Independence and its Properties
Intuitively, we should consider two events independent when knowing that one of them happened doesn’t affect the probability of the other. This motivates our next definition.
Definition 6.1 (Independence of two events) Two positive probability events \(A\) and \(B\) are independent if
\[
P(A|B) = P(A) \text{ and } P(B|A) = P(B).
\tag{6.1}\]
We could have just required one of the two conditions \(P(A | B) = P(A)\) and \(P(B | A) = P(B)\). We’ll soon see that one implies the other. We opt to require both because this emphasizes that independence is symmetric. If \(A\) is independent of \(B\), then \(B\) is independent of \(A\).
Due to the multiplication rule (Corollary 5.1), it turns out that two events are independent exactly when the probability of them both happening is equal to the product of their individual probabilities.
Proposition 6.1 (Independence of two events) Two positive probability events \(A\) and \(B\) are independent if and only if
\[
P(A \cap B) = P(A)P(B).
\tag{6.2}\]
When either \(A\) or \(B\) has zero probability, then both the left and right hand side of Equation 6.2 are zero (see the note after Corollary 5.1 for why). Because of this, we also (somewhat informally) consider probability zero events to be independent of all other events.
The multiplication rule (Corollary 5.1) tells us that \[P(A \cap B) = P(B)P(A|B) = P(A)P(B|A).\]
If we know that \(A\) and \(B\) are independent, then in particular, \(P(A | B) = P(A)\).
- As desired, the first equality becomes \(P(A \cap B) = P(B) P(A)\).
- For free, we can also see how one condition of independence implies the other. The second equality becomes \(P(B) P(A) = P(A) P(B | A)\), and dividing both sides by \(P(A)\) gives \(P(B | A) = P(B)\). Reversing the roles of \(A\) and \(B\) would show that \(P(B | A) = P(B)\) implies \(P(A | B) = P(A)\).
Conversely, if we know that \(P(A \cap B) = P(A) P(B)\), then the first equality becomes \[ P(A) P(B) = P(B) P(A | B). \] Dividing both sides by \(P(B)\) shows that \(P(A | B) = P(A)\). We’ve seen that this also implies that \(P(B | A) = P(B)\), so \(A\) and \(B\) are independent.
Our next example helps build intuition regarding when events are and aren’t independent.
Example 6.1 (Rolling two dice and independence) Recall our experiment of rolling two fair dice, which has \(36\) equally likely outcomes in its sample space:
Consider the following four events, which we’ve highlighted in Figure 6.1:
\(A\): the first (white) die lands on \(6\) (marked with red hatchings).
\(B\): the second (black) die lands on \(5\) (marked with blue hatchings).
\(C\): the rolled numbers sum to \(11\) (marked with orange background).
\(D\): the rolled numbers sum to \(7\) (marked with green background).
We will use Proposition 6.1 to check whether or not certain pairs of these events are independent.
\(A\) and \(B\): Intuitively, we should expect that the number that one die lands on has nothing to do with the number the other lands on, so the events \(A\) and \(B\) should be independent. Indeed, \[
P(A \cap B) = \frac{1}{36} = \frac{6}{36} \times \frac{6}{36} = P(A)P(B).
\]
\(A\) and \(C\): Knowing that the first die landed on \(6\) should make it more likely that the sum of the rolled die is \(11\), so we shouldn’t expect these two events to be independent. Indeed they are not, because \[
P(A \cap C) = \frac{1}{36} \neq \frac{6}{36} \times \frac{2}{36} = P(A)P(C).
\]
\(A\) and \(D\): It feels like knowing that the first die landed on \(6\) should inform to some extent regarding whether the sum of the rolled die will be \(7\). However, the computation \[
P(A \cap D) = \frac{1}{36} = \frac{6}{36} \times \frac{6}{36} = P(A)P(D)
\] shows that these events are actually independent. Knowing \(A\) happened doesn’t change the probability of \(D\), and vice-versa. The identical argument implies that \(B\) and \(D\) are also independent. When we later discuss what independence means for collections of more than two events, we will see that these events do still have a relationship with one another.
\(C\) and \(D\): If we know the sum of the rolled die is \(11\) then the sum of the rolled die can certainly not be \(7\), and vice-versa. For this reason we shouldn’t expect \(C\) and \(D\) to be independent. We can quickly verify that they are indeed not: \[
P(C \cap D) = \frac{0}{36} \neq \frac{2}{36} \times \frac{6}{36} = P(C)P(D).
\]
One may expect that, if \(A\) and \(B\) are independent, then \(B\) not happening also has no influence on \(A\), i.e., \(A\) and \(B^c\) are independent. This is the subject of our next proposition, which tells us that independence is preserved when we take complements.
Proposition 6.2 (Complements preserve independence) If \(A\) and \(B\) are independent, then:
\(A\) and \(B^c\) are independent,
\(A^c\) and \(B\) are independent,
\(A^c\) and \(B^c\) are independent.
- By Proposition 6.1, it suffices to show that \(P(A \cap B^c)\) satisfies Equation 6.2. Recall that \(A\) can be written as the disjoint union of \(A \cap B\) and \(A \cap B^c\) (see Figure 4.5). We can use the independence of \(A\) and \(B\) along with Axiom 3 to see that
\[\begin{align*}
P(A) &= P(A \cap B) + P(A \cap B^{c}) & \text{(Axiom 3)}\\
&= P(A)P(B) + P(A \cap B^{c}). & \text{(independence)}
\end{align*}\]
\(\textcolor{white}{0000}\) Subtracting over \(P(A)P(B)\), we see see that
\[\begin{align*}
P(A \cap B^{c}) &= P(A) - P(A)P(B) & \\
&= P(A)(1 - P(B)) & \text{(factor)}\\
&= P(A)P(B^c). & \text{(complement rule)}
\end{align*}\]
Apply the same argument as above but swap the roles of \(A\) and \(B\).
From our first argument we know that \(A\) and \(B\) being independent implies that \(A\) and \(B^c\) are independent. From our second argument we know that \(A\) and \(B^c\) being independent implies that \(A^c\) and \(B^c\) are independent.
Our next example briefly shows how you may use Proposition 6.2.
Example 6.2 (Rolling two dice, complements, and independence) Recall the experiment of rolling two fair die from Example 6.1. What’s the probability that first die does not land on \(6\) and the sum of the dice also won’t be \(7\)?
We showed in Example 6.1 that the event \(A\) of the first die landing on \(6\) and the event \(D\) of the rolled numbers summing to \(7\) are independent. Therefore Proposition 6.2 tells us that \(A^c\) and \(D^c\) are also independent. Applying Proposition 6.1 we find that \[\begin{align*}
P(A^c \cap D^c) &= P(A^c)P(D^c) & \text{(independence)}\\
&= (1-P(A))(1 - P(D)) & \text{(complement rule)}\\
&= \left(1 - \frac{6}{36}\right) \left(1 - \frac{6}{36}\right) & (\text{plug in probabilities})\\
&= \frac{25}{36}. & \text{(simplify)}
\end{align*}\]
More often than not, we work with more than just two independent events. Like for the case of two events, we consider a whole collection of events independent when some of the events happening doesn’t influence whether another one will happen or not.
Definition 6.2 (Independence of many events) We call a (potentially infinite) collection of positive probability events \(A_1, A_2, \dots\) independent if the probability of any one of them happening does not change given that some other finite number of them have happened.
More concretely, for any two events, \[
\begin{aligned}
P(A_i| A_j) &= P(A_i) & i &\neq j,
\end{aligned}
\tag{6.3}\] and for any three events, \[
\begin{aligned}
P(A_i| A_j, A_k) &= P(A_i) & i&\neq j \neq k,
\end{aligned}
\tag{6.4}\] and for any four events, \[
\begin{aligned}
P(A_i| A_j, A_k, A_{\ell}) &= P(A_i) & i \neq j \neq k \neq \ell,
\end{aligned}
\tag{6.5}\]
and so on.
As was the case for two events, a collection of events is independent exactly when the probability of some of them happening simultaneously is the product of the individual probabilities.
Theorem 6.1 (Independence of many events) The (potentially infinite) collection of positive probability events \(A_1, A_2, \dots\) are all independent if and only if the probability some finite number of them happening simultaneously is the product of each event’s individual probability. That is, for any two events,
\[
\begin{aligned}
P(A_i \cap A_j) &= P(A_i)P(A_j) & i &\neq j,
\end{aligned}
\tag{6.6}\]
and for any three events,
\[
\begin{aligned}
P(A_i \cap A_j \cap A_k) &= P(A_i)P(A_j)P(A_j) & i &\neq j \neq k,
\end{aligned}
\tag{6.7}\]
and so on.
For the same reason as in Proposition 6.1, when we add probability zero events to a collection of independent events we still informally consider the collection to be independent.
Because the statement is an if and only if statement, we have to prove both the forward and reverse directions.
Forward: To show the forward direction, we need to show that if probabilities factor as in Equation 6.6 and Equation 6.7, then the conditional probabilities satisfy Equation 6.3, Equation 6.4 and so on. This is proved by something called strong mathematical induction. We’ll walk through and illustrate the argument. To start we show that Equation 6.3 holds. It suffices to show that for \(i \neq j\) we must have \(P(A_j | A_i) = P(A_j)\). This follows by the multiplication rule: \[\begin{align*}
P(A_i)P(A_j) &= P(A_i \cap A_j) & \text{(assumed factoring)} \\
&= P(A_i) P(A_j | A_i). & \text{(multiplication rule)}
\end{align*}\] Dividing both sides by \(P(A_i)\) gives that \(P(A_j | A_i) = P(A_j)\). Now we can show that Equation 6.4 holds. That is, for \(i \neq j \neq k\) we must have that \(P(A_k | A_i, A_j) = P(A_k)\). We use the same argument, but at the end use that we’ve shown that Equation 6.3 holds. \[\begin{align*}
P(A_i)P(A_k)P(A_j) &= P(A_i \cap A_k \cap A_j) & \text{(assumed factoring)}\\
&= P(A_i) P(A_j | A_i)P(A_k | A_i, A_j) & \text{(multiplication rule)} \\
&= P(A_i) P(A_j) P(A_k | A_i, A_j). & \text{(previous result)}
\end{align*}\] Dividing both sides by \(P(A_i)P(A_j)\), we get that \(P(A_k | A_i, A_j) = P(A_k)\). To show that Equation 6.5 holds, we would apply the same argument but make use of both Equation 6.3 and Equation 6.4 at the end. Then we could use all three of Equation 6.3, Equation 6.4, and Equation 6.5 to prove the next statement, and so on.
Reverse: To show the reverse direction we need to show that if the conditional probabilities satisfy equations like Equation 6.3 and Equation 6.4 then we can factor the unconditional probabilities like in Equation 6.6 and Equation 6.7 and so on. This implied immediately by the multiplication rule (Theorem 5.1). For simplicity, we just show that we can factor in the case of three events, but the same argument applies when we consider any arbitrary finite number of events. For \(i\neq j \neq k\): \[\begin{align*}
P(A_i, A_j, A_k) &= P(A_i) P(A_j|A_i)P(A_k|A_i, A_j) & \text{(multiplication rule)}\\
&= P(A_i) P(A_j)P(A_k). & \text{(def. of independence)}
\end{align*}\]
Verifying that many events are collectively independent can be tricky, and we have to be careful when doing it. As the next example illustrates, pairwise independence of a collection of events does not imply independence of the whole collection.
Example 6.3 (Rolling two dice and independence of many events) Recall Example 6.1, where we showed that the events \(A\) of rolling a \(6\) with the first die, \(B\) of rolling a \(5\) with the second die, and \(D\) of the rolled numbers summing to \(7\) are pairwise independent.
However, \(A\), \(B\), and \(D\) are not collectively independent. If \(A\) and \(B\) both happen, then we know \(D\) can’t possibly happen because the rolled numbers sum to \(11\) and therefore cannot posssibly sum to \(7\). Concretely,
\[\begin{align*}
P(D | A, B) &= \frac{P(D, A, B)}{P(A, B)} & \text{(def. of conditional probability)}\\
&= \frac{0/36}{1/36} & \text{(equally likely events)}\\
&= 0 & \text{(simplify)}
\end{align*}\]
so,
\[
P(D | A, B) = 0 \neq \frac{6}{36} = P(D),
\]
violating the definition of independence (Definition 6.2).
Similarly we could have observed that
\[
P(A, B, D) = 0 \neq \frac{6}{36} \times \frac{6}{36} \times \frac{6}{36} = P(A)P(B)P(D),
\]
so independence of these events would violate Theorem 6.1.
So why doesn’t pairwise independence imply independence of the entire collection? When a collection is pairwise independent, any one event doesn’t inform us about the other events individually, but it may still inform us about them collectively. In this case, knowing that \(D\) has happened doesn’t change the probability of \(A\) happening or the probability of \(B\) happening, but it does change the probability of \(A \and B\) happening.
Just as taking complements preserves independence when there are two events, set operations generally preserve independence when there are many independent events. We give a formal statement of this fact in Proposition 6.3, which we state without proof.
Proposition 6.3 (Independence and set operations) Consider a (potentially infinite) collection \(A_1, A_2, \dots\) of independent events and suppose we make (potentially infinite) new events \(E_1, E_2, \dots\) by performing set operations on \(A_1, A_2, \dots\). If each \(A_i\) is used to construct at most one of the \(E_j\), then the \(E_1, E_2, \dots\) are also an independent collection of events.
Suppose that \(A_1, A_2, A_3, A_4, A_5\) are all independent. Then Proposition 6.3 tells us that the events \[
E_1 = (A_1 \cap A_2)^c, \qquad E_2 = A_3^c, \qquad E_3 = A_4 \cup A_5
\] also form an independent collection. Proposition 6.3 would not guarantee however that
\[
E_1 = (A_1 \cap A_2)^c, \qquad E_2 = A_3^c \cup A_4, \qquad E_3 = A_4 \cup A_5
\] are independent, because now \(A_4\) is used to construct both \(E_2\) and \(E_3\).
Examples
In this section we work through a handful of problems that leverage independence between different events. We often we will refer to some physical actions (e.g., coin flips, die rolls) as independent and leave it to you to infer what specific events are independent of one another.
Although independence is a natural assumption in many settings, it can lead to results that may initially feel counterintuitive. The gambler’s fallacy is a classic example of this.
Example 6.4 (Gambler’s Fallacy) You and your friend are repeatedly flipping a fair coin and making bets on whether it comes up heads or tails. Since the coin flips have no infuence on each other, it’s safe to assume that they’re all independent.
Miraculously, the fist nine flips all come up heads. “It’s super unlikely for a fair coin to come up heads ten times in a row,” your friend says before the tenth flip. “The next flip has got to be a tails. I’m willing to bet a lot of money on it.”
On the one hand, your friend is right. It’s super unlikely for the game to start with ten consecutive heads. Letting \(A_i\) be the event that the \(i\)th flip is heads, we can compute this probability to be less than a tenth of a percent:
\[\begin{align*}
P(10 \text{ heads to start}) &= P(A_1, \dots, A_{10}) & \text{(same event)}\\
&= P(A_1) \dots P(A_{10}) & \text{(independence)} \\
&= \underbrace{\frac{1}{2} \times \dots \times \frac{1}{2}}_{10 \text{ times}} & \text{(fair coin)}\\
&= \frac{1}{1024}. & \text{(simplify)}
\end{align*}\]
But, regardless of whether this event is unlikely or not, the fact is that the first nine flips have already come up heads and they have no influence on the tenth one! Your friend should really be considering the conditional probability that the next flip is heads, given that we saw nine heads to start the game. Because of independence, they will quickly realize that this probability is \(1/2\):
\[\begin{align*}
P(10 \text{th flip is heads} |9 \text{ heads to start}) &= P(A_{10} | A_1, \dots, A_9) \\
&= P(A_{10}) & \text{(independence)} \\
&= \frac{1}{2} & \text{(fair coin)}\\
\end{align*}\]
Because the previous flips are independent of the current flip, they have no influence on it!
Although it may seem to give counterintuitive results at first, independence is a really powerful computational tool. We illustrate this by re-solving Cardano’s problem (Example 1.1), but this time relying on independence.
Example 6.5 Suppose we roll a fair die \(k\) times and the die rolls are independent of one another. What’s the probability that we see at least one \(6\)?
Let \(A_i\) be the event that the \(i\)th roll is not a \(6\). Using independence (Definition 6.2) and the complement rule (Proposition 4.1), we can compute that
\[\begin{align*}
P(\text{at least one six}) &= 1 - P(\text{no sixes}) & \text{(complement rule)}\\
&= 1 - P(A_1, \dots, A_k) & \text{(same event)}\\
&= 1 - P(A_1) \dots P(A_k) & \text{(independence)}\\
&= 1- \underbrace{\frac{5}{6} \times \dots \times \frac{5}{6}}_{k \text{ times}} & \text{(plug in probabilities)}\\
&= 1 - (5/6)^k
\end{align*}\]
This agrees with our earlier answer from Example 2.6.
To finish the section, we give a few more examples of how independence enables us to tackle complicated problems.
Example 6.6 (Quality control for medical devices) Suppose, like in Example 4.5, we’re manufacturing batches of \(n=50\) medical devices. If each device we make has some failure probability \(p\), how low does \(p\) need to be for us to ensure that no devices fail with probability at least \(99\%\)? We’ll imagine that failures are one-off mishaps on the assembly line that have nothing to do with each other, i.e., the failures of different devices are independent.
Let \(A_i\) be the event that the \(i\)th device does not fail. The complement rule (Proposition 4.1) tells us that \(P(A_i) = 1 - p\). Using this we can compute that
\[\begin{align*}
P(\text{no device fails}) &= P(A_1, A_2, \dots A_{50}) & \text{(same event)}\\
&= P(A_1) \dots P(A_{50}) & \text{(independence)}\\
&= (1-p)^{50} & \text{(earlier computation)}
\end{align*}\]
Some standard algebra tells us that we’ll have \(P(\text{no device fails}) = (1-p)^{50} \geq 0.99\) if and only if
\[ p \leq 1 - (0.99)^{1/50} \approx 0.000201.\]
This is almost identical to the worst case analysis we did with the union bound in Example 4.5, which turns out to be not so bad after all!
Example 6.7 (Winning a pass-line bet conditional on come-out roll) Coming soon!
Conditional Independence
Sometimes events are independent conditional on some other event having happened.
Definition 6.3 (Conditional independence) We say that a collection of (potentially infinite) positive probability events \(A_1, A_2, \dots\) are conditionally independent given another positive probability event \(B\) if they are independent under the conditional probability function \(P(\cdot |B)\).
Let’s take a closer look at what it means for two events \(A_1\) and \(A_2\) to be independent under the conditional probability function \(P(\cdot |B)\) (see Theorem 5.2 for a refresher on conditional probability functions). According to Definition 6.1, this happens when:
\[
P(A_1 | B, A_2) = P(A_1| B) \text{ and } P(A_2 | B, A_1) = P(A_2| B).
\]
Applying Proposition 6.1 with the conditional probability function \(P(\cdot |B)\) tells us that this is equivalent to the conditional probability of both events happening being equal to the product of the individual conditional probabilities of each event:
\[
P(A_1 \cap A_2 | B) = P(A_1 | B)P(A_2 | B).
\]
Our next example shows that conditional independence does not imply unconditional independence.
Example 6.8 (COVID-19 antigen testing with one nurse) Suppose we select a random New York resident around the end March 2020 and want to test to see if they have COVID-19. We send them to a pop-up testing site where a nurse is administering an antigen test (i.e., a rapid test). Randomly, these tests can give incorrect readings (i.e., the test comes back positive when the person doesn’t have COVID-19 or negative when they do) due to manufacturing error and/or improper administration. To account for the possibility of an incorrect reading, the nurse administers two tests.
We consider the following five events
\(C_1\): The first test reading is correct.
\(C_2\): The second test reading is correct.
\(T_1\): The first test comes back positive.
\(T_2\): The second test comes back positive.
\(I\): The person is infected with COVID-19.
And we ask four questions:
Is it reasonable to think that \(C_1\) and \(C_2\) are independent?
Is it reasonable to think that \(T_1\) and \(T_2\) are independent?
Is it reasonable to think that \(C_1\) and \(C_2\) are conditionally independent given \(I\)?
Is it reasonable to think that \(T_1\) and \(T_2\) are conditionally independent given \(I\)?
We answer these questions below:
Yes. It’s certainly possible that slip-ups in administration or mistakes in manufacturing could be due to random, one-off incidences, and whether or not one test is correct tells us nothing about whether the other is.
No. If one of the tests comes back positive, that should suggest an increased chance that the person actually has COVID-19, making it more likely that the other test will also be positive. We haven’t provided you with enough information or the tools to show this explicitly yet, but we will in the next chapter.
Yes. If slip-ups in administration or mistakes in manufacturing are due to random, one-off incidences, then the person having COVID-19 doesn’t change that whether or not one test is correct tells us nothing about whether the other is.
Yes. Conditional on the person having COVID-19, the tests coming back positive is the same as the tests giving a correct reading. We’ve already discussed why the tests’ correctness could be conditionally independent given that the person has COVID-19.
Frequetism: For a frequentist, selecting a random person is essential. If the person wasn’t random and was pre-specified, then there would be no randomness in whether or not they had COVID-19: they would either have it or they don’t. If they have it, conditioning on \(I\) is identical to conditioning on the whole sample space, which does nothing. If they don’t have it, then conditioning on \(I\) is like conditioning on \(\emptyset\) which is impossible. As such, when a frequentist makes a statement like, “there’s a 20% chance that person has COVID-19,” they’re typically imagining that the person was randomly selected by some mechanism, and that people selected by said mechanism have COVID-19 \(20\%\) of the time. It’s seldom a statement about a specific individual. If the person they are speaking of has specific relevant characteristics (e.g., age, membership to a certain racial or demographic group) then a frequentist can imagine the person was randomly sampled from the population of people that share those characteristics.
Bayesianism: A Bayesian, on the other hand, can do this analysis even if we pre-specify a specific person. A Bayesian can simply assign a degree of belief as to whether the person has COVID-19 or not. For a Bayesian, a “randomly sampled” person is essentially a person who, in their eyes, is indistinguishable from the population they come from. In such settings, it’s reasonable for a Bayesian to set their belief about a person having COVID-19 to be the proportion of people in the population who have COVID-19. If the person has relevant characteristics that distinguish them from the population, then a Bayesian can restrict the population to be people that share those characteristics.
In our above example, we saw two independent events that were also conditionally independent. This is not always the case. Our next example shows that unconditional independence does not imply conditional independence either!
Example 6.9 (Rolling two dice and conditional independence) Recall the experiment of rolling two fair die from Example 6.1. We established in Example 6.1 that the event \(A\) of rolling a \(6\) with the first die and the event \(B\) of rolling a \(5\) with the second die are independent. We will show, however, that they are not independent given \(C\), the event of the rolled die summing to \(11\).
Conditional on \(C\) and \(B\) happening, \(A\) must happen:
\[\begin{align*}
P(A | C, B) &= \frac{P(A, C, B)}{P(C, B)} & \text{(def. of conditional probability)}\\
&= \frac{1/36}{1/36} & \text{(equally likely outcomes)}\\
&= 1 & \text{(simplify)}
\end{align*}\]
But conditional on just \(C\) happening, \(A\) only happens with probability \(1/2\):
\[\begin{align*}
P(A | C) &= \frac{P(A, C)}{P(C)} & \text{(def. of conditional probability)}\\
&= \frac{1/36}{2/36} & \text{(equally likely outcomes)}\\
&= \frac{1}{2} & \text{(simplify)}.
\end{align*}\]
Therefore \[
P(A | C, B) \neq P(A|C),
\] so \(A\) and \(B\) are not conditionally independent given \(C\).
Alternatively, we could’ve done similar computations to show that \[
P(A, B| C) = \frac{1}{2} \neq \frac{1}{2} \times \frac{1}{2} = P(A|C)P(B|C),
\] which implies the same result.
In fact, as our next example shows, even if events are conditionally independent given \(B\), they may not be conditionally independent given \(B^c\).
Example 6.10 (COVID-19 antigen testing with multiple nurses) Recall the COVID-19 testing example from Example 6.8, but now suppose that there are three nurses at our pop-up testing site. Two of these three nurses are experts, and their tests come back with the correct reading much more often than the third nurse, who is inexperienced and often administers the test incorrectly.
Our selected resident waits in line and is randomly helped by whichever nurse frees up first. Let \(N_1\) and \(N_2\) be the event that one of the two expert nurses administer the tests and \(N_3\) be the event that the other nurse does. We ask two questions:
Is it reasonable to think that the events \(C_1\) and \(C_2\) are conditionally independent given the event \(N_1\)?
Is it reasonable to think that the events \(C_1\) and \(C_2\) are conditionally independent given the event \(N_1^c\)?
We answer these questions below:
Yes. Since the same expert nurse is administering both tests, this is like our setting in Example 6.8, and it’s reasonable to think that correctness of one test has no influence on the correctness of the other.
No. Given \(N_1^c\) we know that either the other expert nurse or the inexperienced nurse is administering the tests, but we’re not sure not which one. If we see that the first test is incorrect, we’ll think it’s more likely that the inexperienced nurse is administering the tests and that it’s therefore more likely for second test to also be incorrect. Therefore, conditional on \(N_1^c\), whether or not one test is correct influences the probability that the other is. Again, we’ll equip you with the tools to show this explicitly in the next chapter.
As we have seen, conditional independence is quite brittle, and we should be cautious when assuming it.
