A Typewriter Sequence: Probability Without Almost-Sure Convergence

A Sequence That Sweeps Across the Unit Interval

Convergence in probability describes how often a random variable differs from its target, while almost-sure convergence asks what happens along individual sample paths. These notions need not agree in the reverse direction: convergence in probability does not, in general, force almost-sure convergence. A vivid example is the typewriter sequence, whose indicator functions sweep across the unit interval in progressively smaller pieces.

At each stage, exactly one small interval is active at any given point. The intervals shrink, so the probability of an active value tends to zero. But as the sequence continues, the sweep revisits every point infinitely many times. Thus, on each sample path, the values keep switching between zero and one.

Definition of the Typewriter Sequence

Use the probability space \(([0,1],\mathcal{B},\lambda)\), where \(\mathcal{B}\) is the Borel sigma-algebra and \(\lambda\) is Lebesgue measure on \([0,1]\). At level \(m=0,1,2,\ldots\), partition the unit interval into \(2^m\) dyadic cells. For \(k=0,\ldots,2^m-1\), define

The last cell includes the endpoint \(1\); this convention makes the cells a partition of all of \([0,1]\), including the endpoints. The cells are disjoint and each has length \(2^{-m}\). Define \(X_n\) by listing all cell indicators at level zero, then all indicators at level one, then all indicators at level two, and so on. Explicitly, the index corresponding to \((m,k)\) is \(n=2^m+k\).

Each \(X_n\) is measurable because its set of points with value one is a Borel interval. At level \(m\), the sequence consists of the indicators of every cell in that level's partition. The indexing matters: the number of indicators per level doubles, so later levels occur at later sequence indices.

Convergence in Probability

For each index \(n=2^m+k\), the random variable \(X_n\) is Bernoulli with success probability equal to the length of its cell. Consequently, its distribution and expectation are

As \(n\) increases, its level \(m\) increases without bound, and the probability of the value one therefore shrinks. To verify convergence in probability to zero, it is essential to check the probability of deviating from zero by any fixed positive tolerance, not merely to observe that the expectations decrease.

Proof. Fix \(\varepsilon>0\). Since \(X_n\) takes only the values zero and one, if \(0<\varepsilon<1\), then \(\{|X_n|>\varepsilon\}=\{X_n=1\}=I_{m,k}\) whenever \(n=2^m+k\). Thus

If \(\varepsilon\geq 1\), the event \(\{|X_n|>\varepsilon\}\) is empty, so its probability is zero. For any \(n\) in level \(m\), \(n\geq 2^m\). Hence \(n\to\infty\) forces \(m\to\infty\), and therefore \(2^{-m}\to0\). The deviation probability tends to zero for every fixed positive \(\varepsilon\), proving \(X_n\xrightarrow{p}0\). \(\square\)

Every Sample Path Fails to Converge

To analyze almost-sure convergence, fix a point \(x\in[0,1]\) and inspect the numerical sequence \(X_1(x),X_2(x),\ldots\). At every level \(m\), the cells form a partition, so \(x\) belongs to exactly one of them. The indicator for that cell equals one at \(x\), while the indicators for all other cells at that level equal zero.

Proof. At each level \(m\), let \(k_m(x)\) be the unique cell index such that \(x\in I_{m,k_m(x)}\). Then \(X_{2^m+k_m(x)}(x)=1\). These indices belong to distinct level blocks, so they form an unbounded sequence of indices. Thus the sample path equals one infinitely often.

For every \(m\geq1\), there are at least two cells at level \(m\). Choose a cell at that level different from the one containing \(x\); its indicator equals zero at \(x\). Choosing one such indicator at each level gives an unbounded sequence of indices along which the sample path equals zero. The path therefore has infinitely many zeros and infinitely many ones. A real sequence with these two properties cannot converge: if it had a limit, the subsequence of ones would force that limit to be one, while the subsequence of zeros would force it to be zero. This contradiction proves nonconvergence for every \(x\). The set of paths converging to zero is empty and has probability zero, so almost-sure convergence fails. \(\square\)

A Related Partition Construction

The essential feature is not the use of powers of two. It is that each stage partitions the space into many small cells, and the sequence lists all cell indicators in a stage before moving to the next. The following variation illustrates this principle using a different number of cells per stage.

Why the Example Matters

The typewriter sequence makes the difference between the two modes of convergence concrete. Convergence in probability controls the chance of a deviation at a given index: at a late index, the active cell has small measure. Almost-sure convergence instead asks whether, for almost every fixed \(x\), the numerical values eventually settle down. In this construction, each individual cell is small, but the sequence keeps sweeping over the whole interval, ensuring that each point is revisited.

There is no contradiction with the implication that almost-sure convergence implies convergence in probability. The example shows the reverse implication can fail; it does not challenge the forward one. In particular, knowing that \(P(X_n\neq0)\to0\) does not establish that, for almost every \(x\), only finitely many of the values \(X_n(x)\) are nonzero. The typewriter sequence has the first property and violates the second at every point.

• Pitfall: “The active interval shrinks to length zero, so for almost every point the sequence must eventually be zero.” This reasoning considers the shrinking size of a single interval but ignores that the interval changes location repeatedly.
• Correction: Check the full sequence of sets. At each level, the intervals cover all of \([0,1]\), so every point belongs to an active interval once per level. The probabilities shrink, but the values along each path do not settle.