Conditional Probability

# Conditional Probability
### Prof. Maria Tackett

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## [Click for PDF of slides](08-conditional-prob.pdf)

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## Conditional probability

The probability an event will occur *given* that another event has already 
occurred is a .vocab[conditional probability]. The conditional probability of
event `$A$` given event `$B$` is:

## Conditional probabilities

Examples come up all the time in the real world:

- *Given* that it rained yesterday, what is the probability that it will rain
today?
- *Given* that a mammogram comes back positive, what is the probability that a
woman has breast cancer?
- *Given* that I've already watched six episodes of How I Met Your Mother tonight,
what is the probability that I'll get any work done this evening?

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## Coffee and mortality

|                           | Did not die| Died|
|:--------------------------|-----------:|----:|
|Does not drink coffee      |        5438| 1039|
|Drinks coffee occasionally |       29712| 4440|
|Drinks coffee regularly    |       24934| 3601|
]

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## Three probabilities

<br>

.question[
.midi[
Define events *A* = died and *B* = non-coffee drinker. Calculate the following for a randomly selected person in the cohort:]
- .vocab[Marginal probability]: `$P(A)$`, `$P(B)$`
- .vocab[Joint probability]: `$P(A \text{ and } B)$`
- .vocab[Conditional probability]: `$P(A | B)$`, `$P(B | A)$`
]

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# Independence

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## The multiplicative rule

We can write the definition of condition probability

`$$P(B) \times P(A | B) = P(A \text{ and } B)$$`
]

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## Defining independence

Events `$A$` and `$B$` are said to be .vocab[independent] when

`$$P(A | B) = P(A) \hspace{10mm} \textbf{OR} \hspace{10mm}
P(B | A) = P(B)$$`

<br>

In other words, knowing that one event has occurred doesn't cause us to "adjust"
the probability we assign to another event.

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## Checking independence

We can use the multiplicative rule to see if two events are independent.

`$$P(A \text{ and } B) = P(A) \times P(B)$$`
]

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## Independent vs. disjoint events

Since for two independent events `$P(A|B) = P(A)$` and `$P(B|A) = P(B)$`, knowing
that one event has occurred tells us nothing more about the probability of the
other occurring.

For two disjoint events `$A$` and `$B$`, knowing that one has occurred
tells us that the other definitely has not occurred: `$P(A \text{ and } B) = 0$`.

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## Checking independence

<br>

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# Bayes' Rule

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## An example

In an introductory statistics course, 50% of students were first years, 30% were
sophomores, and 20% were upperclassmen.

80% of the first years didn’t get enough sleep, 40% of the sophomores didn’t get
enough sleep, and 10% of the upperclassmen didn’t get enough sleep.

.question[
What is the probability that a randomly selected student in this class didn’t
get enough sleep?
]

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## Bayes' Rule

As we saw before, the two conditional probabilities `$P(A | B)$` and `$P(B | A)$` 
are not the same. But are they related in some way?

Yes they are (!) using .vocab[Bayes' rule]:

`$$\begin{align}P(A | B) &= \frac{P(A \text{ and } B)}{P(B)}\\[10pt]
&= \frac{P(B | A)P(A)}{P(B)}
\end{align}$$`
]

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## Bayes' Rule (continued)

Putting together a few rules of probability...

`$$\begin{align}P(A | B) &= \frac{P(A \text{ and } B)}{P(B)}\\[10pt]
&= \frac{P(B | A)P(A)}{P(B)}\\[15pt]
&= \frac{P(B | A)P(A)}{P(B | A)P(A) + P(B | A^c)P(A^c)}\end{align}$$`

Let's took at an example to see how this works.

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# Diagnostic Testing

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## Definitions

Suppose we're interested in the performance of a diagnostic test. Let `$D$` be
the event that a patient has the disease, and let `$T$` be the event that the
test is positive for that disease.

- .vocab[Prevalence]: `$P(D)$`
- .vocab[Sensitivity]: `$P(T | D)$`
- .vocab[Specificity]: `$P(T^c | D^c)$`
- .vocab[Positive predictive value]: `$P(D | T)$`
- .vocab[Negative predictive value]: `$P(D^c | T^c)$`

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## Rapid self-administered HIV tests

.pull-left[
From the FDA package insert for the Oraquick ADVANCE Rapid HIV-1/2 Antibody
Test, 
- Sensitivity, `$P(T | D)$`, is 99.3%
- Specificity, `$P(T^c | D^c)$`, is 99.8%
From CDC statistics in 2016, 14.3/100,000 Americans aged 13 or older are HIV+.
]

.question[
Suppose a randomly selected American aged 13+ has a positive test result. What is the probability they have HIV?
]

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## Using Bayes' Rule

`\begin{align*}
P(D | T) &= \frac{P(D \text{ and } T)}{P(T)}\\
&= \frac{P(T | D)P(D)}{P(T)}\\[5pt]
&= \frac{P(T | D)P(D)}{P(T | D)P(D) + P(T | D^c)P(D^c)}\\[5pt]
&= \frac{P(T | D)P(D)}{P(T | D)P(D) + (1 - P(T^c | D^c))(1 - P(D))}
\end{align*}`

<br>

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## Work through example

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## A discussion

Think about the following questions:

- Is this calculation surprising?
- What is the explanation?
- Was this calculation actually reasonable to perform?
- What if we tested in a different population, such as high-risk individuals?
-  What if we were to
test a random individual in a country where the prevalence of HIV is approximately 25%?