Caution
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Bayesian Statistics: Techniques and Models
Exponential Distribution, Homework
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---
title: "Homework on Predictive distributions and mixture models - M4L12HW1"
subtitle: "Bayesian Statistics: Techniques and Models"
categories:
- Bayesian Statistics
keywords:
- Exponential Distribution
- Homework
---
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::: {#exr-Mixture-Models-1}
[**Mixture Models**]{.column-margin} Consider the Poisson process model we fit in the quiz for Lesson 10 which estimates calling rates of a retailer's customers. The data are attached below.
Re-fit the model and use your posterior samples to simulate predictions of the number of calls by a new 29 year old customer from Group 2 whose account is active for 30 days. What is the probability that this new customer calls at least three times during this period? Round your answer to two decimal places.
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::: {.solution .column-screen-inset-right .callout-tip collapse="true"}
#### Solution:
```{r}
#| label: q1
library("rjags")
calls_dat = read.csv(file="data/callers.csv", header=TRUE)
head(calls_dat)
calls_mod_string = " model {
for (i in 1:length(calls)) {
calls[i] ~ dpois( days_active[i] * lam[i] )
log(lam[i]) = b0 + b[1]*age[i] + b[2]*isgroup2[i]
}
b0 ~ dnorm(0.0, 10.0^2)
for(j in 1:2){
b[j] ~ dnorm(0.0, 10.0^2)
}
} "
set.seed(102)
calls_data_jags = as.list(calls_dat)
calls_params = c("b0",'b')
calls_mod = jags.model(textConnection(calls_mod_string), data=calls_data_jags, n.chains=3)
update(calls_mod, 1e3)
calls_mod_sim = coda.samples(model=calls_mod, variable.names=calls_params,n.iter=5e3)
calls_mod_csim = as.mcmc(do.call(rbind, calls_mod_sim))
x1 = c(29,1) # days active,isgroup2,age
head(calls_mod_csim)
loglam1 = calls_mod_csim[,"b0"] + calls_mod_csim[,c(1,2)] %*% x1
lam1 = exp(loglam1)
(n_sim = length(lam1))
y1 = rpois(n=n_sim, lambda=lam1*30)
sum(y1 >= 3)/length(y1)
```
0.03
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::: {#exr-Mixture-Models-2}
[**Mixture Models**]{.column-margin} Suppose we fit a single component normal distribution to the data whose histogram is shown below.
If we use a noninformative prior for μ and $σ^2$ and plot the fit distribution evaluated at the posterior means (in blue), what would the fit look like? Is this model appropriate for these data?
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::: {.solution .column-screen-inset-right .callout-tip collapse="true"}
#### Solution:
- [x]  A single normal distribution does not allow bi-modality. Consequently, the fit places a lot of probability in a region with no data. It is not appropriate.
- [ ]  The single normal fit nicely captures the features of the data. It is appropriate.
- [ ]  The single normal fit accommodates the bi-modality in the dat, but fails to capture the imbalance in the two components. It is not appropriate.
- [ ]  The single normal fit ignores the smaller component, fitting the cluster of points with most data. Consequently, the model places almost no probability in the region of the smaller component.
Without covariates, we would need a more flexible model to capture the bi-modality of these data. A two component mixture of normals would be appropriate.
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::: {#exr-Mixture-Models-3}
[**Mixture Models**]{.column-margin} Which of the following histograms shows data that might require a mixture model to fit?
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::: {.solution .column-screen-inset-right .callout-tip collapse="true"}
#### Solution:
- [ ] 
- [X] 
- [ ] 
- [ ] 
These data were actually drawn from a two component mixture of normal distributions, one with high variance.
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::: {#exr-Mixture-Models-4}
[**Mixture Models**]{.column-margin}
The Dirichlet distribution with parameters $\alpha_1 = \alpha_2 = \ldots = \alpha_K = 1$ is uniform over its support, the values for which the random vector contains a valid set of probabilities. If $\theta$ contains five probabilities corresponding to five categories and has a $\text{Dirichlet}(1,1,1,1,1)$ prior, what is the effective sample size of this prior?
**Hint**: If $\theta$ has a $\text{Dirichlet}(\alpha_1, \alpha_2, \ldots, \alpha_K)$ prior, and the counts of multinomial data in each category are $x_1, x_2, \ldots, x_K$, then the posterior of $\theta$ is $\text{Dirichlet}(\alpha_1 + x_1, \alpha_2 + x_2, \ldots, \alpha_K + x_K)$. The data sample size is clearly $\sum_{k=1}^K x_k$.
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::: {.solution .column-screen-inset-right .callout-tip collapse="true"}
#### Solution:
based on the hint, what remains is $\sum_{k=1}^K \alpha_k= 5$
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::: {#exr-Mixture-Models-5}
[**Mixture Models**]{.column-margin} Recall that in the Bayesian formulation of a mixture model, it is often convenient to introduce latent variables $z_i$ which indicate "population" membership of $y_i$ (the "population" may or may not have meaning in the context of the data). One possible hierarchical formulation is given by:
$$
\begin{aligned}
y_i \mid z_i, \theta & \overset{\text{ind}}{\sim} f_{z_i}(y \mid \theta) &i = 1, \ldots, n
\\ \text{Pr}(z_i = j \mid \omega) &= \omega_j &j=1, \ldots, J
\\ \omega &\sim Dirichlet(\omega)
\\ \theta &\sim \mathbb{P}r(\theta)
\end{aligned}
$$
where $f_{j} (y \mid \theta)$ is a probability density for $y$ for mixture component $j$ and $w = (w_1, w_2, \ldots, w_J)$ is a vector of prior probabilities of membership.
What is the full conditional distribution for $z_i$?
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::: {.solution .column-screen-inset-right .callout-tip collapse="true"}
#### Solution:
- [x] $\text{Pr}(z_i = j \mid \cdots ) = \frac{ f_{j} (y_i \mid \theta)\, w_j } { \sum_{\ell=1}^J f_{\ell} (y_i \mid \theta)\, w_\ell }\, , \quad j=1, \ldots, J$
- [ ] $\text{Pr}(z_i = j \mid \cdots ) = \frac{ w_j^{I_{(z_i=j))}} (1-w_j)^{1-I_{(z_i=j))}} } { \sum_{\ell=1}^J w_j^{I_{(z_i=j))}} (1-w_j)^{1-I_{(z_i=j))}} }\, , \quad j=1, \ldots, J$
- [ ] $\text{Pr}(z_i = j \mid \cdots ) = w_j \, , \quad j=1, \ldots, J$
- [ ] $\text{Pr}(z_i = j \mid \cdots ) = \frac{ f_{j} (y_i \mid \theta) } { \sum_{\ell=1}^J f_{\ell} (y_i \mid \theta) }\, , \quad j=1, \ldots, J$
Notice the resemblance with this expression and the discrete version of Bayes' theorem. This expression is often used in classification problems, where we need to infer membership of $y_i$ to different candidate populations.
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