Does $\sup_{n \ge m} Y_n < \infty \ \text{a.s.}$ imply $\sup_{n \ge 1} Y_n < \infty \ \text{a.s. ?}$

Question

Suppose $Y_1, Y_2, ...$ are independent random variables and $\exists m \ge 1$ s.t.

$$\sup_{n \ge m} Y_n < \infty \ \text{a.s.}$$

Does this mean

$$\sup_{n \ge 1} Y_n < \infty \ \text{a.s. ?}$$

I was thinking say $m = 2$ and $Y_1 = \infty$, then $\sup_{n \ge m} Y_n < \infty$, but $\sup_{n \ge 1} Y_n = \infty ?$

I have a feeling either the independence or the 'a.s.' is needed here.

My guess is that $\sup_{n \ge 1} Y_n = \infty $ but $\sup_{n \ge 1} Y_n < \infty \ \text{a.s. ?}$

Answer 1

Assuming that $(\sup\limits_n Y_n)(x)=\sup\limits_{n}\{Y_n(x)\}$ then indeed, there is nothing guaranteeing that the first random variable $Y_1$ should be finite a.s and therefore there is no reason that your claim should be true in general. The example you posited disproves it, but you need to fix values for all the later $Y_n$ as well (i.e. with $n\geq2$). For example, setting $Y_n=0$ whenever $n\geq2$, would do the trick.

On the other hand, if the $Y_n$ were an increasing sequence of random variables, then the result would be true.

Answer 2

I was thinking say $m = 2$ and $Y_1 = \infty$, then $\sup_{n \ge m} Y_n < \infty$, but $\sup_{n \ge 1} Y_n = \infty ?$

You almost answered you own question. You need further assumption for $Y_n$ with $n\geq 2$ though. Say, $Y_n\equiv 1$ for each $n$. Then the $Y_n$'s are independent. (Why?) Then you can check that the quoted statement is not true.