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

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  • $\begingroup$ 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 $\{(1,2),(2,2),(2,1)\}=A_2=\{(1,1)\}$. $\endgroup$
    – drhab
    Nov 27, 2019 at 7:43

1 Answer 1

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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 $I$.

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  • $\begingroup$ What does $\left \{ 1,2,3,4,5,6 \right \}^2$stand for? Is the cartesian product? $\endgroup$
    – ron jacobs
    Nov 26, 2019 at 13:47
  • $\begingroup$ Yes, it is the same set as $\Omega$ in your question. $\endgroup$
    – drhab
    Nov 26, 2019 at 13:48
  • $\begingroup$ Singletons must be borel sets then. Why are singletons open sets? $\endgroup$
    – ron jacobs
    Nov 26, 2019 at 14:02
  • $\begingroup$ 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. $\endgroup$
    – drhab
    Nov 26, 2019 at 14:06
  • $\begingroup$ Are they closed then? $\endgroup$
    – ron jacobs
    Nov 26, 2019 at 14:07

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