Definition of Random Variable (from Durrett)

CantorSet
Messages
44
Reaction score
0
Hi everyone,

I'm confused about Durrett's formal definition of a random variable, as well his formal notions of probability spaces in general. I always try to make abstract definitions concrete through simple examples, but I can't wrap my head around this one:

Durrett defines:

X is a random variable if X: \Omega \rightarrow \Re and for every Borel set B \subset R, we have X^{-1}(B) \in F, where F is the sigma-algebra on \Omega.

Well, I tried to make this concrete with a simple example. Suppose we have a box with 100 balls labeled from 1 to 100. We draw 3 balls from this box without replacement. Let X be the sum of these three balls. We know that X is a random variable and it takes on values from 3 to 297.

To interpret this example in the context of Durrett's formal definition, I guess \Omega is the set of all subsets of size three from 1 to 100. So \Omega = \{\{1,2,3\} , \{1,2,4\}, ...\} The sigma-algebra F is all collections of these subsets. Then, our random variable X basically only maps the subsets of size 3 to the integers from 3 to 279. So far so good.

But the definition says for every Borel set B \subset R, which we can take as B = (0,1), a perfectly legit Borel set, and we would have X^{-1}(B) \in F. But the problem is the random variable X only maps to discrete points, not intervals. So based on this definition, X wouldn't be a random variable. But I know it is.

Any suggestions on where I may be interpreting things wrong here?
 
Last edited:
Physics news on Phys.org
X^(-1)(B) would in this case be the empty set, which is contained in any sigma-algebra.
 
Namaste & G'day Postulate: A strongly-knit team wins on average over a less knit one Fundamentals: - Two teams face off with 4 players each - A polo team consists of players that each have assigned to them a measure of their ability (called a "Handicap" - 10 is highest, -2 lowest) I attempted to measure close-knitness of a team in terms of standard deviation (SD) of handicaps of the players. Failure: It turns out that, more often than, a team with a higher SD wins. In my language, that...
Hi all, I've been a roulette player for more than 10 years (although I took time off here and there) and it's only now that I'm trying to understand the physics of the game. Basically my strategy in roulette is to divide the wheel roughly into two halves (let's call them A and B). My theory is that in roulette there will invariably be variance. In other words, if A comes up 5 times in a row, B will be due to come up soon. However I have been proven wrong many times, and I have seen some...

Similar threads

Replies
9
Views
2K
Replies
30
Views
4K
Replies
10
Views
2K
Replies
5
Views
2K
Replies
6
Views
2K
Back
Top