0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 '19 at 7:43
1
$\begingroup$

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 $I$.

$\endgroup$
  • $\begingroup$ What does $\left \{ 1,2,3,4,5,6 \right \}^2$stand for? Is the cartesian product? $\endgroup$ – ron jacobs Nov 26 '19 at 13:47
  • $\begingroup$ Yes, it is the same set as $\Omega$ in your question. $\endgroup$ – drhab Nov 26 '19 at 13:48
  • $\begingroup$ Singletons must be borel sets then. Why are singletons open sets? $\endgroup$ – ron jacobs Nov 26 '19 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 '19 at 14:06
  • $\begingroup$ Are they closed then? $\endgroup$ – ron jacobs Nov 26 '19 at 14:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.