Lassen National Park!

Written for STAT 131.

One of the joys/horrors of TA-ing is that you always encounter things that stumps you as well as your students (especially in probability). Here is a very basic looking problem that stumped me from Introduction to Probability by Joe Blitzstein and Jessica Hwang (Ch2 Q37):

Two different dieseases cause a certain weird symptom: anyone who has either or both of these dieseases will experience the symptom. Let $D_1$ be the event of having the first disease, $D_2$ be the event of having the second, and $W$ be the event of having the weird symptom. Suppose $D_1$ and $D_2$ are independent with $P(D_j) = p_j$ and that a person with neither disease will have the weird symptom with probability $w_0$. Let $q_j =1-p_j$.

### (a) Find $P(W)$

We can use the Law of Total Probability (LOTP).

So let $D_1 \cup D_2$ be the event that you have at least one disease.
Let $D_1^c \cap D_2^c$ be the probability that you have neither.

Then

$P(W) = P(W \mid D_1 \cup D_2) P(D_1 \cup D_2) + P(W \mid D_1^c \cap D_2^c) P(D_1^c \cap D_2^c)$

So using the laws of probability you’ve learned: $P(D_1 \cup D_2) = p_1 + p_2 - p_1p_2$ And $P(D_1^c \cap D_2^c)=(1-p_1 )(1-p_2) = q_1q_2$.

Anyone who has at least one of the diseases will have weird symptoms with probability=1. Thus $P(W \mid D_1 \cup D_2) =1$.
We are given that probability that you have a weird disease without having the disease is $w_0$. Thus we have $P(W) = 1 ( p_1 + p_2 - p_1p_2) + w_0 q_1 q_2$.

### (b)

#### Find $P(D_i \mid W)$

We need to find the prob. that you have disease $i$ given weird symptoms Note that using the formula for conditional probability: $P(Di \mid W) = \frac{P(D_i \cap W)} {P(W)} = \frac{P( W \mid Di) P(Di)} {P(W)}$

We already know $P( W \mid Di)= 1$.

Using result from (a), for $i=1,2$ we have:

$P(Di \mid W) = \frac{P(D_i)} {P(W)} = \frac{p_i} {( p_1 + p_2 - p_1p_2) +w_0 q_1q_2} .$

#### Find $P(D_1 \cap D_2 \mid W)$

We need to find the prob. that you have both diseases given weird symptoms We can solve things in a similar fashion: $P(D_1 \cap D_2 \mid W) =\frac{P( W \mid D_1 \cap D_2) P(D_1 \cap D_2)} {P(W)} = \frac {P(D_1 D_2)} {P(W)} = \frac{ p_1 p_2 }{( p_1 + p_2 - p_1p_2) +w_0 q_1q_2}.$

### (c) Are $D_1, D_2$ conditionally independent on $W$?

Note that for conditional independence to be possible we must have $P(D_1 \cap D_2\mid W) = P(D_1\mid W) P(D_2\mid W).$
But we have

$P(D_1\mid W) P(D_2\mid W) = \frac{p_1}{P(W)} \frac{p_2}{P(W)}.$

This is not generally equal to

$P(D_1 \cap D_2 \mid W) = \frac{p_1 p_2}{P(W)}.$

unless $w_0=1$(see part (d)).

Thus by contradiction, these events are not conditionally independent.

### (d) What if $w_0=0$?

If $w_0=0$, we have $P(D_1 \cap D_2 \mid W) = \frac{p_1 p_2}{p_1 + p_2 - p_1p_2}.$

Which is still NOT equal to $P(D_1\mid W) P(D_2\mid W) = \frac{p_1} {p_1 + p_2 - p_1p_2} \times \frac{p_2} {p_1 + p_2 - p_1p_2}.$ Thus they are not conditionally independent.

#### Explanation: Two independent events may NOT be conditionally independent

This is a gnarly example of the counter-intuitive nature of probability. This is related to the Berkson’s Paradox!

Note that one way to look at independence of two events $A, B$ is to think about things this way:

… knowing the outcome of event B has no impact on the likelihood of A happening.

So if two events are dependent then…

… knowing the outcome of event B changes the likelihood of A happening.

In this case, we are given that someone has weird symptoms: $W$. Then two events we are considering are: $D_1 \mid W$and $D_2 \mid W$. Since we are given $W$, they must have $D_1$and $D_2$since $w_0=0$.

Suppose you know that they don’t have $D_1$. Then they MUST have $D_2$ since they have weird symptoms. Hence knowing $D_1 \mid W$ impacts the likelihood of $D_2 \mid W$.

The only way $D_1 \mid W$ and $D_2 \mid W$ are independent is if $w_0=1$. In that case $w_0 q_1q_2 = 1-p_1 -p_2 +p_1 p_2$ and $P(W) = (p_1 + p_2 - p_1p_2) + (1-p_1 -p_2 +p_1 p_2) =1$. Then knowing $W$won’t change anything since everyone has weird symptoms.