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 $\omega=1,2$;
$(X_n)=\frac1{2𝑛}$ when $\omega=3,4$;
$(X_n)=0$ when $\omega=5,6$.
As far as I understood, the supremum is $\frac1{𝑛}$ and for checking it I have to analyse the value of the sequence for each 𝜔. Therefore I can see that for $\omega=1,2$, the maximum value that the sequence take is $\frac1{𝑛}$, while for $\omega=3,4,5,6$ the value is lower then $\frac1{𝑛}$. Therefore $sup=\frac1{𝑛}$.
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 $\omega=1,2,3,4,5,6$
Best Answer
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: