index.Rmd

---
title       : Conditional Probability
subtitle    : Statistical Inference
author      : Brian Caffo, Jeff Leek, Roger Peng
job         : Johns Hopkins Bloomberg School of Public Health
logo        : bloomberg_shield.png
framework   : io2012        # {io2012, html5slides, shower, dzslides, ...}
highlighter : highlight.js  # {highlight.js, prettify, highlight}
hitheme     : tomorrow      # 
url:
  lib: ../../librariesNew
  assets: ../../assets
widgets     : [mathjax]            # {mathjax, quiz, bootstrap}
mode        : selfcontained # {standalone, draft}
---

## Conditional probability, motivation

- The probability of getting a one when rolling a (standard) die
  is usually assumed to be one sixth
- Suppose you were given the extra information that the die roll
  was an odd number (hence 1, 3 or 5)
- *conditional on this new information*, the probability of a
  one is now one third

---

## Conditional probability, definition

- Let $B$ be an event so that $P(B) > 0$
- Then the conditional probability of an event $A$ given that $B$ has occurred is
  $$
  P(A ~|~ B) = \frac{P(A \cap B)}{P(B)}
  $$
- Notice that if $A$ and $B$ are independent, then
  $$
  P(A ~|~ B) = \frac{P(A) P(B)}{P(B)} = P(A)
  $$

---

## Example

- Consider our die roll example
- $B = \{1, 3, 5\}$
- $A = \{1\}$
$$
  \begin{eqnarray*}
P(\mbox{one given that roll is odd})  & = & P(A ~|~ B) \\ \\
  & = & \frac{P(A \cap B)}{P(B)} \\ \\
  & = & \frac{P(A)}{P(B)} \\ \\ 
  & = & \frac{1/6}{3/6} = \frac{1}{3}
  \end{eqnarray*}
$$



---

## Bayes' rule

$$
P(B ~|~ A) = \frac{P(A ~|~ B) P(B)}{P(A ~|~ B) P(B) + P(A ~|~ B^c)P(B^c)}.
$$
  

---

## Diagnostic tests

- Let $+$ and $-$ be the events that the result of a diagnostic test is positive or negative respectively
- Let $D$ and $D^c$ be the event that the subject of the test has or does not have the disease respectively 
- The **sensitivity** is the probability that the test is positive given that the subject actually has the disease, $P(+ ~|~ D)$
- The **specificity** is the probability that the test is negative given that the subject does not have the disease, $P(- ~|~ D^c)$

---

## More definitions

- The **positive predictive value** is the probability that the subject has the  disease given that the test is positive, $P(D ~|~ +)$
- The **negative predictive value** is the probability that the subject does not have the disease given that the test is negative, $P(D^c ~|~ -)$
- The **prevalence of the disease** is the marginal probability of disease, $P(D)$

---

## More definitions

- The **diagnostic likelihood ratio of a positive test**, labeled $DLR_+$, is $P(+ ~|~ D) / P(+ ~|~ D^c)$, which is the $$sensitivity / (1 - specificity)$$
- The **diagnostic likelihood ratio of a negative test**, labeled $DLR_-$, is $P(- ~|~ D) / P(- ~|~ D^c)$, which is the $$(1 - sensitivity) / specificity$$

---

## Example

- A study comparing the efficacy of HIV tests, reports on an experiment which concluded that HIV antibody tests have a sensitivity of 99.7% and a specificity of 98.5%
- Suppose that a subject, from a population with a .1% prevalence of HIV, receives a positive test result. What is the probability that this subject has HIV?
- Mathematically, we want $P(D ~|~ +)$ given the sensitivity, $P(+ ~|~ D) = .997$, the specificity, $P(- ~|~ D^c) =.985$, and the prevalence $P(D) = .001$

---

## Using Bayes' formula

$$
\begin{eqnarray*}
  P(D ~|~ +) & = &\frac{P(+~|~D)P(D)}{P(+~|~D)P(D) + P(+~|~D^c)P(D^c)}\\ \\
 & = & \frac{P(+~|~D)P(D)}{P(+~|~D)P(D) + \{1-P(-~|~D^c)\}\{1 - P(D)\}} \\ \\
 & = & \frac{.997\times .001}{.997 \times .001 + .015 \times .999}\\ \\
 & = & .062
\end{eqnarray*}
$$

- In this population a positive test result only suggests a 6% probability that the subject has the disease 
- (The positive predictive value is 6% for this test)

---

## More on this example

- The low positive predictive value is due to low prevalence of disease and the somewhat modest specificity
- Suppose it was known that the subject was an intravenous drug user and routinely had intercourse with an HIV infected partner
- Notice that the evidence implied by a positive test result does not change because of the prevalence of disease in the subject's population, only our interpretation of that evidence changes

---

## Likelihood ratios

- Using Bayes rule, we have
  $$
  P(D ~|~ +) = \frac{P(+~|~D)P(D)}{P(+~|~D)P(D) + P(+~|~D^c)P(D^c)} 
  $$
  and
  $$
  P(D^c ~|~ +) = \frac{P(+~|~D^c)P(D^c)}{P(+~|~D)P(D) + P(+~|~D^c)P(D^c)}.
  $$

---

## Likelihood ratios

- Therefore
$$
\frac{P(D ~|~ +)}{P(D^c ~|~ +)} = \frac{P(+~|~D)}{P(+~|~D^c)}\times \frac{P(D)}{P(D^c)}
$$
ie
$$
\mbox{post-test odds of }D = DLR_+\times\mbox{pre-test odds of }D
$$
- Similarly, $DLR_-$ relates the decrease in the odds of the
  disease after a negative test result to the odds of disease prior to
  the test.

---

## HIV example revisited

- Suppose a subject has a positive HIV test
- $DLR_+ = .997 / (1 - .985) \approx 66$
- The result of the positive test is that the odds of disease is now 66 times the pretest odds
- Or, equivalently, the hypothesis of disease is 66 times more supported by the data than the hypothesis of no disease

---

## HIV example revisited

- Suppose that a subject has a negative test result 
- $DLR_- = (1 - .997) / .985  \approx .003$
- Therefore, the post-test odds of disease is now $.3\%$ of the pretest odds given the negative test.
- Or, the hypothesis of disease is supported $.003$ times that of the hypothesis of absence of disease given the negative test result