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**Chebyshev's inequality: if $X$ has finite mean $\mu$ and variance ${\sigma }^2$, then for $a>0$,Apply Markov's inequality to the nonnegative random variable $Y=(X-\mu {)}^2$ with threshold $a^2$. Since $\mathbb{E}[Y]={\sigma }^2$ and $\{|X-\mu |\ge a\}=\{Y\ge a^2\}$, the desired statement follows immediately.

Represent the removed half as the first $m=500$ cards in a uniformly random ordering. Every unordered half is then equally likely. Let $X_1,\dots ,X_{500}$ be their values. Each has mean zero and variance one, and every distinct pair has covariance $-1/999$ by the same ordered-pair argument as in (ii).

The general covariance expansion from (b), applied without imposing independence, givesThis is the finite population correction for a half-sample; treating the draws as independent would give the wrong variance.

Finally $\{Z\ge 50\}\subseteq \{|Z|\ge 50\}$, so Chebyshev's inequality yieldsThe cards returned before this sampling step restore the original population; no conditioning on the earlier two observed values is imposed.

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 $e^{-1}$.

For a nonnegative integer $\ell$, this Coulomb potential energy eigenstate has principal quantum number $\ell +1$ and maximal magnetic quantum number. Put $b=(\ell +1)a$. Including the spherical volume element, its radial probability density is a gamma distribution proportional to $r^{2\ell +2}{e}^{-2r/b}$; its polar density is proportional to ${\sin }^{2\ell +1}\theta$, while its azimuth is uniform. Radial and polar variables are independent. Their moments satisfyThe radial identities follow by successive integration by parts; the angular identity follows from ${\int }_0^{\pi }{\sin }^{j+2}\theta d\theta =(j+1){\int }_0^{\pi }{\sin }^j\theta d\theta /(j+2)$.

Set $h=(2\ell +3{)}^{-1/3}$. Chebyshev's inequality and Markov's inequality give $\mathbb{P}(|r/\mu -1|>h)\le h$ and $\mathbb{P}(|\cos \theta |>h)\le h$. By the union bound, outside an event of probability at most $2h$, the point lies at distance $O(\mu h)$ from the equatorial circle of radius $\mu$. The quantum state therefore concentrates near a classical circular orbit in relative units as $\ell$ grows. The absolute orbit radius and absolute transverse width both grow; this statement does not assert localization in a fixed narrow tube.

Probability inequalities bound event probabilities from limited information, such as moments, without requiring a complete probability distribution. Markov's inequality uses a nonnegative random variable's expected value; Chebyshev's inequality uses a variance to bound deviations from a mean. The attraction is a guarantee that applies to many possible probability distributions. Their bounds can be conservative for a particular model, and distributions attaining equality show when the available information really cannot give more.

The second moment method uses a random variable's expected value and second moment to establish that a counted or measured quantity is positive with substantial probability. It is useful when a positive average alone could be caused by extremely large values on very rare outcomes. Controlling the variance rules out too much of that concentration in rare outcomes.

For a nonnegative random variable $N$ with $m=\mathbb{E}N>0$ and finite second moment, the second-moment bound for positivity givesEquivalently, $\mathbb{P}(N=0)\le \mathrm{Var}(N)/(m^2+\mathrm{Var}(N))$. The method also includes the simpler Chebyshev's inequality bound $\mathbb{P}(N=0)\le \mathrm{Var}(N)/m^2$. Thus a sequence of such random variables with $\mathrm{Var}(N)/(\mathbb{E}N{)}^2\to 0$ has $N>0$ with probability tending to one. These statements require no independence assumption. The variance argument and its use for random counts are developed in Zhao's lecture notes. ^[ref].

For an original toy count, let $N$ count successes in ten independent Bernoulli trials with success probability $p=1/10$. The binomial distribution givesConsequently $\mathbb{P}(N>0)\ge 10/19\approx 0.526316$. The exact probability is $1-(9/10{)}^{10}\approx 0.651322$. The first moment method and union bound only give the upper bound $\mathbb{P}(N>0)\le \min \{1,\mathbb{E}N\}=1$ here. First and second moments therefore answer complementary existence questions.

For a sum of indicator random variables, the variance of a sum isIndependence makes the covariances vanish, but the formula itself also covers dependent counts. For the same ten success probabilities, let every trial share a single Bernoulli random variable $B$: then $N=10B$, with mean one, variance nine, and $\mathbb{P}(N>0)=1/10$. The second-moment lower bound is exactly $1/10$. Ignoring the positive covariances would produce the false independent-count conclusion. Zhao's discussion of sums of indicators explains why overlapping counted structures require these pairwise terms. ^[ref].

The count $K_n$ has binomial distribution with mean $n/4$ and variance $3n/16$. The information content per symbol is an affine function of $K_n/n$, so its variance isChebyshev's inequality therefore gives the finite-sample lower boundThe bound is often loose, but it proves that every fixed positive tolerance eventually captures nearly all the probability. Direct summation of the binomial distribution's probability masses gives $P(T_{0.1}^{(640)})\approx 0.999779$. A finite sample does not need to have a nearly full typical set: the 40-symbol example above captures only about $0.639$.

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.

On a cyclic sequence of $n$ independent Bernoulli random variables with success probability $p$, count all runs of $k$ consecutive successes, allowing overlaps and taking $1\le k\le n/2$. The count has expected value $n{p}^k$. Its variance is the displayed expression: each starting indicator has variance $p^k(1-p^k)$; two starts at cyclic distance $j<k$ have joint probability $p^{k+j}$; nonoverlapping starts are independent. There are $2n$ ordered overlapping pairs at each such distance, because the two ranges do not meet when $k\le n/2$.

For $p\le 1/2$, summing a geometric series gives $\mathrm{Var}X\le n{p}^k(1+2p/(1-p))\le 3n{p}^k$. Markov's inequality therefore proves absence with high probability when $n{p}^k\to 0$, and Chebyshev's inequality proves existence and relative concentration when $n{p}^k\to \infty$. This remains valid for varying probabilities eventually at most one half.

Let $X$ have a symmetric probability distribution with $\mathbb{P}(|X|>x)=1/(x\log x)$ for $x\ge e$, mass $1-1/e$ at zero, and no mass with $0<|X|<e$. Then $\mathbb{E}|X|=\infty$, but independent copies satisfy the weak law of large numbers with limit zero. Indeed $n\mathbb{P}(|X|>n)=1/\log n\to 0$. Their symmetric truncations $Y_{n,j}=X_j{1}_{\{|X_j|\le n\}}$ have mean zero and second moment $O(n/\log n)$, so Chebyshev's inequality gives $n^{-1}{\sum }_{j\le n}{Y}_{n,j}\to 0$ in probability. A union bound shows that the original and truncated sums differ with probability tending to zero. This also gives a differentiable characteristic function at zero whose derivative cannot be interpreted as an absolutely convergent expectation.