Conditional probability

# Conditional probability

---

Click the link below to create the repository for lecture notes #11.
  - [https://classroom.github.com/a/6nsBBc4y](https://classroom.github.com/a/6nsBBc4y)

---

## 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:

`\begin{align*}
P(A | B) = \frac{P(A \cap B)}{P(B)}
\end{align*}`

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* the roulette wheel just landing on green, what is the probability it
  lands on green in the next spin?

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## Three probabilities

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

.question[
Define the events `$A$` = died and `$B$` = non-coffee drinker. Calculate the 
following probabilities for a randomly selected person in the cohort:
]

- .vocab[Marginal probability]: `$P(A)$`, `$P(B)$`
- .vocab[Joint probability]: `$P(A \cap B)$`
- .vocab[Conditional probability]: `$P(A | B)$`, `$P(B | A)$`

---

## Marginal Probability

|                           | Did not die| Died|   Sum|
|:--------------------------|-----------:|----:|-----:|
|Does not drink coffee      |        5438| 1039|  6477|
|Drinks coffee occasionally |       29712| 4440| 34152|
|Drinks coffee regularly    |       24934| 3601| 28535|
|Sum                        |       60084| 9080| 69164|

.question[
Define the events `$A$` = died and `$B$` = non-coffee drinker. Calculate the 
following probabilties for a randomly selected person in the cohort:
]

- .vocab[Marginal probability]: `$P(A)$`

`$$P(A) = \frac{9080}{69164} = 0.131$$`

---

## Joint Probability

.question[
Define the events `$A$` = died and `$B$` = non-coffee drinker. Calculate the 
following probabilties for a randomly selected person in the cohort:
]

- .vocab[Joint probability]: `$P(A \text{ and } B)$`

`$$P(A \text{ and } B) = \frac{1039}{69164} = 0.015$$`

---

## Conditional Probabilities

.question[
Define the events `$A$` = died and `$B$` = non-coffee drinker. Calculate the 
following probabilties for a randomly selected person in the cohort:
]

- .vocab[Conditional probability]: `$P(A | B)$`, `$P(B | A)$`

`$$P(A | B) = \frac{1039}{6477} = 0.160$$`
--

`$$P(B | A) = \frac{1039}{9080} = 0.114$$`

---
class: center, middle, inverse

# Independence

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## The multiplicative rule

We can write the definition of condition probability

`\begin{align*}
P(A | B) &= \frac{P(A \cap B)}{P(B)}\\
P(B) \times P(A | B) &= P(A \cap B)
\end{align*}`

---

## Defining independence

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

`\begin{align*}
P(A | B) = P(A)
\end{align*}`
or
`\begin{align*}
P(B | A) = P(B)
\end{align*}`

That is, when knowing that one event has occurred doesn't cause us to "adjust"
the probability we assign to another event.

We can use the multiplicative rule to see that two events are said to be
independent when the joint probability of two events exactly equals the marginal
probability of their product:

`\begin{align*}
P(A \cap B) = P(A) \times P(B)
\end{align*}`

---

## 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: `$P(A \cap B) = 0$`.

So, two events which are disjoint are **not** independent!

---

## Checking independence

As you take more statistical science courses, you will learn the tools needed
to formally assess whether these two events are independent!

---

# Bayes' Rule

---

## The law of total probability

Suppose we partition the sample space into mutually exclusive events 
`$B_1, B_2, \cdots, B_k$`. These are disjoint events that comprise the entirely
of the sample space.

The .vocab[law of total probability] states that the probability of event `$A$` is

`\begin{align*}
P(A) = P(A \cap B_1) + P(A \cap B_2) + \cdots + P(A \cap B_k)
\end{align*}`

By applying the definition of conditional probability, we can obtain

`\begin{align*}
P(A) = P(B_1)P(A | B_1) + P(B_2)P(A | B_2) + \cdots + P(B_k)P(A | B_k)
\end{align*}`

---

## 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?
]

`\begin{align*}
P(A) &= P(A\text{ and } B_1) + P(A\text{ and } B_2) + P(A \text{ and } B_3)\\
 &= P(A|B_1) P(B_1) + P(A|B_2) P(B_2) + P(A|B_3) P(B_3)\\
 &= .80 \times 0.50 + 0.40 \times 0.30 + 0.10 \times 0.20\\
 &= 0.54
\end{align*}`

---

## 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?

We can use .vocab[Bayes' rule] to "reverse" the order of condition. By 
definition, we have:

`\begin{align*}
P(A | B) &= \frac{P(A \cap B)}{P(B)}\\
&= \frac{P(B | A)P(A)}{P(B)}
\end{align*}`

---

## Bayes' Rule (continued)

By using the rules of probability we've learned so far, we have

`\begin{align*}
P(A | B) &= \frac{P(A \cap B)}{P(B)}\\
&= \frac{P(B | A)P(A)}{P(B)}\\
&= \frac{P(B | A)P(A)}{P(B | A)P(A) + P(B | A^c)P(A^c)}
\end{align*}`

Note how we used the law of total probability in the denominator.

---

# Diagnostic testing

---

## 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]: `$P(D^c | T^c)$`

---

## Rapid self-administered HIV tests

- 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+.
]

<br/>

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

---

## Using Bayes' Rule

`\begin{align*}
P(D | T) &= \frac{P(D \cap T)}{P(T)}\\ 
&= \frac{P(T | D)P(D)}{P(T)}\\
&= \frac{P(T | D)P(D)}{P(T | D)P(D) + P(T | D^c)P(D^c)}\\
&= \frac{P(T | D)P(D)}{P(T | D)P(D) + (1 - P(T^c | D^c))(1 - P(D))}\\
&= \frac{sens. \times prev.}{sens. \times prev. + (1 - spec.) \times (1 - prev.)}
\end{align*}`

---

## Using Bayes' Rule

`\begin{align*}
P(D | T) &= \frac{sens. \times prev.}{sens. \times prev. + (1 - spec.) \times (1 - prev.)}
\end{align*}`

```r
sens <- 0.993; spec <- 0.998; prev <- 14.3/100000
prob <- (sens * prev) / ( (sens * prev) + ((1 - spec) * (1 - prev)) )
prob
```

```
#> [1] 0.0663016
```
]

- 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?

- The prevalence of HIV in Botswana is approximately 25%. What if we were to
  test a random individual in Botswana?