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assignments/search.json

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     "href": "template9.html#r2-model",
     "title": "Notebook for Assignment 9",
     "section": "3.3 R2 Model",
-    "text": "3.3 R2 Model\nEstimate the model, amend the code below\n\nfit2 <- brm(Gmat ~ ., data = studentstd_Gmat,\n            seed = SEED,\n            normalize = FALSE,\n            prior=c(prior(R2D2(mean_R2 = 0.5, prec_R2 = 1, cons_D2 = 1,\n                               autoscale = TRUE),class=b)),\n            backend = \"cmdstanr\")\n\nStart sampling\n\n\nRunning MCMC with 4 sequential chains...\n\nChain 1 Iteration:    1 / 2000 [  0%]  (Warmup) \nChain 1 Iteration:  100 / 2000 [  5%]  (Warmup) \nChain 1 Iteration:  200 / 2000 [ 10%]  (Warmup) \nChain 1 Iteration:  300 / 2000 [ 15%]  (Warmup) \nChain 1 Iteration:  400 / 2000 [ 20%]  (Warmup) \nChain 1 Iteration:  500 / 2000 [ 25%]  (Warmup) \nChain 1 Iteration:  600 / 2000 [ 30%]  (Warmup) \nChain 1 Iteration:  700 / 2000 [ 35%]  (Warmup) \nChain 1 Iteration:  800 / 2000 [ 40%]  (Warmup) \nChain 1 Iteration:  900 / 2000 [ 45%]  (Warmup) \nChain 1 Iteration: 1000 / 2000 [ 50%]  (Warmup) \nChain 1 Iteration: 1001 / 2000 [ 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Iteration:  200 / 2000 [ 10%]  (Warmup) \nChain 4 Iteration:  300 / 2000 [ 15%]  (Warmup) \nChain 4 Iteration:  400 / 2000 [ 20%]  (Warmup) \nChain 4 Iteration:  500 / 2000 [ 25%]  (Warmup) \nChain 4 Iteration:  600 / 2000 [ 30%]  (Warmup) \nChain 4 Iteration:  700 / 2000 [ 35%]  (Warmup) \nChain 4 Iteration:  800 / 2000 [ 40%]  (Warmup) \nChain 4 Iteration:  900 / 2000 [ 45%]  (Warmup) \nChain 4 Iteration: 1000 / 2000 [ 50%]  (Warmup) \nChain 4 Iteration: 1001 / 2000 [ 50%]  (Sampling) \nChain 4 Iteration: 1100 / 2000 [ 55%]  (Sampling) \nChain 4 Iteration: 1200 / 2000 [ 60%]  (Sampling) \nChain 4 Iteration: 1300 / 2000 [ 65%]  (Sampling) \nChain 4 Iteration: 1400 / 2000 [ 70%]  (Sampling) \nChain 4 Iteration: 1500 / 2000 [ 75%]  (Sampling) \nChain 4 Iteration: 1600 / 2000 [ 80%]  (Sampling) \nChain 4 Iteration: 1700 / 2000 [ 85%]  (Sampling) \nChain 4 Iteration: 1800 / 2000 [ 90%]  (Sampling) \nChain 4 Iteration: 1900 / 2000 [ 95%]  (Sampling) \nChain 4 Iteration: 2000 / 2000 [100%]  (Sampling) \nChain 4 finished in 0.7 seconds.\n\nAll 4 chains finished successfully.\nMean chain execution time: 0.7 seconds.\nTotal execution time: 3.0 seconds.\n\n\nVisualise Posteriors\n\n# Plot marginal posteriors\ndraws2 <- as_draws_df(fit2, variable=paste0('b_',predictors)) |>\n  set_variables(predictors)\np <- mcmc_areas(draws2,\n                prob_outer=0.98, area_method = \"scaled height\") +\n  xlim(c(-3,3))\n\nScale for x is already present.\nAdding another scale for x, which will replace the existing scale.\n\np <- p + scale_y_discrete(limits = rev(levels(p$data$parameter)))\n\nScale for y is already present.\nAdding another scale for y, which will replace the existing scale.\n\np\n\n\n\n\n\n\n\n\nFind prior predictive R2. This should be equal to the prior you set on R2 if all predictors are scaled to have unit variance. Since binary predictors were not standardised, the prior predictive might look slightly different, let’s compute it for illustration\n\nppR2_2<-numeric()\nsims <- 4000\n\nfor (i in 1:sims) {\n  sigma2 <- rstudent_t(1,3,0,3)^2\n  R2 <- rbeta(1,1,2)\n  tau2 <- R2/(1-R2)\n  psi <- as.numeric(rdirichlet(1,rep(1,dim(X)[2])))\n  beta <- rnorm(dim(X)[2])*sqrt(sigma2*tau2*psi)\n  mu <- as.matrix(X)%*%as.vector(beta)\n  muvar <- var(mu)\n  ppR2_2[i] <- muvar/(muvar+sigma2)\n}\n\nNow, find the predictive R2 and compare to the posterior.\n\n# Prior vs posterior R2\ndata <- data.frame(Prior=ppR2_2,Posterior=bayes_R2(fit2, summary=FALSE)) \nnames(data) <- c(\"Prior\",\"Posterior\")\nmcmc_hist(data,\n          breaks=seq(0,1,length.out=100),\n          facet_args = list(nrow = 2)) +\n  facet_text(size = 13) +\n  scale_x_continuous(limits = c(0,1), expand = c(0, 0),\n                     labels = c(\"0\",\"0.25\",\"0.5\",\"0.75\",\"1\")) +\n  theme(axis.line.y = element_blank()) +\n  xlab(\"Bayesian R^2\")\n\n\n\n\n\n\n\n\nSummaries of R2 and LOO-R2\n\n# Bayes-R2 and LOO-R2\nbayes_R2(fit2) |> as.data.frame() |> tt()\n\n\n\n    \n\n    \n    \n      \n        \n        \n              \n                Estimate\n                Est.Error\n                Q2.5\n                Q97.5\n              \n        \n        \n        \n                \n                  0.2527685\n                  0.03639153\n                  0.180655\n                  0.3217405\n                \n        \n      \n    \n\n\nloo_R2(fit2) |> as.data.frame() |> tt()\n\n\n\n    \n\n    \n    \n      \n        \n        \n              \n                Estimate\n                Est.Error\n                Q2.5\n                Q97.5\n              \n        \n        \n        \n                \n                  0.2153771\n                  0.03328053\n                  0.1482607\n                  0.278233",
+    "text": "3.3 R2 Model\nEstimate the model, amend the code below\n\nfit2 <- brm(Gmat ~ ., data = studentstd_Gmat,\n            seed = SEED,\n            normalize = FALSE,\n            prior=c(prior(R2D2(mean_R2 = 0.5, prec_R2 = 1, cons_D2 = 1,\n                               autoscale = TRUE),class=b)),\n            backend = \"cmdstanr\")\n\nStart sampling\n\n\nRunning MCMC with 4 sequential chains...\n\nChain 1 Iteration:    1 / 2000 [  0%]  (Warmup) \nChain 1 Iteration:  100 / 2000 [  5%]  (Warmup) \nChain 1 Iteration:  200 / 2000 [ 10%]  (Warmup) \nChain 1 Iteration:  300 / 2000 [ 15%]  (Warmup) \nChain 1 Iteration:  400 / 2000 [ 20%]  (Warmup) \nChain 1 Iteration:  500 / 2000 [ 25%]  (Warmup) \nChain 1 Iteration:  600 / 2000 [ 30%]  (Warmup) \nChain 1 Iteration:  700 / 2000 [ 35%]  (Warmup) \nChain 1 Iteration:  800 / 2000 [ 40%]  (Warmup) \nChain 1 Iteration:  900 / 2000 [ 45%]  (Warmup) \nChain 1 Iteration: 1000 / 2000 [ 50%]  (Warmup) \nChain 1 Iteration: 1001 / 2000 [ 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(Sampling) \nChain 4 finished in 0.8 seconds.\n\nAll 4 chains finished successfully.\nMean chain execution time: 0.7 seconds.\nTotal execution time: 3.1 seconds.\n\n\nVisualise Posteriors\n\n# Plot marginal posteriors\ndraws2 <- as_draws_df(fit2, variable=paste0('b_',predictors)) |>\n  set_variables(predictors)\np <- mcmc_areas(draws2,\n                prob_outer=0.98, area_method = \"scaled height\") +\n  xlim(c(-3,3))\n\nScale for x is already present.\nAdding another scale for x, which will replace the existing scale.\n\np <- p + scale_y_discrete(limits = rev(levels(p$data$parameter)))\n\nScale for y is already present.\nAdding another scale for y, which will replace the existing scale.\n\np\n\n\n\n\n\n\n\n\nFind prior predictive R2. This should be equal to the prior you set on R2 if all predictors are scaled to have unit variance. Since binary predictors were not standardised, the prior predictive might look slightly different, let’s compute it for illustration\n\nppR2_2<-numeric()\nsims <- 4000\n\nfor (i in 1:sims) {\n  sigma2 <- rstudent_t(1,3,0,3)^2\n  R2 <- rbeta(1,1,2)\n  tau2 <- R2/(1-R2)\n  psi <- as.numeric(rdirichlet(1,rep(1,dim(X)[2])))\n  beta <- rnorm(dim(X)[2])*sqrt(sigma2*tau2*psi)\n  mu <- as.matrix(X)%*%as.vector(beta)\n  muvar <- var(mu)\n  ppR2_2[i] <- muvar/(muvar+sigma2)\n}\n\nNow, find the predictive R2 and compare to the posterior.\n\n# Prior vs posterior R2\ndata <- data.frame(Prior=ppR2_2,Posterior=bayes_R2(fit2, summary=FALSE)) \nnames(data) <- c(\"Prior\",\"Posterior\")\nmcmc_hist(data,\n          breaks=seq(0,1,length.out=100),\n          facet_args = list(nrow = 2)) +\n  facet_text(size = 13) +\n  scale_x_continuous(limits = c(0,1), expand = c(0, 0),\n                     labels = c(\"0\",\"0.25\",\"0.5\",\"0.75\",\"1\")) +\n  theme(axis.line.y = element_blank()) +\n  xlab(\"Bayesian R^2\")\n\n\n\n\n\n\n\n\nSummaries of R2 and LOO-R2\n\n# Bayes-R2 and LOO-R2\nbayes_R2(fit2) |> as.data.frame() |> tt()\n\n\n\n    \n\n    \n    \n      \n        \n        \n              \n                Estimate\n                Est.Error\n                Q2.5\n                Q97.5\n              \n        \n        \n        \n                \n                  0.2527685\n                  0.03639153\n                  0.180655\n                  0.3217405\n                \n        \n      \n    \n\n\nloo_R2(fit2) |> as.data.frame() |> tt()\n\n\n\n    \n\n    \n    \n      \n        \n        \n              \n                Estimate\n                Est.Error\n                Q2.5\n                Q97.5\n              \n        \n        \n        \n                \n                  0.2147699\n                  0.03301587\n                  0.148098\n                  0.2784193",
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       "Templates",
       "Notebook for Assignment 9"