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The floor function gives the deterministic boundfor every outcome, including $X=1$. Thus all three convergence assertions hold: there is pointwise, and hence almost sure convergence, as well as convergence in probability and convergence in L2. In detail, for $\epsilon >0$ the probability $\mathbb{P}(|X-X_n|>\epsilon )$ is zero as soon as $2^{-n}\le \epsilon$; alsoThe uniform distribution even gives the exact second moment of the error by integrating over the dyadic partition:

The definition of convergence in probability isThe variables are defined on a common probability space so that these difference events make sense.

The normalized sum has mean one and variance $(5n-2)/n^2$. For every fixed $ϵ>0$, the Chebyshev inequality givesThis proves the requested limit, and in fact proves convergence in probability of $S_n/n$ to one. It is a finite covariance range weak law: adjacent dependence is harmless here because only linearly many covariance terms survive.

By linearity of expectation, $\mathbb{E}{N}_n=n{p}_n$. The variance of a sum and the negative covariance just proved giveApply Chebyshev's inequality to the empty-box indicators in uniform allocation:This proves the required concentration of the empty-box fraction. Since $p_n\to e^{-1}$, it also yields convergence in probability of $N_n/n$ to $e^{-1}$.

Let $W_i=U_i^2$. Applying the same measurable function separately to independent and identically distributed random variables preserves those two properties, andIn particular, these bounded random variables are integrable random variables. The weak law of large numbers givesIf $c>1/3$, then $\{{\overline{W}}_n>c\}\subseteq \{|{\overline{W}}_n-1/3|⩾c-1/3\}$, whose probability tends to zero. If $0⩽c<1/3$, then $\{{\overline{W}}_n⩽c\}\subseteq \{|{\overline{W}}_n-1/3|⩾1/3-c\}$, whose probability also tends to zero. Taking a complement in the second case provesThe positive gap from $1/3$ is essential to this argument. Convergence in probability alone does not determine the answer at the excluded threshold $c=1/3$.

Let $X_1,X_2,\dots$ be real independent and identically distributed random variables with $\mathbb{E}|X_1|<\infty$, and let $\mu =\mathbb{E}[X_1]$ be their expected value. Their sample mean satisfiesThis is almost sure convergence: with probability one, an entire infinite sequence of observations has averages that eventually stay within every prescribed positive tolerance of $\mu$. The variance need not be finite. The strong law of large numbers implies the weak law of large numbers, which asserts convergence in probability at each increasing sample size.

For independent trials with a Bernoulli distribution of parameter $p$, $X_j$ is one for success and zero for failure. Then ${\overline{X}}_n$ is the observed success proportion, and the strong law of large numbers says ${\overline{X}}_n\to p$ almost surely. This turns a theoretical probability into a quantity one can estimate by repetition. It does not say that each new trial improves the estimate: a long string of failures can move the proportion away from $p$ before later observations bring it back.

The strong law of large numbers gives the limiting location of the sample mean, while the central limit theorem describes its fluctuations when the variance is finite and positive. The two assertions can hold together:The first error shrinks to zero; the second has been enlarged enough to retain a nontrivial probability distribution. The second assertion is convergence in distribution, not a claim that the enlarged error along one sequence approaches a particular normal random variable.

The assumptions make a practical difference. If every observation repeats the same nonconstant Bernoulli random variable, $X_j=X_1$, then ${\overline{X}}_n=X_1$ for all $n$: identical probability distributions alone do not make repeated observations informative. For independent observations with a Standard Cauchy distribution, the Cauchy sample mean has an unchanged distribution, so it does not concentrate around a constant. Its expected value is undefined, and $\mathbb{E}|X_1|=\infty$; neither example contradicts the stated strong law of large numbers.

A brief proof outline in the case of finite variance uses the Kolmogorov maximal inequality for the centered partial sums. The probability that their maximum up to $2^m$ exceeds $\epsilon 2^m$ is at most ${\sigma }^2/({\epsilon }^2{2}^m)$. These bounds are summable, so the Borel-Cantelli lemma controls all sufficiently large such blocks and hence every intermediate sample size. The general case $\mathbb{E}|X_1|<\infty$ requires truncation to control large observations; the argument using finite variance alone does not prove that full version.

Drive two copies of a random affine recursion with the same coefficient pairs, independent of their initial coupling. Then $X_n-Y_n=A_n(X_{n-1}-Y_{n-1})$. If $\rho =\mathbb{E}|A_1|<1$ and $\mathbb{E}|X_0-Y_0|<\infty$, independence of random variables givesThis proves convergence in L1 and hence convergence in probability of the difference. If one copy starts in a stationary distribution, Slutsky's theorem shows that the other copy converges to that distribution. The argument compares paths before taking their one-time probability distributions; it does not require the paths ever to become exactly equal.

Uniform convergence in probability means that the supremum of the absolute discrepancy on the entire index set tends to zero in probability. It controls evaluation at random indices in that set, unlike pointwise convergence in probability. On a fixed compact index set it is a special case of uniform convergence on compacts in probability.

Let $X_1,\dots ,X_n$ be an IID sample of real-valued random variables with finite expected value $\mu$ and finite variance ${\sigma }^2$, where $n\ge 1$. Its sample mean $\overline{X}=n^{-1}{\sum }_i{X}_i$ satisfiesLinearity of expectation gives the first identity. The variance of a sum and vanishing cross covariances give $\mathrm{Var}({\sum }_i{X}_i)=n{\sigma }^2$; scaling by $1/n$ scales the variance by $1/n^2$. Pairwise independent random variables with the same expected value and variance suffice for these identities, even without full independence. Grinstead and Snell state the corresponding finite-variance independent-trials result.^[ref]

The standard deviation of the sample mean is $\sigma /\sqrt{n}$. Four times as many observations halve this scale; they do not divide it by four. In the fair Bernoulli distribution example, $\sigma =1/2$, so sample sizes $4$ and $64$ give standard deviations $1/4$ and $1/16$, respectively. The common-coin averaging example shows why merely repeating observations with a common marginal law does not justify this calculation.

Chebyshev's inequality turns the variance identity into a probability bound: for $\epsilon >0$,This proves convergence in probability of the sample mean to $\mu$ in this finite-variance setting, a form of the weak law of large numbers. It is an upper bound, so values above one convey no improvement over the universal probability bound one. Cauchy sample mean stability lies outside these hypotheses.

If independent and identically distributed random variables have characteristic function $\phi (t)=1+iat+o(|t|)$ at zero, then their sample mean has convergence in distribution to $a$: its characteristic function is $\phi (t/n{)}^n\to e^{iat}$. This is also convergence in probability by convergence in distribution to a constant implies convergence in probability.

Let $X_1,X_2,\dots$ be real independent and identically distributed random variables with $\mathbb{E}|X_1|<\infty$ and expected value $\mu =\mathbb{E}[X_1]$. The weak law of large numbers states that their sample mean has convergence in probability to $\mu$: for every fixed $\epsilon >0$,Thus a large sample is unlikely to give an average with a specified fixed-size error. The strong law of large numbers makes the stronger statement of almost sure convergence along an entire sequence. Under the displayed assumptions both laws apply; their names distinguish the conclusions, not two competing predictions.

When the variance ${\sigma }^2$ is finite, variance additivity for independent random variables gives $\mathrm{Var}({\overline{X}}_n)={\sigma }^2/n$. The Chebyshev inequality immediately supplies a useful quantitative form and a brief proof:For independent trials with a Bernoulli distribution of parameter $p=0.2$, the expected value is $0.2$ and the variance is $0.16$. At $n=1000$, the chance of an error of at least $0.1$ is therefore at most $0.016$. The exact binomial distribution gives a much smaller probability, as the figure under the strong law of large numbers shows. The bound is conservative, but it is valid for every sample size under these assumptions and does not require a normal approximation.

The weak law of large numbers addresses a fixed tolerance $\epsilon$. The central limit theorem instead resolves errors on the shrinking scale $\sigma /\sqrt{n}$ and gives their approximate normal distribution. In particular, the weak law of large numbers alone supplies neither a bell-shaped error law nor a universal rate of approach. For example, a Pareto distribution with minimum one and shape $3/2$ has expected value three but infinite variance: both the weak law of large numbers and strong law of large numbers apply, while the finite-variance statement of the central limit theorem does not.