Understanding the supremum of a r.v. with an example

Question

Mentioning the definition of Supremum offered in supremum and limsup of random variables , we have: For each $\omega \in \Omega$, $$(\sup_{n \in \mathbb{N}} X_n)(\omega) := \sup\{X_n(\omega):n \in \mathbb{N}\}$$

However, I'm really struggling on understanding the definition, therefore I tried to imagine an expressive example:

Event: throw a die.

$(X_n)=\frac1n$ when $\omega=1,2$;

$(X_n)=\frac1{2}$ when $\omega=3,4$;

$(X_n)=0$ when $\omega=5,6$.

As far as I understood, the supremum is $\frac1{}$ and for checking it I have to analyse the value of the sequence for each . Therefore I can see that for $\omega=1,2$, the maximum value that the sequence take is $\frac1{}$, while for $\omega=3,4,5,6$ the value is lower then $\frac1{}$. Therefore $sup=\frac1{}$.

Could you tell me if I understood it well? Or could you provide expressful example?

And what is the corresponding Infimum of this sequence? Is it right that it is $0$ for $\omega=1,2,3,4,5,6$

Answer 1

I preassume that $0\notin\mathbb N$ and that you meant to say that $X_n=0$ if $\omega=5,6$.

Then in your example:

• $(\sup X_n)(\omega)=\sup\{X_n(\omega)\mid n\in\mathbb N\}=\sup\{\frac1n\mid n\in\mathbb N\}=1$ if $\omega\in\{1,2\}$
• $(\sup X_n)(\omega)=\sup\{X_n(\omega)\mid n\in\mathbb N\}=\sup\{\frac1{2n}\mid n\in\mathbb N\}=\frac12$ if $\omega\in\{3,4\}$
• $(\sup X_n)(\omega)=\sup\{X_n(\omega)\mid n\in\mathbb N\}=\sup\{0\}=0$ if $\omega\in\{5,6\}$

Answer 2

$\sup_n X_n(\omega)$ depends only on $\omega$ and not on $n$. So your answer is not correct. The correct answer is $\sup_n X_n(\omega)=1$ if $\omega =1$ or $2$, $\sup_n X_n(\omega)=\frac 1 2$ if $\omega =3$ and $0$ if $\omega =5$.

For $\omega =4$ there is an ambiguity in the definition. I suppose you wanted to say $\omega=5,6$ in the last line. In that case the supremum is $0$ for both $5$ and $6$.