0\) is the prior mean of the distribution and \(\sigma^{2}_{0}\) is the variance parameter
  • \(\beta_1\) a normal prior with \(\mathcal{N}(\mu_1, \sigma^{2}_{1})\)
  • \(\beta_2\) a normal prior with \(\mathcal{N}(\mu_2, \sigma^{2}_{2})\)
  • \(\in\) an Inverse Gamma distribution with \(IG(\kappa_0,\theta_0)\), where \(\kappa_0\) is the shape parameter of the distribution and \(\theta_0\) the rate parameter

Next, we need to specify actual values for the hyperparameters of the prior distributions. Let’s say we found a previous study and based on this study the following hyperparameters can be specified:

  • intercept\(\sim \mathcal{N}(-35, 20)\)
  • \(\beta_1 \sim \mathcal{N}(.8, 5)\)
  • \(\beta_2 \sim \mathcal{N}(0, 10)\)
  • \(\in \sim IG(.5, .5)\)

It is a good idea to plot these distribution to see how they look. To do so, one easy way is to sample a lot of values from one of these distributions and make a density plot out of it, see the code below. Replace the ‘XX’ with the values of the hyperparematers.

par(mfrow=c(2,2))
plot(density(rnorm(n=100000,mean=XX, sd=sqrt(XX))), main= "prior intercept") # the rnorm function uses the standard devation instead of variance, that is why we use the sqrt
plot(density(rnorm(n=100000,mean=XX, sd=sqrt(XX))), main= "effect Age")
plot(density(rnorm(n=100000,mean=XX, sd=sqrt(XX))), main= "effect Age^2")
Answer
par(mfrow=c(2,2))
plot(density(rnorm(n=100000,mean=-35, sd=sqrt(20))), main= "prior intercept") # the rnorm function uses the standard devation instead of variance, that is why we use the sqrt
plot(density(rnorm(n=100000,mean=.8, sd=sqrt(5))), main= "effect Age")
plot(density(rnorm(n=100000,mean=0, sd=sqrt(10))), main= "effect Age^2")

We can also plot what the our expected delay would be (like we did in the Blavaan regression assignment) given these priors. With these priors the regression formula would be: \(delay=-35+ .8*age + 0*age^2\). These are just the means of the priors and do not yet qualify the different levels of uncertainty. Replace the ‘XX’ in the following code with the prior means.

years<- 20:80
delay<- XX+XX*years+XX*years^2
plot(years, delay, type= "l")
Answer
years<- 20:80
delay<--35+.8*years+0*years^2
plot(years, delay, type= "l")

We can also plot what the our expected delay would be (like we did in the Blavaan regression assignment) given these priors. With these priors the regression formula would be: \(delay=-35+ .8*age + 0*age^2\). These are just the means and do not yet qualify the different levels of uncertainty. Feel free to change some of the regression coefficients and see how this chances the curve.

years<- 20:80
delay<--35+.8*years+0*years^2
plot(years, delay, type= "l")

Step 2: Run the model and check for convergence

Specify the regression model you want to analyze and run the regression with the blavaan() function. The priors for the intercept, the regression coefficients and the variance are specified in the regression model below.

To run a multiple regression with Blavaan, you first specify the model, then fit the model and finally acquire the summary (similar to the frequentist model in Lavaan). The model is specified as follows:

  1. A dependent variable we want to predict.
  2. A “~”, that we use to indicate that we now give the other variables of interest. (comparable to the ‘=’ of the regression equation).
  3. The different independent variables separated by the summation symbol ‘+’.
  4. We insert that the dependent variable has a variance and that we want an intercept.
  5. Finally, we can use the prior() command to specify any prior we want. Be careful, Blavaan uses precision instead of variance in the normal distribution. The precision is the reciprocal of the variance, so a variance of 5 corresponds to a precision of .2 and a variance of 10 corresponds to a precision of .1.

For more information on the basics of (b)lavaan and how to specify other priors, see their website

model.regression <- '#the regression model
                    diff ~ prior("dnorm(0.8, 0.2)")*age + prior("dnorm(0, 0.1)")*age2

                    #show that dependent variable has variance
                    diff ~~ prior("dgamma(.5, .5)")*diff

                    #we want to have an intercept
                    diff ~ prior("dnorm(-35, 0.05)")*1'

2. Does the trace-plot exhibit convergence?

First we run the anlysis with only a short burnin period of 250 samples and then take another 500 samples. In addition Blavaan needs an adaptation period, which is by default a 1000 samples. We do not change this default.

fit.bayesfewsample <- blavaan(model.regression, data=data, n.chains=3, burnin =250, sample=500, test="none", seed=c(21,08,2018)) 
# the test="none" input stops the calculations of some posterior checks, we do not need at this moment and speeds up the process. 
# the seed command is simply to guarantee the same exact result when running the sampler multiple times. You do not have to set this.
 #Obtain  MCMC taceplot plots
plot(fit.bayesfewsample, pars=1:4, plot.type="trace")

It seems like the trace (caterpillar) plots are not neatly converged into one each other (we ideally want one fat caterpillar, like the one for psi[1,1,1]). This indicates we need more samples.

We can check if the chains convergenced by having a look at the convergence diagnostics. Two of these diagnostics of interest include the Gelman and Rubin diagnostic and the Geweke diagnostic.

  • The Gelman-Rubin Diagnostic shows the PSRF values (using the within and between chain variability). You should look at the Upper CI/Upper limit, which are all should be close to 1. If they aren’t close to 1, you should use more iterations. Note: The Gelman and Rubin diagnostic is also automatically given in the summary of blavaan under the column PSRF.
  • The Geweke Diagnostic shows the z-scores for a test of equality of means between the first and last parts of each chain, which should be <1.96. A separate statistic is calculated for each variable in each chain. In this way it check whether a chain has stabalized. If this is not the case, you should increase the number of iterations. In the plots you should check how often values exceed the boundary lines of the z-scores. Scores above 1.96 or below -1.96 mean that the two portions of the chain significantly differ and full chain convergence was not obtained.

To obtain the Gelman and Rubin diagnostic use:

mcmc.list <- blavInspect(fit.bayesfewsample, what="mcmc")
gelman.diag(mcmc.list)
## Potential scale reduction factors:
## 
##              Point est. Upper C.I.
## beta[1,2,1]        1.24       1.71
## beta[1,3,1]        1.22       1.63
## psi[1,1,1]         1.01       1.03
## alpha[1,1,1]       1.16       1.49
## 
## Multivariate psrf
## 
## 1.2
gelman.plot(mcmc.list)

To obtain the Geweke diagnostic use:

geweke.plot(mcmc.list)

These statistics confirm that the chains have not converged. Therefore we run the same analysis with more samples.

fit.bayes <- blavaan(model.regression, data=data, n.chains=3, burnin =20000, sample=40000, test="none",  seed=c(21,08,2018))

Obtain the trace plots again.

plot(fit.bayes, pars=1:4, plot.type="trace")
## Generating plots...

Obtain the Gelman and Rubin diagnostic again.

mcmc.list <- blavInspect(fit.bayes, what="mcmc")
gelman.diag(mcmc.list)
## Potential scale reduction factors:
## 
##              Point est. Upper C.I.
## beta[1,2,1]        1.01       1.01
## beta[1,3,1]        1.00       1.01
## psi[1,1,1]         1.00       1.00
## alpha[1,1,1]       1.00       1.01
## 
## Multivariate psrf
## 
## 1
gelman.plot(mcmc.list)

Obtain Geweke diagnostic again.

geweke.plot(mcmc.list)

Now we see that the Gelman and Rubin diagnostic (PRSF) is close to 1 for all parameters and the the Geweke diagnostic is not >1.96.

3. Does convergence remain after doubling the number of iterations?

As is recommended in the WAMBS checklist, we double the amount of iterations to check for local convergence.

fit.bayesdouble <- blavaan(model.regression, data=data, n.chains=3, burnin =40000, sample=80000, test="none",  seed=c(21,08,2018))

You should again have a look at the above-mentioned convergence statistics, but we can also compute the relative bias to inspect if doubling the number of iterations influences the posterior parameter estimates (\(bias= 100*\frac{(initial converged model- model with double iteration)}{model with double iteration}\)). In order to preserve clarity we just calculate the bias of the two regression coefficients.

You should combine the relative bias in combination with substantive knowledge about the metric of the parameter of interest to determine when levels of relative deviation are negligible or problematic. For example, with a regression coefficient of 0.001, a 5% relative deviation level might not be substantively relevant. However, with an intercept parameter of 50, a 10% relative deviation level might be quite meaningful. The specific level of relative deviation should be interpreted in the substantive context of the model. Some examples of interpretations are:

  • if relative deviation is < |5|%, then do not worry;
  • if relative deviation > |5|%, then rerun with 4x nr of iterations.

Question: calculate the relative bias. Are you satisfied with number of iterations, or do you want to re-run the model with even more iterations?

Answer

To get the relative bias simply save the means of the regression coefficients for the two different analyses and compute the bias.

estimates<-colMeans(as.matrix(mcmc.list)[,c("beta[1,2,1]","beta[1,3,1]")])
mcmc.list.doubleiter <- blavInspect(fit.bayesdouble, what="mcmc")
estimatesdoubleiter<-colMeans(as.matrix(mcmc.list.doubleiter)[,c("beta[1,2,1]","beta[1,3,1]")])
round(100*((estimates-estimatesdoubleiter)/estimatesdoubleiter), 2)
## beta[1,2,1] beta[1,3,1] 
##       -0.03        0.05

The relative bias is small enough (<5%) not worry about it.

 

4. Does the posterior distribution histogram have enough information?

By having a look at the posterior distribution histogram plot(fit.bayes, pars=1:4, plot.type="histogram"), we can check if it has enough information.

Question: What can you conclude about distribution histograms?

Answer
par(mfrow=c(2,2))
plot(fit.bayes, pars=1:4, plot.type="histogram")
## Generating plots...

The histograms look smooth and have no gaps or other abnormalities. Based on this, adding more iterations is not necessary. However, if you arenot satisfied, you can improve the number of iterations again. Posterior distributions do not have to be symmetrical, but in this example they seem to be.

If we compare this with histograms based on the first analysis (with very few iterations, this difference becomes clear:

plot(fit.bayesfewsample, pars=1:4, plot.type="histogram")
## Generating plots...

 

5. Do the chains exhibit a strong degree of autocorrelation?

To obtain information about autocorrelation the following syntax can be used:

par(mfrow=c(2,2))
plot(fit.bayes, pars=1:4, plot.type="autocorr", col='blue')
## Generating plots...

Question: What can you conclude about these autocorrelation plots?

Answer

These results show that autocorrelation is quite stong after a few lags. This means it is important to make sure we ran the analysis with a lot of samples, because with a high autocorrelation it will take longer until the whole parameter space has been identified. For more informtation on autocorrelation check this paper.

 

6. Do the posterior distributions make substantive sense?

We plot the posterior distributions and see if they are unimodel (one peak), if they are clearly centered around one value, if they give a realistic estimate and if they make substantive sense compared to the our prior believes (priors). Using the as.matrix() command we bind the three MCMC chains we have of the different parameters into one chain. Here we plot the posteriors of the regression coefficients. If you want you can also plot the mean and the 95% Posterior HPD Intervals.

MCMCbinded<-as.matrix(mcmc.list)
par(mfrow=c(2,2))
plot(density(MCMCbinded[,'beta[1,2,1]']),main= "Regression coefficient age")
plot(density(MCMCbinded[,'beta[1,3,1]']),main= "Regression coefficient age^2")
plot(density(MCMCbinded[,'alpha[1,1,1]']),main= "Intercept")
plot(density(MCMCbinded[,'psi[1,1,1]']),main= "Variance")

Question: What is your conclusion; do the posterior distributions make sense?

Answer

Yes, we see a clear negative intercept, which makes sense since a value of age = 0 for Ph.D is impossible. We also have have plausible ranges of values for the regression coefficients and a positive variance.

 

step 3: Understanding the exact influence of the priors

First we check the results of the analysis with the priors we used so far.

summary(fit.bayes)
## blavaan (0.3-2) results of 40000 samples after 21000 adapt/burnin iterations
## 
##   Number of observations                           333
## 
##   Number of missing patterns                         1
## 
##   Statistic                               
##   Value                                   
## 
## Parameter Estimates:
## 
## 
## Regressions:
##                    Estimate    Post.SD  HPD.025    HPD.975       PSRF
##   diff ~                                                             
##     age                 2.132    0.207      1.725      2.548    1.001
##     age2               -0.021    0.003     -0.026     -0.015    1.001
##     Prior       
##                 
##   dnorm(0.8,0.2)
##     dnorm(0,0.1)
## 
## Intercepts:
##                    Estimate    Post.SD  HPD.025    HPD.975       PSRF
##    .diff              -35.978    4.175    -44.202    -27.809    1.001
##     Prior       
##  dnorm(-35,0.05)
## 
## Variances:
##                    Estimate    Post.SD  HPD.025    HPD.975       PSRF
##    .diff              196.938   15.397    167.755    227.528    1.000
##     Prior       
##    dgamma(.5,.5)

7. Do different specification of the variance priors influence the results?

To understand how the prior on the residual variance impacts the posterior results, we compare the previous model with a model where different hyperparameters for the Inverse Gamma distribution are specified. To see what the priors in Blavaan are we can use the dpriors() command.

dpriors()
##              nu           alpha          lambda            beta 
## "dnorm(0,1e-3)" "dnorm(0,1e-2)" "dnorm(0,1e-2)" "dnorm(0,1e-2)" 
##          itheta            ipsi             rho           ibpsi 
##  "dgamma(1,.5)"  "dgamma(1,.5)"    "dbeta(1,1)" "dwish(iden,3)" 
##             tau           delta 
##   "dnorm(0,.1)"  "dgamma(1,.5)"

We see that the observed variable precision parameter itheta has a default prior of dgamma(1,.5). We change the hyperparameters in the regression model that was specified in step 2 using the prior() command. After rerunning the analysis we can then calculate a bias to see impact. So far we have used the –\(\in \sim IG(.5, .5)\) prior, but we can also use the –\(\in \sim IG(.01, .01)\) prior and see if doing so makes a difference. to quantify this difference we again calculate a relative bias.

Question: Are the results robust for different specifications of the prior on the residual variance?

ParametersEstimate with IG(.01, .01)Estimate with IG (.5, .5)Bias
Intercept
Age
Age2
Residual variance
Answer
model.regression.difIG <- '#the regression model
                    diff ~ prior("dnorm(0.8, 0.2)")*age + prior("dnorm(0, 0.1)")*age2

                    #show that dependent variable has variance
                    diff ~~ prior("dgamma(.01, .01)")*diff # if you dont specify the prior it will use the default prior. 

                    #we want to have an intercept
                    diff ~ prior("dnorm(-35, 0.05)")*1'

fit.bayes.difIG <- blavaan(model.regression.difIG, data=data, n.chains=3, burnin =20000, sample=40000, test="none",  seed=c(21,08,2018))
summary(fit.bayes.difIG)
## blavaan (0.3-2) results of 40000 samples after 21000 adapt/burnin iterations
## 
##   Number of observations                           333
## 
##   Number of missing patterns                         1
## 
##   Statistic                               
##   Value                                   
## 
## Parameter Estimates:
## 
## 
## Regressions:
##                    Estimate    Post.SD  HPD.025    HPD.975       PSRF
##   diff ~                                                             
##     age                 2.137    0.211      1.735      2.558    1.002
##     age2               -0.021    0.003     -0.026     -0.015    1.001
##     Prior       
##                 
##   dnorm(0.8,0.2)
##     dnorm(0,0.1)
## 
## Intercepts:
##                    Estimate    Post.SD  HPD.025    HPD.975       PSRF
##    .diff              -36.087    4.232    -44.355    -27.833    1.002
##     Prior       
##  dnorm(-35,0.05)
## 
## Variances:
##                    Estimate    Post.SD  HPD.025    HPD.975       PSRF
##    .diff              197.533   15.431    168.298    228.342    1.000
##     Prior       
##  dgamma(.01,.01)
ParametersEstimate with IG(.01, .01)Estimate with IG (.5, .5)Bias
Intercept-36.09-35.98\(100\cdot \frac{-36.09–35.98}{-36.09} = 0.30\%\)
Age2.142.13\(100\cdot \frac{2.14-2.13}{2.14} = 0.47\%\)
Age2-0.021-0.0210
Residual variance197.53196.94\(100\cdot \frac{ 197.53-196.94}{ 197.53} = 0.29\%\)

Yes, the results are robust, because there is only a really small amount of relative bias for the residual variance.

 

8. Is there a notable effect of the prior when compared with non-informative priors?

The default Blavaan priors are non-informative, so we can run the analysis without any specified priors and compare them to the model we have run thus far, using the relative bias, to see if there is a large influence of the priors.

Question: What is your conclusion about the influence of the priors on the posterior results?

ParametersEstimates with default priorsEstimate with informative priorsBias
Intercept
Age
Age2
Residual variance
Answer
model.regression.defaultprior <- '#the regression model
                    diff ~ age + age2

                    #show that dependent variable has variance
                    diff ~~ diff # if you dont specify the prior it will use the default prior. 

                    #we want to have an intercept
                    diff ~ 1'

fit.bayes.defaultpriors<- blavaan(model.regression.defaultprior, data=data, n.chains=3, burnin =20000, sample=40000, test="none",  seed=c(21,08,2018))
summary(fit.bayes.defaultpriors)
## blavaan (0.3-2) results of 40000 samples after 21000 adapt/burnin iterations
## 
##   Number of observations                           333
## 
##   Number of missing patterns                         1
## 
##   Statistic                               
##   Value                                   
## 
## Parameter Estimates:
## 
## 
## Regressions:
##                    Estimate    Post.SD  HPD.025    HPD.975       PSRF
##   diff ~                                                             
##     age                 2.290    0.555       1.24      3.467    1.018
##     age2               -0.022    0.006     -0.034     -0.011    1.017
##     Prior     
##               
##  dnorm(0,1e-2)
##  dnorm(0,1e-2)
## 
## Intercepts:
##                    Estimate    Post.SD  HPD.025    HPD.975       PSRF
##    .diff              -39.310   11.678    -63.651    -16.725    1.018
##     Prior     
##  dnorm(0,1e-3)
## 
## Variances:
##                    Estimate    Post.SD  HPD.025    HPD.975       PSRF
##    .diff              196.539   15.328     167.07    226.983    1.000
##     Prior     
##   dgamma(1,.5)
ParametersEstimates with default priorsEstimate with informative priorsBias
Intercept-39.31-35.98\(100\cdot \frac{-36.09–35.98}{-36.09} = 0.30\%\)
Age2.292.13\(100\cdot \frac{2.29-2.13}{2.29} = 6.98\%\)
Age2-0.022-0.021\(100\cdot \frac{-0.022–0.021}{-0.022} = 4.55\%\)
Residual variance196.53196.94\(100\cdot \frac{ 197.53-196.94}{ 197.53} = 0.21\%\)

The informative priors have quite some influence (up to 7%) on the posterior results of the regression coefficients. This is not a bad thing, just important to keep in mind.

 

Question: Which results do you use to draw conclusion on?

Answer

This really depends on where the priors come from. If for example your informative priors come from a reliable source, you should use them. The most important thing is that you choose your priors accurately, and have good arguments to use them. If not, you shouldn’t use really informative priors and use the results based on the non-informative priors.

 

9. Are the results stable from a sensitivity analysis?

If you still have time left, you can adjust the hyperparameters of the priors upward and downward and re-estimating the model with these varied priors to check for robustness.

From the original paper:

“If informative or weakly-informative priors are used, then we suggest running a sensitivity analysis of these priors. When subjective priors are in place, then there might be a discrepancy between results using different subjective prior settings. A sensitivity analysis for priors would entail adjusting the entire prior distribution (i.e., using a completely different prior distribution than before) or adjusting hyperparameters upward and downward and re-estimating the model with these varied priors. Several different hyperparameter specifications can be made in a sensitivity analysis, and results obtained will point toward the impact of small fluctuations in hyperparameter values. [.] The purpose of this sensitivity analysis is to assess how much of an impact the location of the mean hyperparameter for the prior has on the posterior. [.] Upon receiving results from the sensitivity analysis, assess the impact that fluctuations in the hyperparameter values have on the substantive conclusions. Results may be stable across the sensitivity analysis, or they may be highly instable based on substantive conclusions. Whatever the finding, this information is important to report in the results and discussion sections of a paper. We should also reiterate here that original priors should not be modified, despite the results obtained.”

For more information on this topic, please also refer to this paper.

10. Is the Bayesian way of interpreting and reporting model results used?

For a summary on how to interpret and report models, please refer to https://www.rensvandeschoot.com/bayesian-analyses-where-to-start-and-what-to-report/

summary(fit.bayes)
## blavaan (0.3-2) results of 40000 samples after 21000 adapt/burnin iterations
## 
##   Number of observations                           333
## 
##   Number of missing patterns                         1
## 
##   Statistic                               
##   Value                                   
## 
## Parameter Estimates:
## 
## 
## Regressions:
##                    Estimate    Post.SD  HPD.025    HPD.975       PSRF
##   diff ~                                                             
##     age                 2.132    0.207      1.725      2.548    1.001
##     age2               -0.021    0.003     -0.026     -0.015    1.001
##     Prior       
##                 
##   dnorm(0.8,0.2)
##     dnorm(0,0.1)
## 
## Intercepts:
##                    Estimate    Post.SD  HPD.025    HPD.975       PSRF
##    .diff              -35.978    4.175    -44.202    -27.809    1.001
##     Prior       
##  dnorm(-35,0.05)
## 
## Variances:
##                    Estimate    Post.SD  HPD.025    HPD.975       PSRF
##    .diff              196.938   15.397    167.755    227.528    1.000
##     Prior       
##    dgamma(.5,.5)

In the current model we see that:

  • The estimate for the intercept is -35.98 [-44.20; -27.81]
  • The estimate for the effect of \(age\) is 2.13 [1.73 ; 2.55]
  • The estimate for the effect of \(age^2\) is -0.021 [-0.026; -0.015]

We can see that none of 95% Posterior HPD Intervals for these effects include zero, which means we are can be quite certain that all of the effects are different from 0.

Remember how we plotted the relation between delay and years based on the prior information? Now, do the same with the posterior estimates.

Answer
years<- 20:80
delay<--35+2.13*years-0.02*years^2
plot(years, delay, type= "l")


References

Depaoli, S., & Van de Schoot, R. (2017). Improving transparency and replication in Bayesian statistics: The WAMBS-Checklist. Psychological Methods, 22(2), 240.

Link, W. A., & Eaton, M. J. (2012). On thinning of chains in MCMC. Methods in ecology and evolution, 3(1), 112-115.

van Erp, S., Mulder, J., & Oberski, D. L. (2017). Prior sensitivity analysis in default Bayesian structural equation modeling.

Van de Schoot, R., & Depaoli, S. (2014). Bayesian analyses: Where to start and what to report. European Health Psychologist, 16(2), 75-84.