Distributions
Example of frequency distribution:
Lithuanian Parliamentary Seats, 2000
Party Abbreviation | Seats Won |
---|---|
ABSK | 51 |
LLS | 33 |
NS | 28 |
TS-LK | 9 |
LVP | 4 |
LKDP | 2 |
LCS | 2 |
LLRA | 2 |
KDS | 1 |
NKS | 1 |
LLS | 1 |
JL/PKS | 1 |
We have produced such a contingency table here:
The Fearon and Laitin (1996) Contingency Table
Ethnic Homogeneity | Ethnic Heterogeneity | |
---|---|---|
Cooperation | Rare | Common |
Conflict | Rare | Rare |
For example, let’s say we are interested in the probability of voting, but voting is conditional on whether or not it is raining:
Parameter and parameter space
To illustrate, let’s consider the case of voter turnout where we ask, “Which registered voters cast ballots?” There are two outcomes for each voter: (0) did not cast a ballot and (1) cast a ballot. We can write the following PMF:
\[p(y_i = 0) = \pi\]
\[p(y_i = 1) = 1 - \pi\]
\[Pr(Y \leq y) = \sum_{i \leq y} p(i)\]
The first PMF we will consider applies to binary variables only and can be written as
\[ Pr(Y = y|p) = \begin{cases} 1-p & \text{for } y = 0, \\ p & \text{for } y = 1. \end{cases} \]
\[ Pr(Y = y|p) = p^y(1-p)^{1-y} \]
where \(y = 0\) or \(y = 1\). If we solve equation (10.4) for \(y = 0\) and \(y = 1\), we get the information provided in equation: \(Pr(Y = 0) = p^0(1-p)^{1-0} = 1-p\), and \(Pr(Y = 1) = p^1(1-p)^{1-1} = p\).
\[ Pr(Y = y|n,p) = \binom{n}{y}p^y(1-p)^{n-y} \]
where \(n > y\), \(n\), and \(y\) are positive integers and \(0 \leq p \leq 1\). The variables \(n\) and \(y\) in equation (10.5) represent the number of cases (or observations) and the number of positive outcomes, respectively.
\[ Pr(Y = y \mid \mu) = \frac{\mu^y}{y! \times e^{-\mu}} \]
Let’s consider a common example one encounters in game theory and expected utility theory.
\[ E_X[X] = \sum_i x_i (Pr(X = x_i)) \]
Any Questions?