Yes, there is a more physically intuitive way of understanding almost-sure convergence!
Let's first return to convergence in probability. Intuitively, this says:
- If you wait a very long time, then with very high probability, $X_n$ will be very close to $X$.
But now let's ask: what if you want not only $X_n$ but also $X_{n+1},\ldots,X_{n+k}$ (for some $k \in \mathbb{N}$) to be close to $X$? In other words, we are concerned with the problem of:
- "If I wait a sufficiently long time, do I have that with very high probability, I will not only be very close to the target $X$, but will also subsequently remain close to the target for some a-priori-specified duration?"
It turns out that this is also covered by convergence in probability. Namely, it is an easy exercise to prove the following:
Lemma. Suppose $X_n$ converges in probability to $X$, and fix any $k \in \mathbb{N}_0$. For every $\varepsilon>0$,
$$ P_k(\varepsilon,n) := \mathbb{P}(X_n,\ldots,X_{n+k} \in [X-\varepsilon,X+\varepsilon]) \to 1 \ \text{ as } n \to \infty\text{.} $$
In other words, even though the definition of convergence in probability just says that $P_0(\varepsilon,n) \to 1$ as $n \to \infty$ for each $\varepsilon$, it actually follows that $P_k(\varepsilon,n) \to 1$ as $n \to \infty$ for each $\varepsilon$, for every $k \in \mathbb{N}$.
But now, suppose $k$ is not a-priori-specified; instead, I need to be able to make allowance for arbitrarily large $k$. In other words, suppose I want the convergence to be "sufficiently reliable that no matter how large the necessary $k$ might turn out to be, I can control how long I need to wait to have high probability that $X_n,\ldots,X_{n+k}$ are all close to $X$".
For this problem, it turns out that convergence in probability will not be enough. Indeed, if you do the exercise of proving the Lemma and then look at your proof, you will already get the indication that just convergence in probability might not be enough: your bound for how long you have to wait for $P_k(\varepsilon,n)$ to be close to $1$ will be an expression that typically tends to $\infty$ as $k \to \infty$!
This is where "almost-sure convergence" comes in!
Alternative definition of almost-sure convergence. We say that $X_n$ converges almost surely to $X$ if for every $\varepsilon>0$,
$$ P_k(\varepsilon,n) \to 1 \ \text{ as } n \to \infty \text{ uniformly across } k \in \mathbb{N}_0\text{.} $$
In other words, "almost-sure convergence" is precisely the stronger version of convergence in probability that makes allowance for arbitrarily large $k$.
Proof that the alternative definition is equivalent to the usual definition. Let
$$ Y_{n,k} := \max\{ |X_n-X|, \ldots, |X_{n+k}-X| \}\text{,} $$
and let
$$ Y_{n,\infty} := \sup\{|X_n-X|, |X_{n+1}-X|, \ldots\ldots\}\text{.} $$
So for each $n$ and $\omega$, we have $Y_{n,k}(\omega) \nearrow Y_{n,\infty}(\omega)$ as $k \to \infty$. Hence the set
$$ E_{n,\infty,\varepsilon} := \{Y_{n,\infty} \leq \varepsilon\} $$
is the intersection over $k \in \mathbb{N}_0$ of the decreasing-in-$k$ sequence of sets
$$ E_{n,k,\varepsilon} := \{Y_{n,k} \leq \varepsilon\}\text{.} $$
Thus
\begin{align*}
\sup_{k \in \mathbb{N}_0} (1-P_k(n,\varepsilon)) &= 1-\mathbb{P}(E_{n,\infty,\varepsilon}) \\
&= 1 - \mathbb{P}(X_n,X_{n+1},\ldots\ldots \in [X-\varepsilon,X+\varepsilon])\text{.}
\end{align*}
Therefore, our alternative definition is equivalent to saying that for every $\varepsilon>0$,
$$ \mathbb{P}(X_n,X_{n+1},\ldots\ldots \in [X-\varepsilon,X+\varepsilon]) \to 1 \ \text{ as } n \to \infty\text{.} $$
It remains to show that this is equivalent to the usual definition. First observe that the set
$$ E_{\infty,\infty,\varepsilon} := \{X_i \in [X-\varepsilon,X+\varepsilon] \ \text{ for all sufficiently large } i\} $$
is the union over $n$ of the increasing-in-$n$ sequence of sets $E_{n,\infty,\varepsilon}$, and that the set
$$ E_{\infty,\infty,0} := \{X_n \to X \ \text{ as } n \to \infty\} $$
is the intersection over $\varepsilon>0$ of the decreasing-in-$\varepsilon$ sequence of sets $E_{\infty,\infty,\varepsilon}$. Hence
$$ \mathbb{P}(E_{\infty,\infty,0}) = \lim_{\downarrow}{}_{\!\varepsilon \to 0} \ \lim_{\uparrow}{}_{\!n \to \infty} \ \mathbb{P}(E_{n,\infty,\varepsilon})\text{.} $$
It follows that $\mathbb{P}(E_{\infty,\infty,0})=1$ if and only if for every $\varepsilon>0$, $\lim_{n \to \infty} \mathbb{P}(E_{n,\infty,\varepsilon})=1$. $\quad\square$
Best Answer
Convergence in the $r$-th mean is often more convenient to prove mathematically than almost sure convergence. Moreover, if you use Markov's inequality, you may get a rate of convergence for convergence in probability, if you can prove a rate for the convergence in $r$-th mean.
A simple illustrative example is the weak law of large numbers: $$ \mathbb{P}\left[\left|\frac{1}{n}\sum_{i=1}^n X_i-E[X_1]\right|>\varepsilon\right] \leq \frac{1}{\varepsilon^2}\mathbb{E}\left[\left|\frac{1}{n}\sum_{i=1}^n X_i-E[X_1]\right|^2\right]$$ The rest of the proof aims to show that $\frac{1}{n}\sum_{i=1}^n X_i-E[X_1]$ converges in the second mean to $0$. This is done as follows: $$ \leq \frac{1}{\varepsilon^2}\mathbb{E}\left[\frac{1}{n^2}\sum_{i,j=1}^n\left| X_i-E[X_1]\right|\left| X_j-E[X_1]\right|\right] $$ Now we can use linearity of the expectation, which we couldn't do when working with $\mathbb{P}$ directly: $$ =\frac{1}{\varepsilon^2}\frac{1}{n^2}\sum_{i,j=1}^n\mathbb{E}\left[\left| X_i-E[X_1]\right|\left| X_j-E[X_1]\right|\right]=\frac{1}{\varepsilon^2}\frac{1}{n}\mathbb{E}\left[\left| X_1-E[X_1]\right|^2\right]=:\frac{\sigma^2}{\varepsilon^2n} $$
Some concluding remarks: