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I have a probability measure space with random variables $X,Y:\Omega \to \mathbb{R}$

Question is how can is show that the sets : $$ A = \{\omega \in \Omega:X(\omega) =Y(\omega)\} $$,

$$ B = \{\omega \in \Omega:X(\omega) >Y(\omega)\} $$

$$ \Gamma = \{\omega \in \Omega:e^{X(\omega)} =Y(\omega)\} $$ belong in $\sigma$-field $\mathcal{F}$

My effort is the below but I am not sure about $\Gamma$

For every $t \in \mathbb{R}$ we have : $$\bigcap_{n=1}^{\infty} \{ \omega: X(\omega)\leq t\}= \{ \omega:\sup X (\omega) \leq t\} =\{\omega: X_{n} (\omega) \leq t \}$$ for every $$n \in \mathbb{N} = \bigcap_{n=1}^{\infty} \{\omega : X_{n}(\omega) \leq t \}$$.

But $$X_{n},n \in \mathbb{N}$$ is a random variable and the class $\mathcal{F}$ is a $\sigma$ - field then : $$ \bigcap_{n=1}^{\infty} \{ \omega : X_{n}(\omega) \leq t \} \in \mathcal{F}$$

The same for $$ \{\omega:Y(\omega)>t \}= \{\omega:X_{n}(\omega)>t \} $$ for every $$n=\mathbb{N}=\bigcap_{n=1}^{\infty}\{ \omega : X_{n}(\omega) >t = \bigcap_{n=1}^{\infty} \{\omega : X_{n}(\omega) \leq t \}^{c} \in \mathcal{F}\}$$ and then $$\{\omega :Y(\omega) \leq t \} \in \mathcal{F}$$ for every $t \in \mathbb{R}$

For $$ \{\omega:Y(\omega)= e^{t} \}= \{\omega:X_{n}(\omega) =e^t \} $$ for every $$n=\mathbb{N}=\bigcap_{n=1}^{\infty}\{ \omega : X_{n}(\omega) \geq e^t = \bigcap_{n=1}^{\infty} \{\omega : X_{n}(\omega) \leq e^t \}^{c} \in \mathcal{F}\}$$ and then $$\{\omega :Y(\omega) \leq e^t \} \in \mathcal{F}$$

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    $\begingroup$ You have used symbols that you did not define. What is your $X_n$? $e^{X}$ is also a random variable so there is no need to prove separately that $\Gamma \in \mathcal F$. $\endgroup$ – Kavi Rama Murthy Jan 4 at 5:22
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I understand that you have a a probability measure space $(\Omega, \mathcal{F}, P)$ and random variables $X,Y:\Omega \to \mathbb{R}$.

You want to prove that the sets : $$ A = \{\omega \in \Omega:X(\omega) =Y(\omega)\} $$ $$ B = \{\omega \in \Omega:X(\omega) >Y(\omega)\} $$ $$ \Gamma = \{\omega \in \Omega:e^{X(\omega)} =Y(\omega)\} $$ belong in $\sigma$-field $\mathcal{F}$.

There is a simple way prove those results. Let us see it.

Proof: Note that $X-Y$ is a random variable, so $(X-Y)^{-1}(\{0\}) \in \mathcal{F}$. Since $$ A = \{\omega \in \Omega:X(\omega) =Y(\omega)\} = (X-Y)^{-1}(\{0\})$$ we have that $A\in \mathcal{F}$.

In a similar way, since $X-Y$ is a random variable, so $(X-Y)^{-1}((0,+\infty))\in \mathcal{F}$. Since $$ B = \{\omega \in \Omega:X(\omega) >Y(\omega)\} = (X-Y)^{-1}((0,+\infty)) $$ we have that $B\in \mathcal{F}$.

Finally, note that, since $X$ is a random variable, so is $e^X$. So just apply the result for $A$ to have that $\Gamma \in \mathcal{F}$.

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  • $\begingroup$ Where $(e^X -Y)^{-1}\in \mathcal{F}$ $\endgroup$ – Mr.Podilatis Jan 4 at 16:13
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    $\begingroup$ @Mr.Podilatis , If you don't want to simply reduce the case for $\Gamma$ to the first case, you can prove it directly as follows. Since $X$ is a random variable, $e^X$ is a random variable. Since $Y$ is a random variable, $e^X-Y$ is a random variable. Since $e^X-Y$ is a random variable, $(e^X-Y)^{-1}(\{0\}) \in \mathcal{F}$. But, $$ \Gamma = \{\omega \in \Omega:e^{X(\omega)} =Y(\omega)\} = (e^X-Y)^{-1}(\{0\}) $$ So $\Gamma \in \mathcal{F}$. $\endgroup$ – Ramiro Jan 4 at 16:23
  • $\begingroup$ I appreciate it mate thank you $\endgroup$ – Mr.Podilatis Jan 4 at 16:24

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