PS 812 - Introduction to Statistical Methods in Political Science
September 8, 2026
tidyverseOverall: An interest in learning and willingness to ask questions.
No formal pre-requisites
Math is not about intrinsic aptitude, it’s about effort
…in mathematics you don’t understand things. You just get used to them.
— John Von Neumann
We don’t do rigor for rigor’s sake
You’re developing your ability to reason and not just memorize
If you’re unsure or confused, always ask - we will match the effort you put in
We all come from different backgrounds. Please have patience with yourself and with others
First and foremost, this course is indebted to Adeline Lo who taught it for the last several years and has provided a wealth of materials.
But it’s also an iteration in the broader project of methods education in political methodology and is indebted to many of those who have taught versions of this course in the past and at other institutions
Figure from Blackwell (2025), Ch. 2
Modeled after Blitzstein and Hwang (2019), Figure 1.1
Probability is a function that maps from the event space \(\mathcal{S}\) to the real numbers.
The probability function \(P\) obeys three axioms (Kolmogorov’s Axioms)
Non-negativity: \(P(A) \geq 0\) for every event \(A \in \mathcal{S}\)
Normalization: \(P(\Omega) = 1\)
Countable additivity: for pairwise disjoint events \(A_1, A_2, A_3, \dotsc\)
\[P\left(\bigcup_{i=1}^{\infty} A_i\right) = \sum_{i=1}^{\infty} P(A_i)\]
A health survey was conducted in a representative sample of adult males in British Columbia of all ages and occupations.
Mr. F. was included in the sample. He was selected by chance from the list of participants.
Which of the following statements is more probable? (check one)
1. Mr. F. has had one or more heart attacks. 2. Mr. F. has had one or more heart attacks and he is over 55 years old.
Split \(B\) into disjoint pieces using \(A\) and \(A^{c}\)
\[B = A \cup (B \cap A^{c})\]
Additivity turns the union into a sum, and \(P(B \cap A^{c}) \geq 0\) by non-negativity
\[P(B) = P(A) + \underbrace{P(B \cap A^{c})}_{\geq 0} \geq P(A)\]
The probability of an event is one minus the probability that it doesn’t happen
\[P(A) = 1 - P(A^{c})\]
Proof
\(A\) and \(A^{c}\) are disjoint and together they make up the whole sample space
\[A \cup A^{c} = \Omega\]
Additivity and then normalization give us
\[P(A) + P(A^{c}) = P(\Omega) = 1\]
Often it’s much easier to compute the probability of the complement – take \(A\) = “at least one” and \(A^{c}\) = “none”
Setting \(A = \Omega\) gives us \(P(\emptyset) = 1 - P(\Omega) = 0\)
The probability of a union of two events is the sum of their respective probabilities minus the probability of their intersection
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Proof
Split both \(A \cup B\) and \(B\) into disjoint pieces using \(A\) and \(A^{c}\)
\[A \cup B = A \cup (B \cap A^{c}) \qquad\qquad B = (A \cap B) \cup (B \cap A^{c})\]
Additivity turns each union into a sum
\[P(A \cup B) = P(A) + P(B \cap A^{c}) \qquad P(B) = P(A \cap B) + P(B \cap A^{c})\]
Solve the second for \(P(B \cap A^{c}) = P(B) - P(A \cap B)\), then substitute into the first.
Rearrange the union rule
\[P(A \cap B) = P(A) + P(B) - P(A \cup B)\]
From earlier, we can split the union into disjoint parts: \(P(A \cup B) = P(A) + P(B \cap A^{c})\)
Non-negativity of \(P(B \cap A^{c})\) bounds the union from below and, symmetrically, splitting on \(B\) and \(B^c\) gives us two bounds:
\[P(A \cup B) \geq P(A) \qquad\qquad P(A \cup B) \geq P(B)\]
Substituting \(P(A \cup B) \geq P(B)\) gives \(P(A \cap B) \leq P(A)\)
Substituting \(P(A \cup B) \geq P(A)\) gives \(P(A \cap B) \leq P(B)\)
Both hold at once, so the intersection is bounded by the smaller of the two
\[P(A \cap B) \leq \min\{P(A),\, P(B)\}\]
That’s an upper bound, how about the lower bound?
Let’s inspect our expression for \(P(A \cap B)\) \[ P(A \cap B) = P(A) + P(B) - P(A \cup B) \]
We have \(P(A) + P(B)\)…but what about \(P(A \cup B)\)
So one lower bound is
\[ P(A \cap B) \geq P(A) + P(B) - 1 \]
But by non-negativity, the probability is also trivially bounded below by \(0\), so we have
\[ P(A \cap B) \geq \max\{0,\, P(A) + P(B) - 1\} \]
How about \(P(A \cup B)\)?
We have a lower bound from earlier
\[ P(A \cup B) \geq \max\{P(A),\, P(B)\} \]
And an upper bound follows from the Union Rule + non-negativity: \(P(A \cap B) \geq 0\)
\[ P(A \cup B) \leq P(A) + P(B) \]
And from before, we also have \(P(A \cup B) \leq 1\)
\[ P(A \cup B) \leq \min\{1,\, P(A) + P(B)\} \]
For any two events \(A, B\), their intersection is bounded by
\[\max\{0,\, P(A) + P(B) - 1\} \;\leq\; P(A \cap B) \;\leq\; \min\{P(A),\, P(B)\}\]
For any two events \(A, B\), their union is bounded by
\[\max\{P(A),\, P(B)\} \;\leq\; P(A \cup B) \;\leq\; \min\{1,\, P(A) + P(B)\}\]
PS 812 - University of Wisconsin-Madison