Clarification - Random Variable

Question

I'm working on Introduction to Probability Theory by Joseph Blitzstein.

Definition 3.2.1(Discrete random variable).
A random variable $X$ is said to be discrete if there is a finite list of values $a_1,a_2,...,a_n$ or an infinite list of values $a_1,a_2,...$ such that $P(X=a_j\ \text{for some}\ j) = 1.$ If $X$ is a discrete r.v., then the finite or countably infinite set of values $x$ such that $P(X=x)>0$ is called the support of $X$.

I am having a problem at this part: $P(X=a_j\ \text{for some}\ j) = 1.$

Does this mean that the probability of X taking on the values $a_i$ is 100% certain?

Shouldn't it be like $\sum_j P(X=a_j) = 1 $ for some $j$ ?

Answer 1

I think Mr Blitzstein, when writing $$P (X=a_j \text{ for some } j)=1$$ really meant that $X$ is always one of $a_1, a_2, \ldots$. So it always is $a_j$ for some $j$.

I'd write $$ P(X \in \{a_1, a_2, \ldots\})=1$$

Answer 2

The notation is the book is intentionally avoiding sums, but in this case it leads to confusion. I prefer the second expression below.

$$ P(X = a_j \text{ for some }a_j) = \sum_{i = 1}^n P\left(X = a_i\right) \,=\, 1 $$

In you formula, if I interpret "for some $j$", it is not correct. The variable X is not equal to a specific $a_j$ with probability one, but it is equal to one of the $a_j$ with probability one. Technically, since your expression sums over $j$, it doesn't make sense to say "for some $j$" at all. In the equality below, the left is a more informal but conventional way of leaving out the iterating variable, translating more explicitly to the expression on the right.

$$ \sum_{j}P\left(X = a_j\right) = \sum_{i = 1}^n P\left(X=a_i\right) $$

It may be helpful to look ahead to a definition of a continuous random variable. A random variable is either continuous or discrete.