Bayesian Statistics

• A test for cystic fibrosis has an accuracy of 99%:

\[ \mbox{Prob}(+|D)=0.99, \mbox{Prob}(-|\mbox{no } D)=0.99, \]

• If we select random person and they test postive what is probability of positive test?

• We write this as \( \mbox{Prob}(D|+)? \)

• cystic fibrosis rate is 1 in 3,900, \( \mbox{Prob}(D)=0.0025 \)

\[ \mbox{Pr}(A|B) = \frac{\mbox{Pr}(B|A)\mbox{Pr}(A)}{\mbox{Pr}(B)} \]

\[ \begin{eqnarray*} \mbox{Prob}(D|+) & = & \frac{ P(+|D) \cdot P(D)} {\mbox{Prob}(+)} \\ & = & \frac{\mbox{Prob}(+|D)\cdot P(D)} {\mbox{Prob}(+|D) \cdot P(D) + \mbox{Prob}(+|\mbox{no } D) \mbox{Prob}(\mbox{no } D)} \\ \end{eqnarray*} \]

\[ \begin{eqnarray*} \mbox{Prob}(D|+) & = & \frac{ P(+|D) \cdot P(D)} {\mbox{Prob}(+)} \\ & = & \frac{\mbox{Prob}(+|D)\cdot P(D)} {\mbox{Prob}(+|D) \cdot P(D) + \mbox{Prob}(+|\mbox{no } D) \mbox{Prob}(\mbox{no } D)} \\ & = & \frac{0.99 \cdot 0.0025}{0.99 \cdot 0.0025 + 0.01 \cdot (.9975)} \\ & = & 0.02 \;\;\; \mbox{not} \; \; \; 0.99 \end{eqnarray*} \]

\( \mbox{Prob}(A|B) \) = % red bottom left, \( \mbox{Prob}(A) \)=%red on top, \( \mbox{Prob}(B|A) \)=% not X, \( \mbox{Prob}(B) \)=total points bottom left,

This is for all players (>500 AB) 2010, 2011, 2012

Average is .275 and SD is 0.027

• Should we trade him?

• What is the SE of our estimate?

• \[ \sqrt{\frac{.450 (1-.450)}{20}}=.111 \]

• Confidence interval?

• .450-.222 to .450+.222 = .228 to .672

Pick a random player, then what is their batting average

\[ \begin{eqnarray*} \theta &\sim& N(\mu, \tau^2) \mbox{ is called a prior}\\ Y | \theta &\sim& N(\theta, \sigma^2) \mbox{ is called a sampling distribution} \end{eqnarray*} \]

Two levels of variability:

• Player to player variability
• Variability due to luck when batting

\[ \begin{eqnarray*} \theta &\sim& N(\mu, \tau^2) \mbox{ is called a prior}\\ Y | \theta &\sim& N(\theta, \sigma^2) \mbox{ is called a sampling distribution} \end{eqnarray*} \]

• \( \theta \) is our players “intrinsic” average value
• \( \mu \) is the average of all players
• \( \tau \) is the SD of all players
• \( Y \) is the observed average
• \( \sigma \) is the variability due to luck at each AB

Here are the equations with our data

\[ \begin{eqnarray*} \theta &\sim& N(.275, .027^2) \\ Y | \theta &\sim& N(\theta, .110^2) \end{eqnarray*} \]

The continuous version of Bayes rule can be used here

\[ \begin{eqnarray*} f_{\theta|Y}(\theta|Y)&=&\frac{f_{Y|\theta}(Y|\theta) f_{\theta}(\theta)}{f_Y(Y)}\\ &=&\frac{f_{Y|\theta}(Y|\theta) f_{\theta}(\theta)}{\int_{\theta}f_{Y|\theta}(Y|\theta)f_{\theta}(\theta)}\\ \end{eqnarray*} \]

We are particularly interested in the \( \theta \) that maximizes \( f_{\theta|Y}(\theta|Y) \).

In our case, these can be shown to be normal so we want the average \( \mbox{E}(\theta|y) \)

We can show the average of this distribution is the following:

\[ \begin{eqnarray*} \mbox{E}(\theta|y) &=& B \mu + (1-B) Y\\ &=& \mu + (1-B)(Y-\mu)\\ B &=& \frac{\sigma^2}{\sigma^2+\tau^2} \end{eqnarray*} \]

In the case of José Iglesias, we have:

\[ \begin{eqnarray*} E(\theta | Y=.450) &=& B \times .275 + (1 - B) \times .450 \\ &=& .275 + (1 - B)(.450 - .260) \\ B &=&\frac{.110^2}{.110^2 + .027^2} = 0.943\\ E(\theta | Y=450) &\approx& .285\\ \end{eqnarray*} \]

The variance can be shown to be:

\[ \begin{eqnarray*} \mbox{var}(\theta|y) &=& \frac{1}{1/\sigma^2+1/\tau^2} &=& \frac{1}{1/.110^2 + 1/.027^2} \end{eqnarray*} \]

In our example the SD is 0.026

Frequentist confidence interval = .450 \( \pm \) 0.220

Empirical Bayes credible interval = .285 \( \pm \) 0.052

Actual = .293

http://www.nytimes.com/newsgraphics/2014/senate-model/comparisons.html

http://election.princeton.edu/

http://fivethirtyeight.com/interactives/senate-forecast/

Note: Kansas, New Hampshire and North Carolina

60% R, 56% I, 62% I, 51% R, 72% I, 61% R, 61% R, Tie, Tie, Tie

75% D, 81% D, 69% D, 60% D, 89% D, 84% D, 99% D, 3 Lean D

Different colors are different pollsters

81% D, 78% D, 73% D, 60% D, 75% D, 71% D, 96% D, Tie, Tie, Lean D

A closer look at NC

Mindless confidence interval:

[ -4.3 -2.1 ]

Using only recent polls

[ -3.8 -1.6 ]

Use, previous North Carolina results, average 0 and SD 5

Two level model: \[ \begin{eqnarray*} \theta &\sim& N(\mu, \tau^2) \\ Y | \theta &\sim& N(\theta, 0.5^2) \end{eqnarray*} \]

Observed \( Y = -2.4 \)

How do we construct a prior? Are there other levels?