Conditional probability

The joint probability \mathbb{P}(P,A)=.40 is computed relative to the complete sample space. Now restrict the reference set to plural forms and ask for the proportion that are accusative.

The phrase among plural forms makes P the conditioning event. We write

\mathbb{P}(A\mid P)

and read it as “the probability of A given P.” The event to the right of the bar is the condition. The event to the left is the target.

Conditioning as rescaling

The plural event contains two outcomes:

P=\{\textit{they},\textit{them}\}.

Their original probabilities are .10 and .40. Those values sum to .50, because the plural event receives half of the mass in the full space.

plural outcome	probability in \Omega_4	probability within P
they	.10	.10/.50=.20
them	.40	.40/.50=.80

Dividing each plural probability by .50 produces a probability measure whose sample space is P. Within this restricted space, them receives probability .80. Since them is the only plural accusative outcome,

\mathbb{P}(A\mid P)=.80.

Definition

For events B,C\in\mathcal F with \mathbb{P}(C)>0, the conditional probability of B given C is defined by

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

The denominator is the probability of the reference event C. The numerator is the part of C for which B also occurs.

In the pronoun probability space,

\begin{aligned} \mathbb{P}(A\mid P) &=\frac{\mathbb{P}(A,P)}{\mathbb{P}(P)}\\ &=\frac{.40}{.50}\\ &=.80. \end{aligned}

This rescaling makes the conditioning event certain within itself:

\mathbb{P}(C\mid C) =\frac{\mathbb{P}(C,C)}{\mathbb{P}(C)} =\frac{\mathbb{P}(C)}{\mathbb{P}(C)} =1.

Reversing target and condition

Now ask for the probability of plurality among accusative forms. The numerator is still the same intersection, since P\cap A=A\cap P. The denominator changes:

\begin{aligned} \mathbb{P}(P\mid A) &=\frac{\mathbb{P}(P,A)}{\mathbb{P}(A)}\\ &=\frac{.40}{.70}\\ &\approx .571. \end{aligned}

Thus \mathbb{P}(A\mid P)=.80, while \mathbb{P}(P\mid A)\approx.57. The two expressions use the same overlap but different reference events.

Base R makes the denominator visible.

Code

p_plural_and_acc <- .40
p_plural <- .50
p_acc <- .70

p_acc_given_plural <- p_plural_and_acc / p_plural
p_plural_given_acc <- p_plural_and_acc / p_acc

c(p_acc_given_plural, p_plural_given_acc)

The output is .8 and approximately .571.

Positive-probability conditions

This elementary ratio definition requires \mathbb{P}(C)>0. If \mathbb{P}(C)=0, the ratio has denominator zero and is undefined.

We will later encounter settings in which conditioning on exact continuous values requires a more careful construction. Nothing from those settings is needed for the finite event calculation here.

The order of the events

Swapping the target event and conditioning event changes the conditional probability. A corpus may show that most passive clauses contain an animate subject. This is a claim about

\mathbb{P}(\text{animate subject}\mid\text{passive}).

It does not by itself show that most clauses with animate subjects are passive, which concerns the reversed conditional.

Read the event on the right of the bar first. It determines the denominator. The target event on the left determines which part of that denominator enters the numerator.

Check your understanding

Compute \mathbb{P}(A^c\mid P) from the table and verify that it sums with \mathbb{P}(A\mid P) to one.
Compute \mathbb{P}(P\mid A^c) and describe the reference event in words.
Explain why \mathbb{P}(A\mid P) and \mathbb{P}(P\mid A) share a numerator but need not be equal.
Give a linguistic pair of events for which reversing the condition changes the scientific question.

--- title: "Conditional probability" --- The joint probability $\mathbb{P}(P,A)=.40$ is computed relative to the complete sample space. Now restrict the reference set to plural forms and ask for the proportion that are accusative. The phrase *among plural forms* makes $P$ the conditioning event. We write $$ \mathbb{P}(A\mid P) $$ and read it as “the probability of $A$ given $P$.” The event to the right of the bar is the condition. The event to the left is the target. ## Conditioning as rescaling The plural event contains two outcomes: $$ P=\{\textit{they},\textit{them}\}. $$ Their original probabilities are $.10$ and $.40$. Those values sum to $.50$, because the plural event receives half of the mass in the full space. | plural outcome | probability in $\Omega_4$ | probability within $P$ | |---|---:|---:| | *they* | $.10$ | $.10/.50=.20$ | | *them* | $.40$ | $.40/.50=.80$ | Dividing each plural probability by $.50$ produces a probability measure whose sample space is $P$. Within this restricted space, *them* receives probability $.80$. Since *them* is the only plural accusative outcome, $$ \mathbb{P}(A\mid P)=.80. $$ ## Definition For events $B,C\in\mathcal F$ with $\mathbb{P}(C)>0$, the [**conditional probability**](https://bruno.nicenboim.me/bayescogsci/ch-intro.html) of $B$ given $C$ is defined by $$ \mathbb{P}(B\mid C) \equiv \frac{\mathbb{P}(B,C)}{\mathbb{P}(C)}. $$ The denominator is the probability of the reference event $C$. The numerator is the part of $C$ for which $B$ also occurs. In the pronoun probability space, $$ \begin{aligned} \mathbb{P}(A\mid P) &=\frac{\mathbb{P}(A,P)}{\mathbb{P}(P)}\\ &=\frac{.40}{.50}\\ &=.80. \end{aligned} $$ This rescaling makes the conditioning event certain within itself: $$ \mathbb{P}(C\mid C) =\frac{\mathbb{P}(C,C)}{\mathbb{P}(C)} =\frac{\mathbb{P}(C)}{\mathbb{P}(C)} =1. $$ ## Reversing target and condition Now ask for the probability of plurality among accusative forms. The numerator is still the same intersection, since $P\cap A=A\cap P$. The denominator changes: $$ \begin{aligned} \mathbb{P}(P\mid A) &=\frac{\mathbb{P}(P,A)}{\mathbb{P}(A)}\\ &=\frac{.40}{.70}\\ &\approx .571. \end{aligned} $$ Thus $\mathbb{P}(A\mid P)=.80$, while $\mathbb{P}(P\mid A)\approx.57$. The two expressions use the same overlap but different reference events. Base R makes the denominator visible. ```{r} #| eval: false p_plural_and_acc <- .40 p_plural <- .50 p_acc <- .70 p_acc_given_plural <- p_plural_and_acc / p_plural p_plural_given_acc <- p_plural_and_acc / p_acc c(p_acc_given_plural, p_plural_given_acc) ``` The output is `.8` and approximately `.571`. ## Positive-probability conditions This elementary ratio definition requires $\mathbb{P}(C)>0$. If $\mathbb{P}(C)=0$, the ratio has denominator zero and is undefined. We will later encounter settings in which conditioning on exact continuous values requires a more careful construction. Nothing from those settings is needed for the finite event calculation here. ## The order of the events Swapping the target event and conditioning event changes the conditional probability. A corpus may show that most passive clauses contain an animate subject. This is a claim about $$ \mathbb{P}(\text{animate subject}\mid\text{passive}). $$ It does not by itself show that most clauses with animate subjects are passive, which concerns the reversed conditional. Read the event on the right of the bar first. It determines the denominator. The target event on the left determines which part of that denominator enters the numerator. ## Check your understanding 1. Compute $\mathbb{P}(A^c\mid P)$ from the table and verify that it sums with $\mathbb{P}(A\mid P)$ to one. 2. Compute $\mathbb{P}(P\mid A^c)$ and describe the reference event in words. 3. Explain why $\mathbb{P}(A\mid P)$ and $\mathbb{P}(P\mid A)$ share a numerator but need not be equal. 4. Give a linguistic pair of events for which reversing the condition changes the scientific question.