# Random variables and $\sigma$-algebras

I came across this problem.

If we throw two dice the sample space is $$\Omega = \{ (i, j), 1 \leq i, j \leq 6 \}$$. The $$\sigma$$-algebra is generated by the events $$A_k = \{ (i,j) : \max(i,j) = k \}$$

Show that $$x_1=max(i,j)$$ is a random variable and $$X_2=i+j$$ isn't a random variable

There's a short explanation which explains the relation within the borel sets of the random variable and their preimages.

"Actually, a function $$X$$ from some measurable space $$(\Omega, \Sigma)$$ to $$\mathbb{R}$$ equipped with the Borel $$\sigma$$-algebra is defined as a random variable if it's measurable; that is, for every Borel set B, the preimage $$X^{-1}(B) = \{\omega : \omega \in \Omega, X(\omega) ∈ B\}$$ is measurable."

"Notice the focus on $$X^{-1}(B)$$ (preimages of Borel sets) rather than $$X^{-1}(x)$$(preimages of individual real numbers)"

1) Whenever you have a random variable, are the events of the random variable always borel sets? I mean if we have that the probability space of the random variable is $$(\Omega_X,F_X,P_X)$$, then for all the random variables, the family of events $$F_X$$ is always the $$\sigma$$-borel algebra?

2) If the preimages $$X_1^{-1}(x)$$ are $$\left \{ 1,2,3,4,5,6 \right \}$$, what are the preimages $$X_2^{-1}(x)$$? And what would be the preimages of $$X_1^{-1}(B)$$ and $$X_2^{-1}(B)$$?

How would you show that $$X_1$$ is a random variable and $$X_2$$ isn't?

Generators of the $$\sigma$$-algebra, first case:

$$A_1=\{(1,1)\}$$

$$A_2=\{(1,2),(2,2),(2,1)\}$$

$$A_3=\{(1,3),(2,3),(3,3),(3,2),(3,1)\}$$

$$A_4=\{(1,4),(2,4),(3,4),(4,4),(4,3),(4,2),(4,1)\}$$

$$A_5=\{(1,5),(2,5),(3,5),(4,5),(5,5),(5,4),(5,3),(5,2),(5,1)\}$$

$$A_6=\{(1,6),(2,6),(3,6),(4,6),(5,6),(6,6),(6,5),(6,4),(6,3),(6,2),(6,1)\}$$

Here $$A_k:=\{(i,j)\in\Omega\mid\max(i,j)=k\}$$

Generators of the $$\sigma$$-algebra, second case

$$B_2=\{(1,1)\}$$

$$B_3=\{(1,2),(2,1)\}$$

$$B_4=\{(1,3),(2,2),(3,1)\}$$

$$B_5=\{(1,4),(2,3),(4,1),(3,2)\}$$

$$B_6=\{(1,5),(2,4),(3,3),(5,1),(4,2)\}$$

$$B_7=\{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\}$$

$$B_8=\{(2,6),(3,5),(4,4),(5,3),(6,2)\}$$

$$B_9=\{(3,6),(4,5),(6,3),(5,4)\}$$

$$B_{10}=\{(4,6),(5,5),(6,4)\}$$

$$B_{11}=\{(5,5),(6,5)\}$$

$$B_{12}=\{(6,6)\}$$

Here $$B_k:=\{(i,j)\in\Omega\mid i+j=k\}$$

• I took the liberty to edit in order to take away confusion. In the original different sets received the same name. For instance the sets $\{(1,2),(2,2),(2,1)\}$ and $\{(1,1)\}$ were both labeled as $A_2$. This leads to confusion and false statements like $\{(1,2),(2,2),(2,1)\}=A_2=\{(1,1)\}$. Nov 27, 2019 at 7:43

1)

If by $$\Omega_X$$ you mean $$\mathbb R$$ and by $$P_X$$ you mean the probability measure prescribed by $$B\mapsto P(X\in B)$$ then: "yes, we usually go for $$\mathcal F_X=\mathcal B(\mathbb R)$$".

$$X$$ is usually by definition a random variable if it is a function that takes real values and is measurable wrt the Borel $$\sigma$$-algebra.

2)

Looking at $$X_1$$ the preimage of $$B\in\mathcal B(\mathbb R)$$ wrt $$X_1$$ is the set:$$X_1^{-1}(B)=\{(i,j)\in\{1,2,3,4,5,6\}^2\mid max(i,j)\in B\}$$

If you take singleton $$B=\{x\}$$ then we get:$$X_1^{-1}(B)=\{(i,j)\in\{1,2,3,4,5,6\}^2\mid max(i,j)=x\}$$

For $$X_2$$ you will get similar expressions where $$\max(i,j)$$ is replaced by $$i+j$$.

You cannot speak of "the preimages of $$X_1^{-1}(B)$$".

The correct wording is that "$$X_1^{-1}(B)$$ is the preimage of $$B$$ under (or with respect to) $$X_1$$".

The $$\sigma$$-algebra generated by the events $$A_k$$ is formally the smallest $$\sigma$$-algebra on $$\Omega$$ that contains these sets

Now observe that every preimage $$X_1^{-1}(B)$$ can be written as a union of these sets. This tells us that $$X_1$$ is measurable wrt to this $$\sigma$$-algebra. That means that $$X_1$$ can be classified as a random variable if $$\Omega$$ is equippes with the $$\sigma$$-algebra. Actually the $$\sigma$$-algebra generated by the $$A_k$$ can be shown to be the collection $$X_1^{-1}(\mathcal B(\mathbb R)):=\{X_1^{-1}(B)\mid B\in\mathcal B(\mathbb R)\}=$$$$\{\{(i,j)\in\Omega\mid \max(i,j)\in B\}\mid B\in\mathcal B(\mathbb R)\}\tag1$$

However it is not possible to write e.g. $$X_2^{-1}(\{4\})=\{(i,j)\mid i+j=4\}$$ as an element of this $$\sigma$$-algebra.

That's why $$X_2$$ cannot be classified as a random variable.

edit:

Working $$(1)$$ out we find that for every $$B\in\mathcal B(\mathbb R)$$ we can find a set $$I=I_B\subseteq\{1,2,3,4,5,6\}$$ such that: $$X_1^{-1}(B)=\bigcup_{i\in I}A_i$$

So actually: $$X_1^{-1}(\mathcal B(\mathbb R))=\left\{\bigcup_{i\in I}A_i\mid I\subseteq\{1,2,3,4,5,6\}\right\}$$

For e.g. preimage $$X_2^{-1}(\{4\})=\{(1,3),(2,2),(3,1)\}$$ we cannot find such a set $$I$$.

• What does $\left \{ 1,2,3,4,5,6 \right \}^2$stand for? Is the cartesian product? Nov 26, 2019 at 13:47
• Yes, it is the same set as $\Omega$ in your question. Nov 26, 2019 at 13:48
• Singletons must be borel sets then. Why are singletons open sets? Nov 26, 2019 at 14:02
• Yes, singletons are Borel sets. They are not open but that is not necessary for being a Borel set. In fact it is quite hard to find sets that are not Borel sets. Note that I added something to my answer which concerns the last sentence in your question. Nov 26, 2019 at 14:06
• Are they closed then? Nov 26, 2019 at 14:07