Joint probability of events

Select one case-coded pronoun form from the population represented by the probability space below. We want the probability that the selected form is both plural and accusative.

Return to the four form sample space

\Omega_4\equiv\{\textit{he},\textit{him},\textit{they},\textit{them}\}

with the probability measure below.

outcome	probability
he	.20
him	.30
they	.10
them	.40

Let P be the plural event and A the accusative event:

P\equiv\{\textit{they},\textit{them}\}

and

A\equiv\{\textit{him},\textit{them}\}.

Joint-probability notation

The joint probability of events B and C is the probability of their intersection. We define the comma notation by

\mathbb{P}(B,C) \equiv \mathbb{P}(B\cap C).

The expression \mathbb{P}(B,C) is read as “the joint probability of B and C.” The comma means that both events occur. We use \mathbb{P}(B,C) for joint probabilities from this point forward; B\cap C remains the underlying event.

The joint probability

Begin with the set operation:

P\cap A=\{\textit{them}\}.

Then apply the probability measure:

\begin{aligned} \mathbb{P}(P,A) &=\mathbb{P}(\{\textit{them}\})\\ &=.40. \end{aligned}

The value \mathbb{P}(P,A)=.40 is relative to the full sample space. The conditional probability of accusative case among plural forms has a different denominator and will be defined on the next page.

Arrange all joint events in a table

The number distinction P versus P^c and the case distinction A versus A^c divide the sample space into four nonoverlapping intersections.

	accusative A	nominative A^c	total
plural P	\mathbb{P}(P,A)=.40	\mathbb{P}(P,A^c)=.10	.50
singular P^c	\mathbb{P}(P^c,A)=.30	\mathbb{P}(P^c,A^c)=.20	.50
total	.70	.30	1

For instance, the upper right cell represents the event

P\cap A^c=\{\textit{they}\},

so \mathbb{P}(P,A^c)=.10. The lower left cell represents

P^c\cap A=\{\textit{him}\},

so \mathbb{P}(P^c,A)=.30.

The four interior cells are mutually exclusive and exhaustive. Their probabilities must sum to one:

.40+.10+.30+.20=1.

Marginal probabilities

The row and column totals are marginal probabilities. To obtain the marginal probability of P, partition the sample space by A and A^c and add the corresponding joint probabilities:

\begin{aligned} \mathbb{P}(P) &=\mathbb{P}(P,A)+\mathbb{P}(P,A^c)\\ &=.40+.10\\ &=.50. \end{aligned}

This equality follows from countable additivity because P\cap A and P\cap A^c are disjoint and their union is P. In this calculation, marginalizing over case means summing over the exhaustive case alternatives while keeping the number event fixed.

Reconstruct the table in base R

The following code stores the four joint probabilities in a matrix and computes its margins.

Code

joint <- matrix(
  c(.40, .30, .10, .20),
  nrow = 2,
  dimnames = list(
    number = c("plural", "singular"),
    case = c("accusative", "nominative")
  )
)

joint
rowSums(joint)
colSums(joint)
sum(joint)

The row sums are .50 and .50, the column sums are .70 and .30, and the four cells sum to one. These are the margins and total shown in the table.

Information in the joint-probability table

A joint table records the probabilities of the four number-by-case combinations. Two probability measures can both have \mathbb{P}(P)=.50 and \mathbb{P}(A)=.70 while assigning different values to \mathbb{P}(P,A). The two marginal probabilities do not determine the four joint probabilities.

For instance, an association between plurality and accusative case is represented by a difference between \mathbb{P}(A\mid P) and \mathbb{P}(A\mid P^c). The two margins do not determine those conditional probabilities.

Marginal and joint probabilities

In the toy model, \mathbb{P}(A)=.70 includes both him and them. Only the .40 assigned to them belongs to the event P\cap A, so \mathbb{P}(P,A)=.40.

To compute a joint probability, identify the intersection denoted by the comma and apply the probability measure to that event.

Check your understanding

Write the event represented by each of the four interior cells in the table.
Compute \mathbb{P}(P^c,A^c) by identifying the corresponding event first.
Explain why \mathbb{P}(P,A)=.40 does not mean that 40\% of plural forms are accusative.
Construct a different four-cell table with the same row and column margins but with \mathbb{P}(P,A)=.35.

--- title: "Joint probability of events" --- Select one case-coded pronoun form from the population represented by the probability space below. We want the probability that the selected form is both plural and accusative. Return to the four form sample space $$ \Omega_4\equiv\{\textit{he},\textit{him},\textit{they},\textit{them}\} $$ with the probability measure below. | outcome | probability | |---|---:| | *he* | $.20$ | | *him* | $.30$ | | *they* | $.10$ | | *them* | $.40$ | Let $P$ be the plural event and $A$ the accusative event: $$ P\equiv\{\textit{they},\textit{them}\} $$ and $$ A\equiv\{\textit{him},\textit{them}\}. $$ ## Joint-probability notation The [**joint probability**](https://bruno.nicenboim.me/bayescogsci/ch-intro.html) of events $B$ and $C$ is the probability of their intersection. We define the comma notation by $$ \mathbb{P}(B,C) \equiv \mathbb{P}(B\cap C). $$ The expression $\mathbb{P}(B,C)$ is read as “the joint probability of $B$ and $C$.” The comma means that both events occur. We use $\mathbb{P}(B,C)$ for joint probabilities from this point forward; $B\cap C$ remains the underlying event. ## The joint probability Begin with the set operation: $$ P\cap A=\{\textit{them}\}. $$ Then apply the probability measure: $$ \begin{aligned} \mathbb{P}(P,A) &=\mathbb{P}(\{\textit{them}\})\\ &=.40. \end{aligned} $$ The value $\mathbb{P}(P,A)=.40$ is relative to the full sample space. The conditional probability of accusative case among plural forms has a different denominator and will be defined on the next page. ## Arrange all joint events in a table The number distinction $P$ versus $P^c$ and the case distinction $A$ versus $A^c$ divide the sample space into four nonoverlapping intersections. | | accusative $A$ | nominative $A^c$ | total | |---|---:|---:|---:| | plural $P$ | $\mathbb{P}(P,A)=.40$ | $\mathbb{P}(P,A^c)=.10$ | $.50$ | | singular $P^c$ | $\mathbb{P}(P^c,A)=.30$ | $\mathbb{P}(P^c,A^c)=.20$ | $.50$ | | total | $.70$ | $.30$ | $1$ | For instance, the upper right cell represents the event $$ P\cap A^c=\{\textit{they}\}, $$ so $\mathbb{P}(P,A^c)=.10$. The lower left cell represents $$ P^c\cap A=\{\textit{him}\}, $$ so $\mathbb{P}(P^c,A)=.30$. The four interior cells are mutually exclusive and exhaustive. Their probabilities must sum to one: $$ .40+.10+.30+.20=1. $$ ## Marginal probabilities The row and column totals are [**marginal probabilities**](https://bruno.nicenboim.me/bayescogsci/ch-intro.html). To obtain the marginal probability of $P$, partition the sample space by $A$ and $A^c$ and add the corresponding joint probabilities: $$ \begin{aligned} \mathbb{P}(P) &=\mathbb{P}(P,A)+\mathbb{P}(P,A^c)\\ &=.40+.10\\ &=.50. \end{aligned} $$ This equality follows from countable additivity because $P\cap A$ and $P\cap A^c$ are disjoint and their union is $P$. In this calculation, *marginalizing over case* means summing over the exhaustive case alternatives while keeping the number event fixed. ## Reconstruct the table in base R The following code stores the four joint probabilities in a matrix and computes its margins. ```{r} #| eval: false joint <- matrix( c(.40, .30, .10, .20), nrow = 2, dimnames = list( number = c("plural", "singular"), case = c("accusative", "nominative") ) ) joint rowSums(joint) colSums(joint) sum(joint) ``` The row sums are $.50$ and $.50$, the column sums are $.70$ and $.30$, and the four cells sum to one. These are the margins and total shown in the table. ## Information in the joint-probability table A joint table records the probabilities of the four number-by-case combinations. Two probability measures can both have $\mathbb{P}(P)=.50$ and $\mathbb{P}(A)=.70$ while assigning different values to $\mathbb{P}(P,A)$. The two marginal probabilities do not determine the four joint probabilities. For instance, an association between plurality and accusative case is represented by a difference between $\mathbb{P}(A\mid P)$ and $\mathbb{P}(A\mid P^c)$. The two margins do not determine those conditional probabilities. ## Marginal and joint probabilities In the toy model, $\mathbb{P}(A)=.70$ includes both *him* and *them*. Only the $.40$ assigned to *them* belongs to the event $P\cap A$, so $\mathbb{P}(P,A)=.40$. To compute a joint probability, identify the intersection denoted by the comma and apply the probability measure to that event. ## Check your understanding 1. Write the event represented by each of the four interior cells in the table. 2. Compute $\mathbb{P}(P^c,A^c)$ by identifying the corresponding event first. 3. Explain why $\mathbb{P}(P,A)=.40$ does not mean that $40\%$ of plural forms are accusative. 4. Construct a different four-cell table with the same row and column margins but with $\mathbb{P}(P,A)=.35$.