# Clarification - Random Variable

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 $$j$$ ?

• 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" – Sheridan Grant Oct 15 at 7:40

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 $$P(X \in \{a_1, a_2, \ldots\})=1$$

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.

• " 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? – Benj Cabalona Jr. Oct 15 at 8:16
• No. That would say that, for example, $P(X = a_6) = 1$, so X is equal to $a_6$ with probability one. – Gijs Oct 15 at 8:20
• 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. – whuber Oct 15 at 13:24