STA 360/602L: Module 3.5b

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# STA 360/602L: Module 3.5b
## The normal model: joint inference for mean and variance (illustration)
### Dr. Olanrewaju Michael Akande

---

## Joint posterior for normal model

- Recall that
.block[
$$
`\begin{split}
\pi(\mu,\tau | Y) \ & \boldsymbol{=} \ \ \mathcal{N}\left(\mu_n, \frac{1}{\kappa_n\tau} \right) \cdot \textrm{Gamma}\left(\frac{\nu_n}{2}, \frac{\nu_n \sigma_n^2}{2}\right)\\
& = \pi(\mu|Y, \tau) \cdot \pi(\tau|Y),
\end{split}`
$$
]

where
.block[
$$
`\begin{split}
\kappa_n & = \kappa_0 + n\\
\mu_n & = \frac{\kappa_0 \mu_0 + n\bar{y}}{\kappa_n} = \frac{\kappa_0}{\kappa_n} \mu_0 + \frac{n}{\kappa_n} \bar{y}\\
\\
\nu_n & = \nu_0 + n\\
\sigma_n^2 & = \frac{1}{\nu_n}\left[\nu_0\sigma_0^2 + s^2(n-1) + \frac{n\kappa_0}{\kappa_n} (\bar{y} - \mu_0)^2 \right] \\
& = \frac{1}{\nu_n}\left[\nu_0\sigma_0^2 + \sum_{i=1}^n (y_i-\bar{y})^2 + \frac{n\kappa_0}{\kappa_n} (\bar{y} - \mu_0)^2 \right]\\
\end{split}`
$$
]

---
## Back to our examples

- .hlight[Pygmalion: questions of interest]
  + Is the average improvement for the accelerated group larger than that for the no growth group?
      - What is `$\Pr[\mu_A > \mu_N | Y_A, Y_N)$`?
  + Is the variance of improvement scores for the accelerated group larger than that for the no growth group?
      - What is `$\Pr[\sigma^2_A > \sigma^2_N | Y_A, Y_N)$`?
      
--

- .hlight[Job training: questions of interest]
  + Is the average change in annual earnings for the training group larger than that for the no training group?
      - What is `$\Pr[\mu_T > \mu_N | Y_T, Y_N)$`?
  + Is the variance of change in annual earnings for the training group larger than that for the no training group?
      - What is `$\Pr[\sigma^2_T > \sigma^2_N | Y_T, Y_N)$`?

---
## Mildly informative priors

- We will focus on the Pygmalion study. Follow the same approach for the job training data.

- Suppose you have no idea whether students would improve IQ on average. Set `$\mu_{0A} = \mu_{0N} = 0$`.

- Suppose you don't have any faith in this belief, and think it is the equivalent of having only 1 prior observation in each group. Set `$\kappa_{0A} = \kappa_{0N} = 1$`.

- Based on the literature, SD of change scores should be around 10 in each group, but still you don't have a lot of faith in this belief. Set 
`$\nu_{0A} = \nu_{0N} = 1$` and `$\sigma^2_{0A} = \sigma^2_{0N} = 100$`.

- Graph priors to see if they accord with your beliefs. Sampling new values of `$Y$` from the priors offers a good check.

---
## Recall the Pygmalion data

- Data:
  
  + Accelerated group (A): 20, 10, 19, 15, 9, 18.
  
  + No growth group (N): 3, 2, 6, 10, 11, 5.
--

- Summary statistics:
  
  + `$\bar{y}_A = 15.2$`; `$s_A = 4.71$`.
  
  + `$\bar{y}_N = 6.2$`; `$s_N = 3.65$`.

---
## Analysis with mildly informative priors

.block[
.small[
$$
`\begin{split}
\kappa_{nA} & = \kappa_{0A} + n_A = 1 + 6 = 7\\
\kappa_{nN} & = \kappa_{0N} + n_N = 1 + 6 = 7\\
\nu_{nA} & = \nu_{0A} + n_A = 1 + 6 = 7\\
\nu_{nN} & = \nu_{0N} + n_N = 1 + 6 = 7\\
\\
\mu_{nA} & = \frac{\kappa_{0A} \mu_{0A} + n_A\bar{y}_A}{\kappa_{nA}} = \frac{ (1)(0) + (6)(15.2) }{7} \approx 13.03\\
\mu_{nN} & = \frac{\kappa_{0N} \mu_{0N} + n_N\bar{y}_N}{\kappa_{nN}} = \frac{ (1)(0) + (6)(6.2) }{7} \approx 5.31\\
\\
\sigma_{nA}^2 & = \frac{1}{\nu_{nA}}\left[\nu_{0A}\sigma_{0A}^2 + s^2_A(n_A-1) + \frac{n_A\kappa_{0A}}{\kappa_{nA}} (\bar{y}_A - \mu_{0A})^2\right] \\
& = \frac{1}{7}\left[(1)(100) + (22.17)(5) + \frac{(6)(1)}{(7)} (15.2- 0)^2\right] \approx 58.41\\
\\
\sigma_{nN}^2 & = \frac{1}{\nu_{nN}}\left[\nu_{0N}\sigma_{0N}^2 + s^2_N(n_N-1) + \frac{n_N\kappa_{0N}}{\kappa_{nN}} (\bar{y}_N - \mu_{0N})^2\right] \\
& = \frac{1}{7}\left[(1)(100) + (13.37)(5) + \frac{(6)(1)}{(7)} (6.2- 0)^2\right] \approx 28.54\\
\end{split}`
$$
]
]

---
## Analysis with mildly informative priors

- So our joint posterior is
.block[
$$
`\begin{split}
\mu_A | Y_A, \tau_A & \sim \ \mathcal{N}\left(\mu_{nA}, \frac{1}{\kappa_{nA}\tau_A} \right) = \mathcal{N}\left(13.03, \frac{1}{7\tau_A} \right)\\
\\
\tau_A | Y_A & \sim \textrm{Gamma}\left(\frac{\nu_{nA}}{2}, \frac{\nu_{nA} \sigma_{nA}^2}{2}\right) = \textrm{Gamma}\left(\frac{7}{2}, \frac{7 (58.41)}{2}\right)\\
\\
\mu_N | Y_N, \tau_N & \sim \ \mathcal{N}\left(\mu_{nN}, \frac{1}{\kappa_{nN}\tau_N} \right) = \mathcal{N}\left(5.31, \frac{1}{7\tau_N} \right)\\
\\
\tau_N | Y_N & \sim \textrm{Gamma}\left(\frac{\nu_{nN}}{2}, \frac{\nu_{nN} \sigma_{nN}^2}{2}\right) = \textrm{Gamma}\left(\frac{7}{2}, \frac{7 (28.54)}{2}\right)\\
\end{split}`
$$
]

---
## Monte Carlo sampling

- To evaluate whether the accelerated group has larger IQ gains than the normal group, we would like to estimate quantities like `$\Pr[\mu_A > \mu_N | Y_A, Y_N)$` which are based on the **marginal posterior** of `$\mu$` rather than the **conditional distribution**.

- Fortunately, this is easy to do by generating samples of `$\mu$` and `$\sigma^2$` from their joint posterior.

---
## Monte Carlo sampling

- Suppose we simulate values using the following Monte Carlo procedure:
.block[
.small[
$$
`\begin{split}
\tau^{(1)} & \sim \textrm{Gamma}\left(\frac{\nu_{n}}{2}, \frac{\nu_{n} \sigma_{n}^2}{2}\right)\\
\mu^{(1)} & \sim \ \mathcal{N}\left(\mu_{n}, \frac{1}{\kappa_{n}\tau^{(1)}} \right)\\
\\
\tau^{(2)} & \sim \textrm{Gamma}\left(\frac{\nu_{n}}{2}, \frac{\nu_{n} \sigma_{n}^2}{2}\right)\\
\mu^{(2)} & \sim \ \mathcal{N}\left(\mu_{n}, \frac{1}{\kappa_{n}\tau^{(2)}} \right)\\
& \ \ \vdots \\
& \ \ \vdots \\
& \ \ \vdots \\
\tau^{(m)} & \sim \textrm{Gamma}\left(\frac{\nu_{n}}{2}, \frac{\nu_{n} \sigma_{n}^2}{2}\right)\\
\mu^{(m)} & \sim \ \mathcal{N}\left(\mu_{n}, \frac{1}{\kappa_{n}\tau^{(m)}} \right)\\
\end{split}`
$$
]
]

---
## Monte Carlo sampling

- Note that we are sampling each `$\mu^{(j)}$`, `$j=1,\ldots,m$`, from its conditional distribution, not from the marginal.

- The sequence of pairs `$\{(\tau, \mu)^{(1)}, \ldots, (\tau, \mu)^{(m)}\}$` simulated using this method are independent samples from the joint posterior `$\pi(\mu,\tau | Y)$`.

- Additionally, the simulated sequence `$\{\mu^{(1)}, \ldots, \mu^{(m)}\}$` are independent samples from the **marginal posterior distribution**.

- While this may seem odd, keep in mind that while we drew the `$\mu$`'s as conditional samples, each was conditional on a different value of `$\tau$`.

- Thus, together they constitute marginal samples of `$\mu$`.

---
## Monte Carlo sampling

It is easy to sample from these posteriors:

```r
aA <- 7/2
aN <- 7/2
bA <- (7/2)*58.41
bN <- (7/2)*28.54
muA <- 13.03
muN <- 5.31
kappaA <- 7
kappaN <- 7
tauA_postsample <- rgamma(10000,aA,bA)
thetaA_postsample <- rnorm(10000,muA,sqrt(1/(kappaA*tauA_postsample)))
tauN_postsample <- rgamma(10000,aN,bN)
thetaN_postsample <- rnorm(10000,muN,sqrt(1/(kappaN*tauN_postsample)))
sigma2A_postsample <- 1/tauA_postsample
sigma2N_postsample <- 1/tauN_postsample
```

---
## Monte Carlo sampling

- Is the average improvement for the accelerated group larger than that for the no growth group?
  + What is `$\Pr[\mu_A > \mu_N | Y_A, Y_N)$`?
  
  ```r
  mean(thetaA_postsample > thetaN_postsample)
  ```
  
  ```
  ## [1] 0.9675
  ```

- Is the variance of improvement scores for the accelerated group larger than that for the no growth group?
  + What is `$\Pr[\sigma^2_A > \sigma^2_N | Y_A, Y_N)$`?
  
  ```r
  mean(sigma2A_postsample > sigma2N_postsample)
  ```
  
  ```
  ## [1] 0.8159
  ```
      
--

- <div class="question">
What can we conclude from this?
</div>

---
## Recall the job training data

- Data:
  
  + No training group (N): sample size `$n_N = 429$`.
  
  + Training group (T): sample size `$n_A = 185$`.

- Summary statistics for change in annual earnings:
  
  + `$\bar{y}_N = 1364.93$`; `$s_N = 7460.05$`
  
  + `$\bar{y}_T = 4253.57$`; `$s_T = 8926.99$`
  
--

- <div class="question">
Using the same approach we used for the Pygmalion data, answer the questions of interest.
</div>

---

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# What's next?

### Move on to the readings for the next module!