# Bayes' Theorem: Disease Probabilities

## What We're Given

* $D \in \{0, 1\}$ = has disease
* $T \in \{0, 1\}$ = test result
* (In the real world, however, we cannot observe $D$, only $T$)
* 99% accuracy:
* $\Pr(T = 1 \mid D = 1) = 0.99$
* $\Pr(T = 0 \mid D = 0) = 0.99$
* Rare disease: 1 in 10000 people has it
* $\Pr(D = 1) = \frac{1}{10000}$

## Basic Bayes: Probability of Disease Given Positive Test

We want

$$
\Pr(D = 1 \mid T = 1)
$$

Which we can compute, using the given information, via Bayes' rule (where the second line is how I like to write it out, a bit more cluttered but makes it clear how to compute the denominator)

$$
\begin{align*}
\Pr(D = 1 \mid T = 1) &= \frac{\Pr(T = 1 \mid D = 1)\Pr(D = 1)}{\Pr(T = 1)} \\
&= \frac{\Pr(T = 1 \mid D = 1)\Pr(D = 1)}{\Pr(T = 1 \mid D = 1)\Pr(D = 1) + \Pr(T = 1 \mid D = 0)\Pr(D = 0)}
\end{align*}
$$

Numerically:

$$
\Pr(D = 1 \mid T = 1) = \frac{(0.99)(1/10000)}{(0.99)(1/10000) + (0.01)(9999/10000)} \approx 0.0098
$$

(Less than 1%)

## Deeper Dive

But now let's think about what's behind this... It's a dangerous, **highly contagious** disease, meaning that (societally) false **negatives** are much, much worse than false **positives**:

* A false **negative**, in this case, means that someone is walking around thinking they **don't** have the disease (because they tested negative), when they actually **do**. This means that they are not quarantining, they are going out to parties and events and etc., spreading the disease.
* A false **positive**, on the other hand, means someone who panics unnecessarily: maybe it means, they go to the hospital, the hospital performs additional tests, and successfully discovers that the person, despite their positive test result, doesn't have the disease.
* So, **consequence-wise**, a **false negative** potentially means a new outbreak of the disease in the society, while a **false positive** means a quick (scary, but hopefully quick) trip to the hospital

### The Catastrophic Case

So, let's focus on the disastrous first case: what's the probability of a **false negative**? First, we can compute the **conditional** probability of a negative test, conditional on someone having the disease? Here we just use our **complement rule** of probability: that $\Pr(E^c) = 1 - \Pr(E)$ for any event $E$:

$$
\Pr(T = 0 \mid D = 1) = 1 - \Pr(T = 1 \mid D = 1) = 0.01
$$

Now that we know this, let's incorporate the **base rate** information---that is, the information we have about the likelihood of having the disease (the thing we conditioned on above):

$$
\Pr(D = 1) = \frac{1}{10000}
$$

So, given these two pieces of information, we can compute the probability of a person in the population being a **false negative case**: having the disease, but not being detected by the test.

$$
\begin{align*}
\Pr(T = 0 \cap D = 1) &= \Pr(T = 0 \mid D = 1)\Pr(D = 1) \\
&= (0.01)(1/10000) = \frac{1}{1000000},
\end{align*}
$$

i.e., **one in a million**.

### The Bad (But Not Catastrophic) Case

Now let's turn to the second, bad but not catastrophic, case: the probability of a **false positive**. As before, we start by computing the **conditional probability** of a positive test result for someone who in fact does **not** have the disease:

$$
\Pr(T = 1 \mid D = 0) = 1 - \Pr(T = 0 \mid D = 0) = 0.01
$$

This time, however, we'll see that the base rate will make a big difference. The base rate in this case---the probability of someone **not** having the disease---is:

$$
\Pr(D = 0) = 1 - \Pr(D = 1) = \frac{9999}{10000}
$$

So, incorporating these two pieces of information, we can compute the likelihood of a **false positive case**: someone in the population who **doesn't** have the disease but **does** test positive:

$$
\begin{align*}
\Pr(T = 1 \cap D = 0) &= \Pr(T = 1 \mid D = 0)\Pr(D = 0) \\
&= (0.01)(9999/10000) = \frac{9999}{1000000}
\end{align*}
$$

In words: for every million people in the population, 9999 of them will have a **false positive** panic: they won't have the disease, but they will **think** they have the disease because of their positive test.

## Putting It Together:

At first, this example is depressing: "Oh no, that's terrible! We're forcing thousands of people to panic, thinking that they have the disease, when they really don't!"

But, walking through it with this false negative / false positive paradigm, we see the real takeaway: that there is always a **tradeoff** between false positives and false negatives. In this case, from a public health perspective for example, it's actually somewhat of a **good** situation: at the "cost" of having several thousand people panic unnecessarily, we **achieve** the benefit of making it very, very unlikely (one in a million, literally) that someone goes undetected in the population with this dangerous, contagious disease.

<table>
<thead>
<tr style="border-top: 0px solid black;">
<th colspan="2" rowspan="2" style="border-top: 0px solid black; border-left: 0px solid black;"></th>
<th colspan="2">True State of the World</th>
</tr>
<tr>
<th>0</th>
<th>1</th>
</tr>
</thead>
<tbody>
<tr>
<td rowspan="2"><b>Prediction</b></td>
<td><b>0</b></td>
<td>True Negative</td>
<td>False Negative</td>
</tr>
<tr>
<td><b>1</b></td>
<td>False Positive</td>
<td>True Positive</td>
</tr>
</tbody>
</table>

### (Compuation of the remaining two cells:)

**True Positive**:

$$
\begin{align*}
\Pr(T = 1 \cap D = 1) &= \Pr(T = 1 \mid D = 1)\Pr(D = 1) \\
&= (0.99)\frac{1}{10000} = \frac{99}{1000000}
\end{align*}
$$

**True Negative**:

$$
\begin{align*}
\Pr(T = 0 \cap D = 0) &= \Pr(T = 0 \mid D = 0)\Pr(D = 0) \\
&= (0.99)\frac{9999}{10000} = \frac{989901}{1000000}
\end{align*}
$$