What does tightness of convergence mean and why do we use this?

Question

In a paper that I am reading it is written:

We have $$(X(t_j))_{j=1}^k \xrightarrow{d} (Y(t_j))_{j=1}^k,$$ where $t_0=0, t_1 < \dotsb < t_k$.

It further is written: In order to show that $X(t) \xrightarrow{d} Y(t)$, we need to show tightness of $Y(t)$.

What does that mean? Why do we need to show that? $t$ is a fixed constant number.

By the way: $X$ and $Y$ are functions of some other random variables, thus random variables themselves, depending on $t$, $t_j$.

More detailed question: We want to show that $$X(t):=\frac{S(tx)-E[S(tx)]}{\sqrt{x}} \xrightarrow{d} W\left(\frac{t^l}{l}\right)$$ for some fixed $l\geq 2$ and $S$ some random variable with Binomial distribution. $W$ denotes the standard Brownian motion. It might have to do something with empirical distribution functions, as $X(t)$ is about the empirical distribution function for $S$.

Answer 1

Tightness (of a family of measures, not of a convergence) is useful to deduce the convergence in distribution (when $x\to\infty$, say) of a whole process $X_x=(X_x(t))_t$ from the convergence in distribution of all its finite-dimensional marginals.

Thus, the somewhat sloppy "In order to show that $X(t) \xrightarrow{d} Y(t)$, we need to show tightness of $Y(t)$" should be understood as "To show that $X_x\xrightarrow{d} Y$ when $x\to\infty$, we need to show tightness of the family of processes $(X_x)_x$".

Answer 2

Tightness is a kind of compactness condition. It means that for every positive $\epsilon$, there is some fixed compact set $K_\epsilon$ such that $P(X_n \in K_\epsilon) > 1-\epsilon$ for all $n$. Without this condition, there might not be any limit, the probability mass "may escape to infinity". An example of a sequence which is not tight is $X_n \sim N(n, 1)$.