Probability mass functions

The preceding page used support to list the possible values of a discrete random variable. Suppose S records the number of syllables in a sampled word token. Knowing that S is discrete does not tell us how likely its support values are.

Consider this constructed probability table.

syllable count s	probability
1	.42
2	.33
3	.17
4	.08

The function assigning a probability to each support value is a probability mass function, abbreviated PMF. We write

p_S:\mathbb Z_{\geq0}\to[0,1], \qquad p_S(s)\equiv\mathbb{P}(S=s).

The capital S denotes the random variable. The lowercase s stands for one possible value. Thus

p_S(2)=.33.

By the PMF definition, the probability measure assigns .33 to the event \{S=2\}. In this constructed model, \operatorname{supp}(S)=\{1,2,3,4\}.

Checking the two PMF requirements

First, a PMF must assign nonnegative mass:

p_S(s)\geq0 \qquad\text{for every support value }s.

Second, the masses across the complete support must sum to one:

\sum_s p_S(s)=1.

For the constructed table,

.42+.33+.17+.08=1.

The rows represent mutually exclusive and exhaustive value events. One word token falls in one syllable count row, and the table includes every value in the modeled support.

Adding mass across selected values

What is the probability of a token with at least three syllables? The relevant event contains the rows for 3 and 4:

\begin{aligned} \mathbb{P}(S\geq3) &=p_S(3)+p_S(4)\\ &=.17+.08\\ &=.25. \end{aligned}

The calculation adds masses because the two value events do not overlap. A token cannot have exactly three syllables and at least four syllables under this representation.

Representing the PMF in base R

Code

syllable_value <- c("1", "2", "3", "4")
syllable_mass <- c(.42, .33, .17, .08)

all(syllable_mass >= 0)
sum(syllable_mass)
sum(syllable_mass[syllable_value %in% c("3", "4")])

The three results are TRUE, 1, and .25. These checks establish that the entered table satisfies the PMF requirements and recover the selected event probability.

Interpreting zero mass carefully

If p_S(7)=0, then the event S=7 has zero probability under the stated model. This may reflect a deliberately restricted support or a model that treats the value as impossible. It need not mean that no seven syllable word can exist in every linguistic population.

The PMF belongs to a represented sampling process. A lexicon weighted by word types and a corpus weighted by word tokens can assign different masses to the same syllable counts.

The masses must cover the complete support

A probability mass function cannot omit possible values without accounting for their probability. Suppose the first three rows above were retained but the final row were dropped. The remaining masses would sum to .92, so the table would not define a PMF on the declared support.

Renormalizing the three values would define a different model conditional on at most three syllables. That change may be useful, but it must be stated as a new reference event rather than treated as harmless deletion.

Check your understanding

Verify that the values .50, .30, and .20 define a PMF over a three value support.
Under the syllable PMF, compute \mathbb{P}(S\leq2).
Explain why the support must be declared before interpreting a mass of zero for a count above 4.
What model change occurs if the final row is dropped and the remaining masses are renormalized?

A PMF assigns mass to separate support values. The next page considers a model whose values instead fill an interval.

--- title: "Probability mass functions" --- The [preceding page](discrete-random-variables.qmd) used support to list the possible values of a discrete random variable. Suppose $S$ records the number of syllables in a sampled word token. Knowing that $S$ is discrete does not tell us how likely its support values are. Consider this constructed probability table. | syllable count $s$ | probability | |---:|---:| | 1 | $.42$ | | 2 | $.33$ | | 3 | $.17$ | | 4 | $.08$ | The function assigning a probability to each support value is a [**probability mass function**](https://bruno.nicenboim.me/bayescogsci/ch-intro.html), abbreviated [**PMF**](https://bruno.nicenboim.me/bayescogsci/ch-intro.html). We write $$ p_S:\mathbb Z_{\geq0}\to[0,1], \qquad p_S(s)\equiv\mathbb{P}(S=s). $$ The capital $S$ denotes the random variable. The lowercase $s$ stands for one possible value. Thus $$ p_S(2)=.33. $$ By the PMF definition, the probability measure assigns $.33$ to the event $\{S=2\}$. In this constructed model, $\operatorname{supp}(S)=\{1,2,3,4\}$. ## Checking the two PMF requirements First, a PMF must assign nonnegative mass: $$ p_S(s)\geq0 \qquad\text{for every support value }s. $$ Second, the masses across the complete support must sum to one: $$ \sum_s p_S(s)=1. $$ For the constructed table, $$ .42+.33+.17+.08=1. $$ The rows represent mutually exclusive and exhaustive value events. One word token falls in one syllable count row, and the table includes every value in the modeled support. ## Adding mass across selected values What is the probability of a token with at least three syllables? The relevant event contains the rows for 3 and 4: $$ \begin{aligned} \mathbb{P}(S\geq3) &=p_S(3)+p_S(4)\\ &=.17+.08\\ &=.25. \end{aligned} $$ The calculation adds masses because the two value events do not overlap. A token cannot have exactly three syllables and at least four syllables under this representation. ## Representing the PMF in base R ```{r} #| eval: false syllable_value <- c("1", "2", "3", "4") syllable_mass <- c(.42, .33, .17, .08) all(syllable_mass >= 0) sum(syllable_mass) sum(syllable_mass[syllable_value %in% c("3", "4")]) ``` The three results are `TRUE`, `1`, and `.25`. These checks establish that the entered table satisfies the PMF requirements and recover the selected event probability. ## Interpreting zero mass carefully If $p_S(7)=0$, then the event $S=7$ has zero probability under the stated model. This may reflect a deliberately restricted support or a model that treats the value as impossible. It need not mean that no seven syllable word can exist in every linguistic population. The PMF belongs to a represented sampling process. A lexicon weighted by word types and a corpus weighted by word tokens can assign different masses to the same syllable counts. ## The masses must cover the complete support A probability mass function cannot omit possible values without accounting for their probability. Suppose the first three rows above were retained but the final row were dropped. The remaining masses would sum to $.92$, so the table would not define a PMF on the declared support. Renormalizing the three values would define a different model conditional on at most three syllables. That change may be useful, but it must be stated as a new reference event rather than treated as harmless deletion. ## Check your understanding 1. Verify that the values $.50$, $.30$, and $.20$ define a PMF over a three value support. 2. Under the syllable PMF, compute $\mathbb{P}(S\leq2)$. 3. Explain why the support must be declared before interpreting a mass of zero for a count above 4. 4. What model change occurs if the final row is dropped and the remaining masses are renormalized? A PMF assigns mass to separate support values. The [next page](continuous-random-variables.qmd) considers a model whose values instead fill an interval.