# Definition of random variable with respect to event space

#### endlessend25

##### New Member
It is given that:

We toss two dice with sample space Ω = {(i, j), 1 ≤ i, j ≤ 6} and the σ-algebra is generated by the events Ak = {(i,j) : max(i,j) = k} (k = 1, . . . , 6). Show that whereas X1(i, j) = max(i, j) is a random variable in the corresponding probability space, X2(i, j) = i + j is not a random variable.

I know that to be a random variable, X1 and X2 need to have a pre-image in the event space Sigma. X1 can take on values from 1 to 6 depending on the highest number in the (i,j) pair. X2 can take on values from 2 (1,1) to 12 (6,6).

So if i follow the rule, one element outcome of X1 and X2 should have an outcome which is an element of Sigma as a pre-image. I understand that each value of X1 (1 up to 6) is a possible outcome of Sigma. But it is less clear for X2, if I have a sum equals to 12, should I say it's not a random variable because there's not 12 in the Sigma even if the Sigma includes (6,6) ?

#### Buckeye

##### Active Member
I think it has to do with the idea that the sum of two dice can be reached through different ways. (Ex: a sum of 4 can be reached by rolling a 1 and 3 or rolling a pair of 2's) I don't believe this happens in the max-min problem. I don't know if I'm on the right track, but that's my attempt.

#### Dason

To build an intuition for these kinds of things it might be beneficial to try smaller problems that are related. In this case think of two-sided die (I'd say coin flips but "2-sided-die" more naturally extends to your case and we would think of the outcomes as 1,2 instead of H,T, or 0,1 as we would for a coin flip) with the sigma-algebra defined analogously.

In that case $$\sigma$$-algebra looks like: $$\{\emptyset, \{(1,1)\}, \{(1,2), (2,1), (2,2)\}, \Omega\}$$ - there aren't any other sets you can generate by intersections or unions.

Using that think about whether the pre-images for your random variables are sets in the defined sigma-algebra.

#### endlessend25

##### New Member
If I get it right:

Omega = {A1, A2, A3, A4, A5, A6} = {1,2,3,4,5,6}

X1 = {1,2,3,4,5,6}. All of the outcomes are inside Omega -> X1 is a random variable

X2 = {2,3,4,5,6,7,8,9,10,11,12}. The outcomes 7 to 12 are not inside Omega -> X2 is not a random variable

But how to say it in a more formal way? Knowing that I know that:

A fonction X : Omega -> R is called a random variable
if X^-1 (B) = {w is an element of Omega : X(w) is an element of B} is an element of Sigma
for any B is a Borel set of R

#### Dason

Well there are some outcomes that map X2 to 12 as elements that exist inside some of the sets in the sigma algebra. For example (6,6) is contained in the set of points mapped to max(i,j)=6. But does (6,6) exist alone in the sigma algebra or is it always with some other sets?

Let's talk about that small example that I proposed earlier. X1 = max(i,j) is a random variable since the preimages for X1=1 are {(1,1)} and the preimage for X1=2 is {(1,2), (2,1), (2,2)} and those exist in the sigma algebra. But if you consider X2 (which we'll let be defined for k=2,3,4) then what would the preimage for i+j=2 look like? It would be {(1,1)} which is in the sigma algebra. But what about X2=3? The preimage for that is {(1,2), (2,1)} and although that is a subset of one of the sets in our sigma algebra it isn't a set in the sigma-algebra itself so the preimage for X2=3 doesn't exist in the sigma-algebra.

#### endlessend25

##### New Member
I understand you explanation if you set that the sigma-algebra is {emptyset, {(1,1)}, {(1,2), (2,1), (2,2)}, Omega}.
But we could say that the X2=3 in your example whose preimage int {(1,2),(2,1)} is in the sigma-algebra if we say the sigma algebra is {emptyset, {(1,1)}, {(1,2),(2,1)}, {(2,2)}}.

It depends on the way you constructed your sigma-algebra, because you have multiple right ways to construct it. There's surely something that I miss.