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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 $j$ ?

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    $\begingroup$ This is one of those instances in which just writing the words is probably better: "X takes on some value in this list with probability one" $\endgroup$ – Sheridan Grant Oct 15 at 7:40
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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 $$ P(X \in \{a_1, a_2, \ldots\})=1$$

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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.

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  • $\begingroup$ " The variable X is not equal to a specific aj with probability one, but it is equal to one of the aj with probability one. " Does this mean that there exist a j such that P( X = a_j) = 1? $\endgroup$ – Benj Cabalona Jr. Oct 15 at 8:16
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    $\begingroup$ No. That would say that, for example, $P(X = a_6) = 1$, so X is equal to $a_6$ with probability one. $\endgroup$ – Gijs Oct 15 at 8:20
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    $\begingroup$ The second expression isn't the idea the book is trying to convey: the book overtly avoids expressing probabilities as sums. The intention is to describe an event (namely, the set of $a_j,$ which may be countably infinite) and to declare that it has unit probability. $\endgroup$ – whuber Oct 15 at 13:24

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