# Set notation

I’ve just started the Mathematical Biostatistics Boot Camp 1 and to help me remember the set notations introduced in the first lecture, I’ll include them here:

The sample space, $$\Omega$$ (upper case omega), is the collection of possible outcomes of an experiment, such as a die roll:

$$!\Omega = \{1, 2, 3, 4, 5, 6\}$$

An event, say E, is a subset of $$\Omega$$, such as the even dice rolls:

$$!E = \{2, 4, 6\}$$

An elementary or simple event is a particular result of an experiment, such as the roll of 4 (represented as a lowercase omega):

$$!\omega = 4$$

A null event or the empty set is represented as $$\emptyset$$.

### Interpretations of set operations

1. Is an Element Of

$$!\omega \in E$$

implies that E occurs when $$\omega$$ occurs; for example if $$\omega = 4$$ occurs, then the event E (even numbers) occurs.

2. Not an element of

$$!\omega \not \in E$$

implies that E does not occur when $$\omega$$ occurs; for example if $$\omega = 5$$, E (even numbers) does not occur.

3. Subset

$$!E \subset F$$

implies that the occurrence of E implies the occurrence of F; for example if $$E = \{2, 4, 6\}$$ and $$F = \{2, 4, 5, 6\}$$.

4. Intersect

$$!E \cap F$$

implies the event that both E and F occur; for example if $$E = \{2, 4, 6\}$$ (even numbers) and $$F = \{2, 3, 5\}$$ (prime numbers), then $$E \cap F = 2$$.

5. Union

$$!E \cup F$$

implies the event that at least one of E or F occur; for example if E = even numbers and F = prime numbers, $$E \cup F$$ implies that I get either an even number of a prime number or both.

6. If intersect = null event

$$!E \cap F = \emptyset$$

implies that E and F are mutually exclusive, or cannot both simultaneously occur; for example if E = even numbers and F = odd numbers.

7. Complement

$$!E^\complement$$

is the event that E does not occur; for example if E are the even numbers then $$E^\complement$$ are the odd numbers.

### Set theory facts

DeMorgan’s laws

$$!(A \cap B)^\complement = A^\complement \cup B^\complement$$ $$!(A \cup B)^\complement = A^\complement \cap B^\complement$$

To remember these, think of distributing the complements to A and B and then invert the cap into a cup or vice vera.

Indeed DeMorgan’s laws holds.

$$!(A^\complement)^\complement = A$$

The complement of A complemented = A

And lastly:

$$!(A \cup B) \cap C = (A \cap C) \cup (B \cap C)$$

The lecturer wasn’t lying.

### Probability and set notation

Here’s how we can use set notation to answer questions on probability (this question appeared on the Coursera biostatistics course).

Consider influenza epidemics for two parent heterosexual families. Suppose that the probability is 17% that at least one of the parents has contracted the disease. The probability that the father has contracted influenza is 12% while the probability that both the mother and father have contracted the disease is 6%. What is the probability that the mother has contracted influenza?

Let M be the mother and F be the father. Recall that the union ($$\cup$$) implies the event that at least one of M or F occurred, therefore:

$$!P(M \cup F) = 17\%.$$

Since the probability that the father contracted influenza is 12%:

$$!P(F) = 12\%.$$

An intersection ($$\cap$$) implies the event that both M and F occur, i.e. the probability that both the mother and father have contracted the disease:

$$!P(M \cap F) = 6\%.$$

We know that:

$$!P(M \cup F) = P(M) + P(F) – P(M \cap F).$$

It is worth noting that if the events are mutually exclusive, $$P(M \cap F) = 0$$, since they both can’t occur together, but in our case the events can occur together. From the above equation, we get:

$$!17 = P(M) + 12 – 6.$$

So the probability of the mother contracting the disease is 11%.

### Conditional probability

The probability of getting a 1 when rolling a fair die is 1/6. If we are given the extra information that the die roll was an odd number (1, 3 or 5), conditional on this new information, the probability of a 1 is now one third. Let’s test this example out:

Let B be an event so that P(B) > 0, i.e. it’s an event that actually occurs. Then the conditional probability of an event A given that B has occurred is:

$$!P(A | B) = \frac{P(A \cap B)}{P(B)}.$$

So using the example above:

$$!B = \{1, 3, 5\}\ and\ A = \{1\}$$

So:

$$!P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A)}{P(B)} = \frac{\frac{1}{6}}{\frac{3}{6}} = \frac{1}{3}$$

### $$\LaTeX$$ links

For set notation in $$\LaTeX$$, see here.

$$\LaTeX$$ symbols.