Show that two random vectors have the same distribution

Question

I'm unsure of a result that looks simple, but I want to make sure I'm not getting it wrong.

Let $(\epsilon_t, t \in \mathbb{Z})$ be a process such that $\epsilon_t$ i.i.d.. I want to show that for any two arrays with the same size $t_1< t_2< ...< t_m$ and $t'_1< t'_2< ...< t'_m$ , the vectors $$(\epsilon_{t_1 - j}, \epsilon_{t_2 - j }, ... \epsilon_{t_m - j }) \hbox{ and } (\epsilon_{t'_1 - j}, \epsilon_{t'_2 - j }, ... \epsilon_{t'_m - j })$$ have the same distribution.

My intuition says that it is true. Supposse that $\epsilon_t \sim F, \,\, \forall\, t$ (i.i.d.). Then all elements of the vector have the same distribution. Thus, the two vector have the same distribution.

Am I right?

Answer 1

The abbreviation "iid" means "independent, identically distributed". I.e. the distribution of your two vectors is in each case the product measures $\mathbb{P}^m$, where $\mathbb{P}$ is the distribution that all your random variables $\epsilon_t$ follow.

So, yes, they have the same distribution.