STA 360/602L: Module 5.3

STA 360/602L: Module 5.3Hierarchical normal models with constant variance: multiple groupsDr. Olanrewaju Michael Akande1 / 15

Comparing multiple groupsSuppose we wish to investigate the mean (and distribution) of test scores for students at J different high schools.
2 / 15

Comparing multiple groups

Suppose we wish to investigate the mean (and distribution) of test scores for students at $J$ different high schools.
In each school $j$ , where $j = 1, \ldots, J$ , suppose we test a random sample of $n_j$ students.

2 / 15

Comparing multiple groups

Suppose we wish to investigate the mean (and distribution) of test scores for students at $J$ different high schools.
In each school $j$ , where $j = 1, \ldots, J$ , suppose we test a random sample of $n_j$ students.
Let $y_{ij}$ be the test score for the $i$ th student in school $j$ , with $i = 1,\ldots, n_j$ , with

$y_{ij} | \theta_j, \sigma^2_j \sim \mathcal{N} \left( \theta_j, \sigma^2_j\right)$

where for each school $j$ , $\theta_j$ is the school-wide average test score, and $\sigma^2_j$ is the school-wide variance of individual test scores.

2 / 15

Comparing multiple groups

Suppose we wish to investigate the mean (and distribution) of test scores for students at $J$ different high schools.
In each school $j$ , where $j = 1, \ldots, J$ , suppose we test a random sample of $n_j$ students.
Let $y_{ij}$ be the test score for the $i$ th student in school $j$ , with $i = 1,\ldots, n_j$ , with

$y_{ij} | \theta_j, \sigma^2_j \sim \mathcal{N} \left( \theta_j, \sigma^2_j\right)$

where for each school $j$ , $\theta_j$ is the school-wide average test score, and $\sigma^2_j$ is the school-wide variance of individual test scores.
This is what we did for the the Pygmalion study and job training data.

2 / 15

School testing exampleOption I: Classical inference for each school can be based on large sample 95% CI: ˉyj±1.96√s2j/nj, where ˉyj is the sample average in school j, and s2j is the sample variance in school j.
3 / 15

School testing example

: Classical inference for each school can be based on large sample 95% CI: $\bar{y}_j \pm 1.96 \sqrt{s^2_j/n_j}$ , where $\bar{y}_j$ is the sample average in school $j$ , and $s^2_j$ is the sample variance in school $j$ .
Clearly, we can overfit the data within schools, for example, what if we only have 4 students from one of the schools? $\bar{y}_j$ can be a good estimate if $n_j$ is large but it may be poor if $n_j$ is small.

3 / 15

School testing example

: Classical inference for each school can be based on large sample 95% CI: $\bar{y}_j \pm 1.96 \sqrt{s^2_j/n_j}$ , where $\bar{y}_j$ is the sample average in school $j$ , and $s^2_j$ is the sample variance in school $j$ .
Clearly, we can overfit the data within schools, for example, what if we only have 4 students from one of the schools? $\bar{y}_j$ can be a good estimate if $n_j$ is large but it may be poor if $n_j$ is small.
: alternatively, we might believe that $\theta_j = \mu$ for all $j$ ; that is, all schools have the same mean. This is the assumption (null hypothesis) in ANOVA models for example. We can also set $\sigma^2_j = \sigma^2$ for all $J$ .

3 / 15

School testing example

: Classical inference for each school can be based on large sample 95% CI: $\bar{y}_j \pm 1.96 \sqrt{s^2_j/n_j}$ , where $\bar{y}_j$ is the sample average in school $j$ , and $s^2_j$ is the sample variance in school $j$ .
Clearly, we can overfit the data within schools, for example, what if we only have 4 students from one of the schools? $\bar{y}_j$ can be a good estimate if $n_j$ is large but it may be poor if $n_j$ is small.
: alternatively, we might believe that $\theta_j = \mu$ for all $j$ ; that is, all schools have the same mean. This is the assumption (null hypothesis) in ANOVA models for example. We can also set $\sigma^2_j = \sigma^2$ for all $J$ .
Option I ignores that the $\theta_j$ 's should be reasonably similar, whereas option II ignores any differences between them.

3 / 15

School testing example

: Classical inference for each school can be based on large sample 95% CI: $\bar{y}_j \pm 1.96 \sqrt{s^2_j/n_j}$ , where $\bar{y}_j$ is the sample average in school $j$ , and $s^2_j$ is the sample variance in school $j$ .
Clearly, we can overfit the data within schools, for example, what if we only have 4 students from one of the schools? $\bar{y}_j$ can be a good estimate if $n_j$ is large but it may be poor if $n_j$ is small.
: alternatively, we might believe that $\theta_j = \mu$ for all $j$ ; that is, all schools have the same mean. This is the assumption (null hypothesis) in ANOVA models for example. We can also set $\sigma^2_j = \sigma^2$ for all $J$ .
Option I ignores that the $\theta_j$ 's should be reasonably similar, whereas option II ignores any differences between them.
It would be nice to find a compromise! Borrowing information across, and shrinking our estimate towards a grand mean could be very useful here.

3 / 15

School testing exampleFor the Pygmalion study and job training data, we focused on using priors that are independent between the groups.
4 / 15

School testing example

For the Pygmalion study and job training data, we focused on using priors that are independent between the groups.
For example, in the conjugate case, we would have

$\begin{split} \pi(\theta_j|\sigma^2_j) & = \mathcal{N}\left(\mu_0, \dfrac{\sigma^2_j}{\kappa_0}\right)\\ \pi(\sigma^2_j) & = \mathcal{IG} \left(\dfrac{\nu_0}{2}, \dfrac{\nu_0\sigma_0^2}{2}\right)\\ \end{split}$

for some hyperparameters (constants), $\mu_0$ , $\kappa_0$ , $\nu_0$ , and $\sigma_0^2$ .

4 / 15

School testing example

For the Pygmalion study and job training data, we focused on using priors that are independent between the groups.
For example, in the conjugate case, we would have

$\begin{split} \pi(\theta_j|\sigma^2_j) & = \mathcal{N}\left(\mu_0, \dfrac{\sigma^2_j}{\kappa_0}\right)\\ \pi(\sigma^2_j) & = \mathcal{IG} \left(\dfrac{\nu_0}{2}, \dfrac{\nu_0\sigma_0^2}{2}\right)\\ \end{split}$

for some hyperparameters (constants), $\mu_0$ , $\kappa_0$ , $\nu_0$ , and $\sigma_0^2$ .
In the semi-conjugate case,

$\begin{split} \pi(\theta_j) & = \mathcal{N}\left(\mu_0, \sigma^2_0\right)\\ \pi(\sigma^2_j) & = \mathcal{IG} \left(\dfrac{\nu_0}{2}, \dfrac{\nu_0\gamma_0^2}{2}\right)\\ \end{split}$

for some hyperparameters (constants), $\mu_0$ , $\sigma^2_0$ , $\nu_0$ , and $\gamma_0^2$ .

4 / 15

Hierarchical normal modelInstead, we can assume that the θj's are drawn from a distribution based on the following: conceive of the schools themselves as being a random sample from all possible schools.
5 / 15

Hierarchical normal model

Instead, we can assume that the $\theta_j$ 's are drawn from a distribution based on the following: conceive of the schools themselves as being a random sample from all possible schools.
For now, assume the variance is constant across schools. The hierarchical normal model assumes normal sampling models both within and between groups:

$\begin{split} y_{ij} | \theta_j, \sigma^2 & \sim \mathcal{N} \left( \theta_j, \sigma^2\right); \ \ \ i = 1,\ldots, n_j\\ \theta_j | \mu, \tau^2 & \sim \mathcal{N} \left( \mu, \tau^2 \right); \ \ \ j = 1, \ldots, J, \end{split}$

which gives us an extra level in the prior on the means, and leads to sharing of information across the groups in estimating the group-specific means.

5 / 15

Hierarchical normal model

Instead, we can assume that the $\theta_j$ 's are drawn from a distribution based on the following: conceive of the schools themselves as being a random sample from all possible schools.
For now, assume the variance is constant across schools. The hierarchical normal model assumes normal sampling models both within and between groups:

$\begin{split} y_{ij} | \theta_j, \sigma^2 & \sim \mathcal{N} \left( \theta_j, \sigma^2\right); \ \ \ i = 1,\ldots, n_j\\ \theta_j | \mu, \tau^2 & \sim \mathcal{N} \left( \mu, \tau^2 \right); \ \ \ j = 1, \ldots, J, \end{split}$

which gives us an extra level in the prior on the means, and leads to sharing of information across the groups in estimating the group-specific means.
We have an extra variance parameter $\tau^2$ . Comparing $\tau^2$ to $\sigma^2$ tells us how much of the variation in $Y$ is due to within-group versus between-group variation.

5 / 15

Hierarchical normal model

Standard semi-conjugate priors are given by
$\begin{split} \pi(\mu) & = \mathcal{N}\left(\mu_0, \gamma^2_0\right)\\ \pi(\sigma^2) & = \mathcal{IG} \left(\dfrac{\nu_0}{2}, \dfrac{\nu_0\sigma_0^2}{2}\right)\\ \pi(\tau^2) & = \mathcal{IG} \left(\dfrac{\eta_0}{2}, \dfrac{\eta_0\tau_0^2}{2}\right).\\ \end{split}$

6 / 15

Hierarchical normal model

Standard semi-conjugate priors are given by

$\begin{split} \pi(\mu) & = \mathcal{N}\left(\mu_0, \gamma^2_0\right)\\ \pi(\sigma^2) & = \mathcal{IG} \left(\dfrac{\nu_0}{2}, \dfrac{\nu_0\sigma_0^2}{2}\right)\\ \pi(\tau^2) & = \mathcal{IG} \left(\dfrac{\eta_0}{2}, \dfrac{\eta_0\tau_0^2}{2}\right).\\ \end{split}$

with
- $\mu_0$ : best guess of average of school averages
- $\gamma^2_0$ : set based on plausible ranges of values of $\mu$

6 / 15

Hierarchical normal model

Standard semi-conjugate priors are given by

$\begin{split} \pi(\mu) & = \mathcal{N}\left(\mu_0, \gamma^2_0\right)\\ \pi(\sigma^2) & = \mathcal{IG} \left(\dfrac{\nu_0}{2}, \dfrac{\nu_0\sigma_0^2}{2}\right)\\ \pi(\tau^2) & = \mathcal{IG} \left(\dfrac{\eta_0}{2}, \dfrac{\eta_0\tau_0^2}{2}\right).\\ \end{split}$

with
- $\mu_0$ : best guess of average of school averages
- $\gamma^2_0$ : set based on plausible ranges of values of $\mu$
- $\tau_0^2$ : best guess of variance of school averages
- $\eta_0$ : set based on how tight prior for $\tau^2$ is around $\tau_0^2$

6 / 15

Hierarchical normal model

Standard semi-conjugate priors are given by

$\begin{split} \pi(\mu) & = \mathcal{N}\left(\mu_0, \gamma^2_0\right)\\ \pi(\sigma^2) & = \mathcal{IG} \left(\dfrac{\nu_0}{2}, \dfrac{\nu_0\sigma_0^2}{2}\right)\\ \pi(\tau^2) & = \mathcal{IG} \left(\dfrac{\eta_0}{2}, \dfrac{\eta_0\tau_0^2}{2}\right).\\ \end{split}$

with
- $\mu_0$ : best guess of average of school averages
- $\gamma^2_0$ : set based on plausible ranges of values of $\mu$
- $\tau_0^2$ : best guess of variance of school averages
- $\eta_0$ : set based on how tight prior for $\tau^2$ is around $\tau_0^2$
- $\sigma_0^2$ : best guess of variance of individual test scores around respective school means
- $\nu_0$ : set based on how tight prior for $\sigma^2$ is around $\sigma_0^2$ .

6 / 15

ExchangeabilityThis model relies heavily on exchangeability across units at each level.
7 / 15

Exchangeability

This model relies heavily on exchangeability across units at each level.
For example, we assume the schools are a random sample from the population of all schools, and the students within schools are a random sample of all the students in each school.

7 / 15

Exchangeability

This model relies heavily on exchangeability across units at each level.
For example, we assume the schools are a random sample from the population of all schools, and the students within schools are a random sample of all the students in each school.
This is not always completely true.

7 / 15

Exchangeability

This model relies heavily on exchangeability across units at each level.
For example, we assume the schools are a random sample from the population of all schools, and the students within schools are a random sample of all the students in each school.
This is not always completely true.
Note: we can allow the variance to vary across schools if desired (and we will soon in fact).

7 / 15

ExchangeabilityTurns out that conditional exchangeability would be enough if we control for relevant variables in our modeling.
8 / 15

Exchangeability

Turns out that conditional exchangeability would be enough if we control for relevant variables in our modeling.
For example, the schools in Chapel Hill/Carrboro are not entirely exchangeable.

8 / 15

Exchangeability

Turns out that conditional exchangeability would be enough if we control for relevant variables in our modeling.
For example, the schools in Chapel Hill/Carrboro are not entirely exchangeable.
For example, Phoenix Academy is for students on long-term out-of-school suspension or who need to make up work due to extended absences (e.g., pregnancy), and Memorial Hospital School is for children battling serious illnesses.

8 / 15

Exchangeability

Turns out that conditional exchangeability would be enough if we control for relevant variables in our modeling.
For example, the schools in Chapel Hill/Carrboro are not entirely exchangeable.
For example, Phoenix Academy is for students on long-term out-of-school suspension or who need to make up work due to extended absences (e.g., pregnancy), and Memorial Hospital School is for children battling serious illnesses.
However, if we condition on school type (public, charter, private, special services, home), the schools may then be exchangeable.

8 / 15

Posterior inference

Recall the model is
$\begin{split} y_{ij} | \theta_j, \sigma^2 & \sim \mathcal{N} \left( \theta_j, \sigma^2\right); \ \ \ i = 1,\ldots, n_j\\ \theta_j | \mu, \tau^2 & \sim \mathcal{N} \left( \mu, \tau^2 \right); \ \ \ j = 1, \ldots, J, \end{split}$

9 / 15

Posterior inference

Recall the model is

$\begin{split} y_{ij} | \theta_j, \sigma^2 & \sim \mathcal{N} \left( \theta_j, \sigma^2\right); \ \ \ i = 1,\ldots, n_j\\ \theta_j | \mu, \tau^2 & \sim \mathcal{N} \left( \mu, \tau^2 \right); \ \ \ j = 1, \ldots, J, \end{split}$
Under our prior specification, we can factor the posterior as follows:

$\begin{split} \pi(\theta_1, \ldots, \theta_J, \mu, \sigma^2,\tau^2 | Y) & \boldsymbol{\propto} p(y | \theta_1, \ldots, \theta_J, \mu, \sigma^2,\tau^2)\\ & \ \ \ \ \times p(\theta_1, \ldots, \theta_J | \mu, \sigma^2,\tau^2)\\ & \ \ \ \ \times \pi(\mu, \sigma^2,\tau^2)\\ \\ & \boldsymbol{=} p(y | \theta_1, \ldots, \theta_J, \sigma^2 )\\ & \ \ \ \ \times p(\theta_1, \ldots, \theta_J | \mu,\tau^2)\\ & \ \ \ \ \times \pi(\mu) \cdot \pi(\sigma^2) \cdot \pi(\tau^2)\\ \\ & \boldsymbol{=} \left\{ \prod_{j=1}^{J} \prod_{i=1}^{n_j} p(y_{ij} | \theta_j, \sigma^2 ) \right\}\\ & \ \ \ \ \times \left\{ \prod_{j=1}^{J} p(\theta_j | \mu,\tau^2) \right\}\\ & \ \ \ \ \times\pi(\mu) \cdot \pi(\sigma^2) \cdot \pi(\tau^2)\\ \end{split}$

9 / 15

Full conditional for grand meanThe full conditional distribution of \mu is proportional to the part of the joint posterior \pi(\theta_1, \ldots, \theta_J, \mu, \sigma^2,\tau^2 | Y) that involves \mu.
10 / 15

Full conditional for grand mean

The full conditional distribution of $\mu$ is proportional to the part of the joint posterior $\pi(\theta_1, \ldots, \theta_J, \mu, \sigma^2,\tau^2 | Y)$ that involves $\mu$ .
That is,

$\begin{split} \pi(\mu | \theta_1, \ldots, \theta_J, \sigma^2,\tau^2, Y) & \boldsymbol{\propto} \left\{ \prod_{j=1}^{J} p(\theta_j | \mu,\tau^2) \right\} \cdot \pi(\mu). \end{split}$

10 / 15

Full conditional for grand mean

The full conditional distribution of $\mu$ is proportional to the part of the joint posterior $\pi(\theta_1, \ldots, \theta_J, \mu, \sigma^2,\tau^2 | Y)$ that involves $\mu$ .
That is,

$\begin{split} \pi(\mu | \theta_1, \ldots, \theta_J, \sigma^2,\tau^2, Y) & \boldsymbol{\propto} \left\{ \prod_{j=1}^{J} p(\theta_j | \mu,\tau^2) \right\} \cdot \pi(\mu). \end{split}$
This looks like the full conditional distribution from the one-sample normal case, so you can show that

$\begin{split} \pi(\mu | \theta_1, \ldots, \theta_J, \sigma^2,\tau^2, Y) & = \mathcal{N}\left(\mu_n, \gamma^2_n \right) \ \ \ \ \textrm{where}\\ \\ \gamma^2_n = \dfrac{1}{ \dfrac{J}{\tau^2} + \dfrac{1}{\gamma_0^2} } ; \ \ \ \ \ \ \ \ \mu_n = \gamma^2_n \left[ \dfrac{J}{\tau^2} \bar{\theta} + \dfrac{1}{\gamma_0^2} \mu_0 \right] \end{split}$

and $\bar{\theta} = \frac{1}{J} \sum\limits^J_{j=1} \theta_j$ .

10 / 15

Full conditionals for group meansSimilarly, the full conditional distribution of each \theta_j is proportional to the part of the joint posterior \pi(\theta_1, \ldots, \theta_J, \mu, \sigma^2,\tau^2 | Y) that involves \theta_j.
11 / 15

Full conditionals for group means

Similarly, the full conditional distribution of each $\theta_j$ is proportional to the part of the joint posterior $\pi(\theta_1, \ldots, \theta_J, \mu, \sigma^2,\tau^2 | Y)$ that involves $\theta_j$ .
That is,

$\begin{split} \pi(\theta_j | \mu, \sigma^2,\tau^2, Y) & \boldsymbol{\propto} \left\{ \prod_{i=1}^{n_j} p(y_{ij} | \theta_j, \sigma^2 ) \right\} \cdot p(\theta_j | \mu,\tau^2) \\ \end{split}$

11 / 15

Full conditionals for group means

Similarly, the full conditional distribution of each $\theta_j$ is proportional to the part of the joint posterior $\pi(\theta_1, \ldots, \theta_J, \mu, \sigma^2,\tau^2 | Y)$ that involves $\theta_j$ .
That is,

$\begin{split} \pi(\theta_j | \mu, \sigma^2,\tau^2, Y) & \boldsymbol{\propto} \left\{ \prod_{i=1}^{n_j} p(y_{ij} | \theta_j, \sigma^2 ) \right\} \cdot p(\theta_j | \mu,\tau^2) \\ \end{split}$
Those terms include a normal for $\theta_j$ multiplied by a product of normals in which $\theta_j$ is the mean, again mirroring the one-sample case, so you can show that

$\begin{split} \pi(\theta_j | \mu, \sigma^2,\tau^2, Y) & = \mathcal{N}\left(\theta_j^\star, \nu_j^\star \right) \ \ \ \ \textrm{where}\\ \\ \nu_j^\star & = \dfrac{1}{ \dfrac{n_j}{\sigma^2} + \dfrac{1}{\tau^2} } ; \ \ \ \ \ \ \ \theta_j^\star = \nu_j^\star \left[ \dfrac{n_j}{\sigma^2} \bar{y}_j + \dfrac{1}{\tau^2} \mu \right] \end{split}$

11 / 15

Full conditionals for group meansOur estimate for each \theta_j is a weighted average of \bar{y}_j and \mu, ensuring that we are borrowing information across all levels through \mu and \tau^2.
12 / 15

Full conditionals for group means

Our estimate for each $\theta_j$ is a weighted average of $\bar{y}_j$ and $\mu$ , ensuring that we are borrowing information across all levels through $\mu$ and $\tau^2$ .
The weights for the weighted average is determined by relative precisions from the data and from the second level model.

12 / 15

Full conditionals for group means

Our estimate for each $\theta_j$ is a weighted average of $\bar{y}_j$ and $\mu$ , ensuring that we are borrowing information across all levels through $\mu$ and $\tau^2$ .
The weights for the weighted average is determined by relative precisions from the data and from the second level model.
The groups with smaller $n_j$ have estimated $\theta_j^\star$ closer to $\mu$ than schools with larger $n_j$ .

12 / 15

Full conditionals for group means

Our estimate for each $\theta_j$ is a weighted average of $\bar{y}_j$ and $\mu$ , ensuring that we are borrowing information across all levels through $\mu$ and $\tau^2$ .
The weights for the weighted average is determined by relative precisions from the data and from the second level model.
The groups with smaller $n_j$ have estimated $\theta_j^\star$ closer to $\mu$ than schools with larger $n_j$ .
Thus, degree of shrinkage of $\theta_j$ depends on ratio of within-group to between-group variances.

12 / 15

Full conditionals for across-group varianceThe full conditional distribution of \tau^2 is proportional to the part of the joint posterior \pi(\theta_1, \ldots, \theta_J, \mu, \sigma^2,\tau^2 | Y) that involves \tau^2.
13 / 15

Full conditionals for across-group variance

The full conditional distribution of $\tau^2$ is proportional to the part of the joint posterior $\pi(\theta_1, \ldots, \theta_J, \mu, \sigma^2,\tau^2 | Y)$ that involves $\tau^2$ .
That is,

$\begin{split} \pi(\tau^2 | \theta_1, \ldots, \theta_J, \mu, \sigma^2, Y) & \boldsymbol{\propto} \left\{ \prod_{j=1}^{J} p(\theta_j | \mu,\tau^2) \right\} \cdot \pi(\tau^2)\\ \end{split}$

13 / 15

Full conditionals for across-group variance

The full conditional distribution of $\tau^2$ is proportional to the part of the joint posterior $\pi(\theta_1, \ldots, \theta_J, \mu, \sigma^2,\tau^2 | Y)$ that involves $\tau^2$ .
That is,

$\begin{split} \pi(\tau^2 | \theta_1, \ldots, \theta_J, \mu, \sigma^2, Y) & \boldsymbol{\propto} \left\{ \prod_{j=1}^{J} p(\theta_j | \mu,\tau^2) \right\} \cdot \pi(\tau^2)\\ \end{split}$
As in the case for $\mu$ , this looks like the one-sample normal problem, and our full conditional posterior is

$\begin{split} \pi(\tau^2 | \theta_1, \ldots, \theta_J, \mu, \sigma^2, Y) & = \mathcal{IG} \left(\dfrac{\eta_n}{2}, \dfrac{\eta_n\tau_n^2}{2}\right) \ \ \ \ \textrm{where}\\ \\ \eta_n = \eta_0 + J ; \ \ \ \ \ \ \ \tau_n^2 & = \dfrac{1}{\eta_n} \left[\eta_0\tau_0^2 + \sum\limits_{j=1}^{J} (\theta_j - \mu)^2 \right].\\ \end{split}$

13 / 15

Full conditionals for within-group varianceFinally, the full conditional distribution of \sigma^2 is proportional to the part of the joint posterior \pi(\theta_1, \ldots, \theta_J, \mu, \sigma^2,\tau^2 | Y) that involves \sigma^2.
14 / 15

Full conditionals for within-group variance

Finally, the full conditional distribution of $\sigma^2$ is proportional to the part of the joint posterior $\pi(\theta_1, \ldots, \theta_J, \mu, \sigma^2,\tau^2 | Y)$ that involves $\sigma^2$ .
That is,

$\begin{split} \pi(\sigma^2 | \theta_1, \ldots, \theta_J, \mu, \tau^2, Y) & \boldsymbol{\propto} \left\{ \prod_{j=1}^{J} \prod_{i=1}^{n_j} p(y_{ij} | \theta_j, \sigma^2 ) \right\} \cdot \pi(\sigma^2)\\ \end{split}$

14 / 15

Full conditionals for within-group variance

Finally, the full conditional distribution of $\sigma^2$ is proportional to the part of the joint posterior $\pi(\theta_1, \ldots, \theta_J, \mu, \sigma^2,\tau^2 | Y)$ that involves $\sigma^2$ .
That is,

$\begin{split} \pi(\sigma^2 | \theta_1, \ldots, \theta_J, \mu, \tau^2, Y) & \boldsymbol{\propto} \left\{ \prod_{j=1}^{J} \prod_{i=1}^{n_j} p(y_{ij} | \theta_j, \sigma^2 ) \right\} \cdot \pi(\sigma^2)\\ \end{split}$
We can again take advantage of the one-sample normal problem, so that our full conditional posterior is

$\begin{split} \pi(\sigma^2 | \theta_1, \ldots, \theta_J, \mu, \tau^2, Y) & = \mathcal{IG} \left(\dfrac{\nu_n}{2}, \dfrac{\nu_n\sigma_n^2}{2}\right) \ \ \ \ \textrm{where}\\ \\ \nu_n = \nu_0 + \sum\limits_{j=1}^{J} n_j ; \ \ \ \ \ \ \ \sigma_n^2 & = \dfrac{1}{\nu_n} \left[\nu_0\sigma_0^2 + \sum\limits_{j=1}^{J}\sum\limits_{i=1}^{n_j} (y_{ij} - \theta_j)^2 \right].\\ \end{split}$

14 / 15

What's next?Move on to the readings for the next module!15 / 15

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STA 360/602L: Module 5.3

Hierarchical normal models with constant variance: multiple groups

Dr. Olanrewaju Michael Akande

Comparing multiple groups

Comparing multiple groups

Comparing multiple groups

Comparing multiple groups

School testing example

School testing example

School testing example

School testing example

School testing example

School testing example

School testing example

School testing example

Hierarchical normal model

Hierarchical normal model

Hierarchical normal model

Hierarchical normal model

Hierarchical normal model

Hierarchical normal model

Hierarchical normal model

Exchangeability

Exchangeability

Exchangeability

Exchangeability

Exchangeability

Exchangeability

Exchangeability

Exchangeability

Posterior inference

Posterior inference

Full conditional for grand mean

Full conditional for grand mean

Full conditional for grand mean

Full conditionals for group means

Full conditionals for group means

Full conditionals for group means

Full conditionals for group means

Full conditionals for group means

Full conditionals for group means

Full conditionals for group means

Full conditionals for across-group variance

Full conditionals for across-group variance

Full conditionals for across-group variance

Full conditionals for within-group variance

Full conditionals for within-group variance

Full conditionals for within-group variance

What's next?

Move on to the readings for the next module!

Comparing multiple groups

Help