# Almost surely

(Redirected from Almost never)

In probability theory, one says that an event happens almost surely (sometimes abbreviated as a.s.) if it happens with probability one. In other words, the set of possible exceptions may be non-empty, but it has probability zero. The concept is precisely the same as the concept of "almost everywhere" in measure theory.

In probability experiments on a finite sample space, there is often no difference between almost surely and surely. However, the distinction becomes important when the sample space is an infinite set, because an infinite set can have non-empty subsets of probability zero.

Some examples of the use of this concept include the strong and uniform versions of the law of large numbers, and the continuity of the paths of Brownian motion.

The terms almost certainly (a.c.) and almost always (a.a.) are also used. Almost never describes the opposite of almost surely: an event that happens with probability zero happens almost never.

## Formal definition

Let $(\Omega ,{\mathcal {F}},P)$  be a probability space. An event $E\in {\mathcal {F}}$  happens almost surely if $P(E)=1$ . Equivalently, $E$  happens almost surely if the probability of $E$  not occurring is zero: $P(E^{C})=0$ . More generally, any event $E\subseteq \Omega$  (not necessarily in ${\mathcal {F}}$ ) happens almost surely if $E^{C}$  is contained in a null set: a subset of some $N\in {\mathcal {F}}$  such that $P(N)=0$ . The notion of almost sureness depends on the probability measure $P$ . If it is necessary to emphasize this dependence, it is customary to say that the event $E$  occurs P-almost surely, or almost surely (P).

## Illustrative examples

In general, an event can happen "almost surely" even if the probability space in question includes outcomes which do not belong to the event, as is illustrated in the examples below.

### Throwing a dart

Imagine throwing a dart at a unit square (i.e. a square with area 1) so that the dart always hits exactly one point of the square, and so that each point in the square is equally likely to be hit.

Now, notice that since the square has area 1, the probability that the dart will hit any particular subregion of the square equals the area of that subregion. For example, the probability that the dart will hit the right half of the square is 0.5, since the right half has area 0.5.

Next, consider the event that "the dart hits a diagonal of the unit square exactly". Since the areas of the diagonals of the square are zero, the probability that the dart lands exactly on a diagonal is zero. So, the dart will almost never land on a diagonal (i.e. it will almost surely not land on a diagonal). Nonetheless the set of points on the diagonals is not empty and a point on a diagonal is no less possible than any other point: the diagonal does contain valid outcomes of the experiment.

### Tossing a coin repeatedly

Consider the case where a (possibly biased) coin is tossed, corresponding to the probability space $(\{H,T\},2^{\{H,T\}},P)$ , where the event $\{H\}$  occurs if heads is flipped, and $\{T\}$  if tails. For this particular coin, assume the probability of flipping heads is $P(H)=p\in (0,1)$  from which it follows that the complement event, flipping tails, has $P(T)=1-p$ .

Suppose we were to conduct an experiment where the coin is tossed repeatedly, with outcomes $\omega _{1},\omega _{2},\ldots$ , and it is assumed each flip's outcome is independent of all the others. That is, they are i.i.d.. Define the sequence of random variables on the coin toss space, $(X_{i})_{i\in \mathbb {N} }$  where $X_{i}(\omega )=\omega _{i}$ . i.e. each $X_{i}$  records the outcome of the $i$ 'th flip.

Any infinite sequence of heads and tails is a possible outcome of the experiment. However, any particular infinite sequence of heads and tails has probability zero of being the exact outcome of the (infinite) experiment. To see why, note that the i.i.d. assumption implies that the probability of flipping all heads over $n$  flips is simply $P(X_{i}=H,\ i=1,2,\dots ,n)=\left(P(X_{1}=H)\right)^{n}=p^{n}$ . Letting $n\rightarrow \infty$  yields zero, since $p\in (0,1)$  by assumption. Note that the result is the same no matter how much we bias the coin towards heads, so long as we constrain $p$  to be greater than 0, and less than 1.

In particular, the event "the sequence contains at least one $T$ " happens almost surely (i.e., with probability 1). However, if instead of an infinite number of flips we stop flipping after some finite time, say a million flips, then the all-heads sequence has non-zero probability. The all-heads sequence has probability $p^{1,000,000}\neq 0$ , while the probability of getting at least one tails is $1-p^{1,000,000}$  and the event is no longer almost sure.

## Asymptotically almost surely

In asymptotic analysis, one says that a property holds asymptotically almost surely (a.a.s.) if, over a sequence of sets, the probability converges to 1. For instance, a large number is asymptotically almost surely composite, by the prime number theorem; and in random graph theory, the statement "$G(n,p_{n})$  is connected" (where $G(n,p)$  denotes the graphs on $n$  vertices with edge probability $p$ ) is true a.a.s. when, for some $\varepsilon >0$

$p_{n}>{\frac {(1+\varepsilon )\ln n}{n}}.$    

In number theory this is referred to as "almost all", as in "almost all numbers are composite". Similarly, in graph theory, this is sometimes referred to as "almost surely".