Probability Notation Question

Question

I am a nOOb studying math and working through 75 of Stats with Julia

(https://people.smp.uq.edu.au/YoniNazarathy/julia-stats/StatisticsWithJulia.pdf)

On page 75 is this statement...

“Ω consists of 6 names… Let now, X : Ω → Z, be the function (i.e. random variable) that counts the number of letters in each name. The question is then finding: 

p(x) := P(X = x) , for k ∈ Z

The function p(x) represents the probability distribution of the random variable X. In this case, since X measures name lengths, X is a discrete random variable, and its probability distribution may be represented by a Probability Mass Function (PMF), such as p(x).”

I am struggling to understand it. I am trying to learn how to pronounce the math notation in English words...

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I can see it reads “X : Ω → Z” is a function that "counts the number of letters"

Is that read as "X is a function run over the sample space Ω (containing 6 names) and returning an integer (the length of each name)

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(this line is the one that I need most help with) ...

“p(x) := P(X = x) , for k ∈ Z” reads …

the probability distribution “p(x)” is defined as …

the probability of the input variable (x) appearing in X … (but it's not in X it's in Ω??)

for all numbers (k) that are an integer (why isn't this denoted as x?)

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I can see from the code in the book the answer looks a bit like ... in a set of names … {“ABC”,”BCA”,”CAB”,”ABCD”,”BCDA”,”ABCDE”} the probability distribution will be …

3 = 3/6

4 = 2/6

5 = 1/6

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Basically I am asking how you should read aloud the math notation in English words. I am very grateful for any help here.

Answer 1

$X$ is a map from the sample space $\Omega$ to the set of integers $\mathbb Z$ (in fact, to the set of natural numbers $\mathbb N$). $X$ assigns to each word (i.e., an element of $\Omega)$ an integer representing the length of the word.

In fact, $\Omega$ is a probability space, which means that it is equipped with a (discrete) probability distribution, assigning equal probability to each of its elements. In this case there are $6$ elements (i.e., there are $6$ words in the list) so each is assigned a probability of $\tfrac16$.

The probability mass function of the random variable $X$ is obtained by applying the function $X$ to the probability distribution given above. Thus $p(x)$ denotes the probability mass function, which is a function on the set of integers, defined by the property that for every integer $x\in\mathbb Z$ we declare $p(x)=\mathbb P(X=x)$. To compute $\mathbb P(X=x)$, we consider the event $\{\omega\in\Omega\colon X(\omega)=x\}$ i.e. the set of all words in $\Omega$ which have length $x$, and let $p(x)$ be the sum of the probability assigned to all words belonging to the event.

Answer 2

$X : \Omega \mapsto\Bbb Z$ , says the "$X$ maps the sample space to the integers".

$p(x) := \mathsf P(X = x) ,\mbox{ for }x \in\Bbb Z$ reads as "the function $p$ for $x$ is defined as the probability function for the event of random variable $X$ realising the value $x$, when $x$ is an integer."

The event of $X=x$, which is just shorthand for $\{\omega\in\Omega: X(\omega)=x\}$, is "the set of outcomes in the probability space that function $X$ maps to the value of the argument $x$".

So properly we should say, $p(x):=\mathsf P(\{\omega\in\Omega:X(\omega)=x\})$

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I can see from the code in the book the answer looks a bit like ... in a set of names … {“ABC”,”BCA”,”CAB”,”ABCD”,”BCDA”,”ABCDE”} the probability distribution will be …

Almost, it will be:

$$p(3) = 3/6\\p(4) = 2/6\\p(5) = 1/6$$

It is kind of important not to claim that 3 equals 1/2.