Gamma-Poisson and Normal-Normal Models

Introduction: Gamma-Poisson

• Recall the Beta-Binomial from last week,

  • y \sim \text{Bin}(n, \pi) (data distribution)

  • \pi \sim \text{Beta}(\alpha, \beta) (prior distribution)

  • \pi|y \sim \text{Beta}(\alpha+y, \beta+n-y) (posterior distribution)

• Beta-Binomial is from a conjugate family (i.e., the posterior is from the same model family as the prior).

• Today, we will learn about other conjugate families, the Gamma-Poisson and the Normal-Normal.

Poisson Data Model

• Suppose we are now interested in modeling the number of spam calls we receive.

  • This means that we are modeling the rate, \lambda.

• We take a guess and say that the value of \lambda that is most likely is around 5,

  • … but reasonably ranges between 2 and 7 calls per day.

• Why can’t we use the Beta distribution as our prior distribution?

Poisson Data Model

• Suppose we are now interested in modeling the number of spam calls we receive.

  • This means that we are modeling the rate, \lambda.

• We take a guess and say that the value of \lambda that is most likely is around 5,

  • … but reasonably ranges between 2 and 7 calls per day.

• Why can’t we use the Beta distribution as our prior distribution?

  • \lambda is the mean of a count \to \lambda \in \mathbb{R}^+ \to \lambda is not limited to [0, 1] \to broken assumption for Beta distribution.

Poisson Data Model

• Suppose we are now interested in modeling the number of spam calls we receive.

  • This means that we are modeling the rate, \lambda.

• We take a guess and say that the value of \lambda that is most likely is around 5,

  • … but reasonably ranges between 2 and 7 calls per day.

• Why can’t we use the Beta distribution as our prior distribution?

  • \lambda is the mean of a count \to \lambda \in \mathbb{R}^+ \to \lambda is not limited to [0, 1] \to broken assumption for Beta distribution.

• Why can’t we use the binomial distribution as our data distribution?

Poisson Data Model

• Suppose we are now interested in modeling the number of spam calls we receive.

  • This means that we are modeling the rate, \lambda.

• We take a guess and say that the value of \lambda that is most likely is around 5,

  • … but reasonably ranges between 2 and 7 calls per day.

• Why can’t we use the Beta distribution as our prior distribution?

  • \lambda is the mean of a count \to \lambda \in \mathbb{R}^+ \to \lambda is not limited to [0, 1] \to broken assumption for Beta distribution.

• Why can’t we use the binomial distribution as our data distribution?

  • Y_i is a count \to Y_i \in \mathbb{N}^+ \to Y_i is not limited to \{0, 1\} \to broken assumption for Binomial distribution.

Poisson Data Model

• We will use the Poisson distribution to model the number of spam calls – Y \in \{0, 1, 2, ...\}.

  • Y is the number of independent events that occur in a fixed amount of time or space.

  • \lambda > 0 is the rate at which these events occur.

• Mathematically,

Y | \lambda \sim \text{Pois}(\lambda),

• with pmf,

f(y|\lambda) = \frac{\lambda^y e^{-\lambda}}{y!}, \ \ \ y \in \{0,1, 2, ... \}

Poisson Data Model

• \lambda defines the mean and the variance

  • The shape of the Poisson pmf depends on \lambda.

Poisson Data Model

• \lambda defines the mean and the variance

  • The shape of the Poisson pmf depends on \lambda.

Poisson Data Model

• We will be taking samples from different days.

  • We assume that the daily number of calls may different from day to day.

  • On each day i,

Y_i|\lambda \sim \text{Pois}(\lambda)

• This has a unique pmf for each day (i),

f(y_i|\lambda) = \frac{\lambda^{y_i} e^{-\lambda}}{y_i!}

• But really, we are interested in the joint information in our sample of n observations.

  • The joint pmf gives us this information.

Poisson Data Model

• The joint pmf for the Poisson,

\begin{align*} f\left(\overset{\to}{y_i}|\lambda\right) &= \prod_{i=1}^n f(y_i|\lambda) \\ &= f(y_1|\lambda) \times f(y_2|\lambda) \times ... \times f(y_n|\lambda) \\ &= \frac{\lambda^{y_1}e^{-\lambda}}{y_1!} \times \frac{\lambda^{y_2}e^{-\lambda}}{y_2!} \times ... \times \frac{\lambda^{y_n}e^{-\lambda}}{y_n!} \\ &= \frac{\left( \lambda^{y_1} \lambda^{y_2} \cdot \cdot \cdot \ \lambda^{y_n} \right) \left( e^{-\lambda} e^{-\lambda} \cdot \cdot \cdot e^{-\lambda}\right)}{y_1! y_2! \cdot \cdot \cdot y_n!} \\ &= \frac{\lambda^{\sum y_i}e^{-n\lambda}}{\prod_{i=1}^n y_i !} \end{align*}

Poisson Data Model

• If the joint pmf for the Poisson is

f\left(\overset{\to}{y_i}|\lambda\right) = \frac{\lambda^{\sum y_i}e^{-n\lambda}}{\prod_{i=1}^n y_i !}

• then the likelihood function for \lambda > 0 is

\begin{align*} L\left(\lambda|\overset{\to}{y_i}\right) &= \frac{\lambda^{\sum y_i}e^{-n\lambda}}{\prod_{i=1}^n y_i !} \\ & \propto \lambda^{\sum y_i} e^{-n\lambda} \end{align*}

• Pease see page 102 in the textbook for full derivations.

Gamma Prior

• If \lambda is a continuous random variable that can take on any positive value (\lambda > 0), then the variability may be modeled with the Gamma distribution with

  • shape hyperparameter s>0

  • rate hyperparameter r>0.

• Thus,

\lambda \sim \text{Gamma}(s, r)

• and the Gamma pdf is given by

f(\lambda) = \frac{r^s}{\Gamma(s)} \lambda^{s-1} e^{-r\lambda}

Gamma Prior

• …then the variability may be modeled with the Gamma distribution with shape s>0 and rate r>0.

Gamma Prior

• Let’s now tune our prior.

• We are assuming \lambda \approx 5, somewhere between 2 and 7.

• We know the mean of the gamma distribution,

E(\lambda) = \frac{s}{r} \approx 5 \to 5r \approx s

• Your turn! Use the plot_gamma() function to figure out what value of s and r we need.

Gamma Prior

• Looking at different values:

Gamma-Poisson Conjugacy

• Let \lambda > 0 be an unknown rate parameter and (Y_1, Y_2, ... , Y_n) be an independent sample from the Poisson distribution.

• The Gamma-Poisson Bayesian model is as follows:

\begin{align*} Y_i | \lambda &\overset{ind}\sim \text{Pois}(\lambda) \\ \lambda &\sim \text{Gamma}(s, r) \\ \lambda | \overset{\to}y &\sim \text{Gamma}\left( s + \sum y_i, r + n \right) \end{align*}

• The proof can be seen in section 5.2.4 of the textbook.

Gamma-Poisson Conjugacy

• Suppose we use Gamma(10, 2) as the prior for \lambda, the daily rate of calls.

• On four separate days in the second week of August (i.e., independent days), we received \overset{\to}y = (6, 2, 2, 1) calls.

• Your turn! Use the plot_poisson_likelihood() function:

plot_poisson_likelihood(y = c(6, 2, 2, 1), lambda_upper_bound = 10)

• Notes:

  • lambda_upper_bound limits the x axis – recall that \lambda \in (0, \infty)!

  • lambda_upper_bound’s default value is 10.

Gamma-Poisson Conjugacy

plot_poisson_likelihood(y = c(6, 2, 2, 1), lambda_upper_bound = 10) + theme_bw()

Gamma-Poisson Conjugacy

• We can see that the average is around 2.75.

• We can also verify that –

mean(c(6, 2, 2, 1))

[1] 2.75

• We know our prior distribution is Gamma(10, 2) and the data distribution is Poi(2.75).

• Thus, the posterior is as follows,

\begin{align*} \lambda | \overset{\to}y &\sim \text{Gamma}\left( s + \sum y_i, r + n \right) \\ &\sim \text{Gamma}\left(10 + 11, 2 + 4 \right) \\ &\sim \text{Gamma}\left(21, 6 \right) \end{align*}

Gamma-Poisson Conjugacy

• Your turn! Use the plot_gamma_poisson() function:

plot_gamma_poisson(shape = 10, rate = 2, sum_y = 11, n = 4)

Gamma-Poisson Conjugacy

• Your turn! Use the plot_gamma_poisson() function:

plot_gamma_poisson(shape = 10, rate = 2, sum_y = 11, n = 4)

Gamma-Poisson Conjugacy

• Your turn! What is different if we had used one of the other priors?

Gamma-Poisson Conjugacy

• Your turn! What is different if we had used Gamma(15, 3) as our prior?

Introduction: Normal-Normal

• So far, we have learned two conjugate families:

  • Beta-Binomial (binary outcomes)

    • y \sim \text{Bin}(n, \pi) (data distribution)

    • \pi \sim \text{Beta}(\alpha, \beta) (prior distribution)

    • \pi|y \sim \text{Beta}(\alpha+y, \beta+n-y) (posterior distribution)

  • Gamma-Poisson (count outcomes)

    • Y_i | \lambda \overset{ind}\sim \text{Pois}(\lambda) (data distribution)

    • \lambda \sim \text{Gamma}(s, r) (prior distribution)

    • \lambda | \overset{\to}y \sim \text{Gamma}\left( s + \sum y_i, r + n \right) (posterior distribution)

• Now, we will learn about another conjugate family, the Normal-Normal, for continuous outcomes.

Introduction

• As scientists learn more about brain health, the dangers of concussions are gaining greater attention.

• We are interested in \mu, the average volume (cm³) of a specific part of the brain: the hippocampus.

• Wikipedia tells us that among the general population of human adults, each half of the hippocampus has volume between 3.0 and 3.5 cm³.

  • Total hippocampal volume of both sides of the brain is between 6 and 7 cm³.

  • Let’s assume that the mean hippocampal volume among people with a history of concussions is also somewhere between 6 and 7 cm³.

• We will take a sample of n=25 participants and update our belief.

The Normal Model

• Let Y \in \mathbb{R} be a continuous random variable.

  • The variability in Y may be represented with a Normal model with mean parameter \mu \in \mathbb{R} and standard deviation parameter \sigma > 0.

• The Normal model’s pdf is as follows,

f(y) = \frac{1}{\sqrt{2\pi\sigma^2}} \exp \left\{ \frac{-(y-\mu)^2}{2\sigma^2} \right\}

The Normal Model

• Use the plot_normal() function to plot the following:

  • Vary \mu:

    • N(-1, 1)

    • N(0, 1)

    • N(1, 1)

  • Vary \sigma:

    • N(0, 1)

    • N(0, 2)

    • N(0, 3)

The Normal Model

• If we vary \mu,

The Normal Model

• If we vary \sigma,

The Normal Model

• Our data model is as follows,

Y_i | \mu \sim N(\mu, \sigma^2)

• The joint pdf is as follows,

f(\overset{\to}y | \mu) = \prod_{i=1}^n f(y_i | \mu) = \prod_{i=1}^n \frac{1}{\sqrt{2 \pi \sigma^2}} \exp \left\{ \frac{-(y_i-\mu)^2}{2\sigma^2} \right\}

• Meaning the likelihood is as follows,

L(\mu|\overset{\to}y) \propto \prod_{i=1}^n \frac{1}{\sqrt{2 \pi \sigma^2}} \exp \left\{ \frac{-(y_i-\mu)^2}{2\sigma^2} \right\} = \exp \left\{ \frac{- \sum_{i=1}^n(y_i-\mu)^2}{2\sigma^2} \right\}

The Normal Model

• Our data model is as follows,

Y_i | \mu \sim N(\mu, \sigma^2)

• Returning to our brain analysis, we will assume that the hippocampal volumes of our n = 25 subjects have a normal distribution with mean \mu and standard deviation \sigma.

  • Right now, we are only interested in \mu, so we assume \sigma = 0.5 cm³

  • This choice suggests that most people have hippocampal volumes within 2 \sigma = 1 cm³.

Normal Prior

• We know that with Y_i | \mu \sim N(\mu, \sigma^2), \mu \in \mathbb{R}.

  • We think a normal prior for \mu is reasonable.

• Thus, we assume that \mu has a normal distribution around some mean, \theta, with standard deviation, \tau.

\mu \sim N(\theta, \tau^2),

• meaning that \mu has prior pdf

f(\mu) = \frac{1}{\sqrt{2 \pi \tau^2}} \exp \left\{ \frac{-(\mu - \theta)^2}{2 \tau^2} \right\}

Normal Prior

• We can tune the hyperparameters \theta and \tau to reflect our understanding and uncertainty about the average hippocampal volume (\mu) among people with a history of concussions.

• Wikipedia showed us that hippocampal volumes tend to be between 6 and 7 cm³ \to \theta=6.5.

• When we set the standard deviation we can check the plausible range of values of \mu:

  • Follow up: why 2?

\theta \pm 2 \times \tau

• If we assume \tau=0.4,

(6.5 \pm 2 \times 0.4) = (5.7, 7.3)

Normal Prior

• Thus, our tuned prior is \mu \sim N(6.5, 0.4^2)

• This range incorporates our uncertainty - it is wider than the Wikipedia range.

Normal-Normal Conjugacy

• Let \mu \in \mathbb{R} be an unknown mean parameter and (Y_1, Y_2, ..., Y_n) be an independent N(\mu, \sigma^2) sample where \sigma is assumed to be known.

• The Normal-Normal Bayesian model is as follows:

\begin{align*} Y_i | \mu &\overset{\text{iid}} \sim N(\mu, \sigma^2) \\ \mu &\sim N(\theta, \tau^2) \\ \mu | \overset{\to}y &\sim N\left( \theta \frac{\sigma^2}{n\tau^2 + \sigma^2} + \bar{y} \frac{n\tau^2}{n\tau^2 + \sigma^2}, \frac{\tau^2 \sigma^2}{n \tau^2 + \sigma^2} \right) \end{align*}

Normal-Normal Conjugacy

• Let’s think about our posterior and some implications,

\mu | \overset{\to}y \sim N\left( \theta \frac{\sigma^2}{n\tau^2 + \sigma^2} + \bar{y} \frac{n\tau^2}{n\tau^2 + \sigma^2}, \frac{\tau^2 \sigma^2}{n \tau^2 + \sigma^2} \right)

• What happens as n increases?

Normal-Normal Conjugacy

• Let’s think about our posterior and some implications,

\mu | \overset{\to}y \sim N\left( \theta \frac{\sigma^2}{n\tau^2 + \sigma^2} + \bar{y} \frac{n\tau^2}{n\tau^2 + \sigma^2}, \frac{\tau^2 \sigma^2}{n \tau^2 + \sigma^2} \right)

• What happens as n increases?

\begin{align*} \frac{\sigma^2}{n\tau^2 + \sigma^2} &\to 0 \\ \frac{n\tau^2}{n\tau^2 + \sigma^2} &\to 1 \\ \frac{\tau^2 \sigma^2}{n \tau^2 + \sigma^2} &\to 0 \end{align*}

Normal-Normal Conjugacy

• Let’s think about our posterior and some implications,

\mu | \overset{\to}y \sim N\left( \theta \frac{\sigma^2}{n\tau^2 + \sigma^2} + \bar{y} \frac{n\tau^2}{n\tau^2 + \sigma^2}, \frac{\tau^2 \sigma^2}{n \tau^2 + \sigma^2} \right)

\begin{align*} \frac{\sigma^2}{n\tau^2 + \sigma^2} &\to 0 \\ \frac{n\tau^2}{n\tau^2 + \sigma^2} &\to 1 \\ \frac{\tau^2 \sigma^2}{n \tau^2 + \sigma^2} &\to 0 \end{align*}

• The posterior mean places less weight on the prior mean and more weight on the sample mean \bar{y}.

• The posterior certainty about \mu increases and becomes more in sync with the data.

Example

• Let us now apply this to our example.

• We have our prior model, \mu \sim N(6.5, 0.4^2).

• Let’s look at the football dataset in the bayesrules package.

data(football)
concussion_subjects <- football %>% 
  filter(group == "fb_concuss")

• What is the average hippocampal volume?

Example

• Let us now apply this to our example.

• We have our prior model, \mu \sim N(6.5, 0.4^2).

• Let’s look at the football dataset in the bayesrules package.

data(football)
concussion_subjects <- football %>% 
  filter(group == "fb_concuss")

• What is the average hippocampal volume?

mean(concussion_subjects$volume)

[1] 5.7346

Example

• We can also plot the density!

concussion_subjects %>% ggplot(aes(x = volume)) + geom_density() + theme_bw()

Example

• Now, we can plug in the information we have (n = 25, \bar{y} = 5.735, \sigma = 0.5) into our likelihood,

L(\mu|\overset{\to}y) \propto \exp \left\{ \frac{-(5.735 - \mu)^2}{2(0.5^2/25)} \right\}

Example

• We are now ready to put together our posterior:

  • Data distribution, Y_i | \mu \overset{\text{iid}} \sim N(\mu, \sigma^2)

  • Prior distribution, \mu \sim N(\theta, \tau^2)

  • Posterior distribution, \mu | \overset{\to}y \sim N\left( \theta \frac{\sigma^2}{n\tau^2 + \sigma^2} + \bar{y} \frac{n\tau^2}{n\tau^2 + \sigma^2}, \frac{\tau^2 \sigma^2}{n \tau^2 + \sigma^2} \right)

• Given our information (\theta=6.5, \tau=0.4, n=25, \bar{y}=5.735, \sigma=0.5), our posterior is

\begin{align*} \mu | \overset{\to}y &\sim N\left( \theta \frac{\sigma^2}{n\tau^2 + \sigma^2} + \bar{y} \frac{n\tau^2}{n\tau^2 + \sigma^2}, \frac{\tau^2 \sigma^2}{n \tau^2 + \sigma^2} \right) \\ &\sim N\left( 6.5 \frac{0.5^2}{25 \cdot 0.4^2 + 0.5^2} + 5.735 \frac{25 \cdot 0.4^2}{25 \cdot 0.4^2 + 0.5^2}, \frac{0.4^2 \cdot 0.5^2}{25 \cdot 0.4^2 + 0.5^2} \right) \\ &\sim N(6.5 \cdot 0.0588 + 5.737 \cdot 0.9412, 0.09^2) \\ &\sim N(5.78, 0.09^2) \end{align*}

Example

• Looking at the posterior, we can see the weights

\begin{align*} \mu | \overset{\to}y &\sim N\left( 6.5 \frac{0.5^2}{25 \cdot 0.4^2 + 0.5^2} + 5.735 \frac{25 \cdot 0.4^2}{25 \cdot 0.4^2 + 0.5^2}, \frac{0.4^2 \cdot 0.5^2}{25 \cdot 0.4^2 + 0.5^2} \right) \\ &\sim N(6.5 \cdot 0.0588 + 5.737 \cdot 0.9412, 0.009^2) \end{align*}

• 95% on the data mean, 6% on the prior mean.

Example

• We can plot the distribution,

Example

• We can summarize the distribution,

summarize_normal_normal(mean = 6.5, sd = 0.4, sigma = 0.5, y_bar = 5.735, n = 25) 

Homework

• 5.3
• 5.5
• 5.6
• 5.9
• 5.10