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Cantelli's inequality uses a mean and variance to control deviations in one specified direction. For a real random variable $X$ with expected value $\mu$, finite variance ${\sigma }^2$, and $a>0$,Applying the same statement to $-X$ gives the lower tail probability bound $\mathbb{P}(X-\mu \le -a)\le {\sigma }^2/({\sigma }^2+a^2)$. No independence or particular probability distribution is required. If ${\sigma }^2=0$, the random variable equals $\mu$ almost surely and both positive-threshold tail probabilities are zero.

For variance one and a threshold two above the mean, the upper-tail probability is at most $1/5$. Applying the two-sided Chebyshev's inequality to contain that upper event would give the weaker bound $1/4$. For the standard normal distribution, the actual upper tail probability at two is about $0.0227501$, much smaller than either bound. Yet the two-point probability distribution in equality in Cantelli's inequality attains $1/5$, showing why the smaller Gaussian value cannot be promised from these two moments alone.

A short explanation is to centre $Y=X-\mu$ and choose a nonnegative shift $b$. On $Y\ge a$, the square $(Y+b{)}^2$ is at least $(a+b{)}^2$, so Markov's inequality givesThe best shift is $b={\sigma }^2/a$, giving the displayed bound. Choosing the shift before squaring exploits the direction of the deviation; squaring the unshifted random variable treats both directions together. This shifted-square proof is also the strategy requested in the University of Utah's 2005 qualifying problem on Cantelli's inequality. ^[ref].

The one-sided improvement should be used for a one-sided question. Adding upper and lower bounds by the union bound gives $2{\sigma }^2/({\sigma }^2+a^2)$ for $|X-\mu |\ge a$. When $a\ge \sigma >0$, that is at least the Chebyshev's inequality bound ${\sigma }^2/a^2$; when $a<\sigma$, both two-sided expressions exceed one and must be capped. The separate one-sided bounds generally cannot be attained simultaneously by the same probability distribution.

For a real random variable $X$ with finite variance ${\sigma }^2$, expected value $\mu$, and a threshold $a>0$, Chebyshev's inequality givesOne can also cap this upper bound at one, since a probability cannot exceed one. If $\sigma >0$, taking $a=k\sigma$ gives $\mathbb{P}(|X-\mu |\ge k\sigma )\le 1/k^2$. For example, at least $3/4$ of the probability lies strictly within two standard deviations of the mean. This statement needs only the finite variance assumption. If the variance is zero, $X=\mu$ almost surely and every positive deviation threshold has zero tail probability.

Consider a random variable uniformly distributed on $\{-2,-1,0,1,2\}$. Its expected value is zero and its variance is $(4+1+0+1+4)/5=2$. At $a=2$ the actual tail probability is $2/5$, while Chebyshev's inequality gives $2/2^2=1/2$. At $a=3$ the actual tail probability is zero, while the bound is $2/9$. A valid bound need not describe the exact shape or bounded support of a particular probability distribution.

For the standard normal distribution, the actual tail probabilities beyond two and three standard deviations are approximately $0.0455003$ and $0.0026998$, whereas the general bounds are $1/4$ and $1/9$. More knowledge of the probability distribution can therefore improve the estimate. The comparison does not diminish the guarantee: the three-point example in equality in Chebyshev's inequality attains it exactly.

The proof is a one-line use of Markov's inequality on the nonnegative random variable $Y=(X-\mu {)}^2$. The event $|X-\mu |\ge a$ equals $Y\ge a^2$, and $\mathbb{E}Y={\sigma }^2$. This explains why the denominator is squared and why a finite variance supplies the necessary information. ^[ref].

One important use is control of a sample mean. For independent and identically distributed random variables with finite variance ${\sigma }^2$, the variance of a sample mean is ${\sigma }^2/n$. HenceFor every fixed tolerance, the bound tends to zero as the sample size grows, proving the finite-variance case of the weak law of large numbers. For ${\sigma }^2=1$ and $\epsilon =0.1$, $n=2000$ makes the bound $0.05$. This is a sufficient bound under the stated assumptions, rather than an optimal required sample size.

An early historical source is Pafnuty Chebyshev's 1867 paper “Des valeurs moyennes”, published in French translation by N. de Khanikof. Its opening theorem bounds deviations of sums using their expected values and second moments, linking averages with control of exceptional outcomes. ^[ref].

Continuous square-integrable martingales on a fixed finite horizon form a complete space under the displayed terminal norm, modulo indistinguishability of stochastic processes. A terminal-norm Cauchy sequence admits a subsequence whose successive differences have norms below $2^{-n}$. The Doob L2 maximal inequality and Markov's inequality make the probabilities that these differences have suprema above $2^{-n/2}$ summable. The Borel-Cantelli first lemma then gives an almost surely uniformly convergent telescoping series, so the limit has continuous paths.

Conditional-expectation contractivity gives convergence in $L^2$ at every fixed time, and passage to the limit in the martingale identities proves that the limit is a martingale. Adaptedness can be retained by taking the pointwise finite subsequence limit when it exists, and zero otherwise. The original Cauchy sequence converges in terminal norm to this limit. This completeness allows the Itô isometry to extend stochastic integration from simple predictable integrands to the full square-integrable integrand space while retaining a continuous martingale version.

For $a>0$ and ${\sigma }^2>0$, place probabilitiesTheir weighted deviations cancel, giving expected value $\mu$; their weighted squared deviations give variance ${\sigma }^2$. Only the upper point belongs to $X-\mu \ge a$, so its tail probability attains Cantelli's inequality. With $a=2$, ${\sigma }^2=1$, $\mu =0$, the locations are two and $-1/2$, with masses $1/5$ and $4/5$. The optimized shift makes the lower point zero before squaring, explaining equality in the Markov's inequality step.

The threshold convention matters: this example has $\mathbb{P}(X-\mu >a)=0$. For a strict threshold, placing the upper point at $a+\epsilon$ instead gives probabilities approaching the same bound as $\epsilon \downarrow 0$, with the prescribed mean and variance retained. Extra restrictions on the support of a probability distribution can change which extremal models are allowed.

For a finite-expected value nonnegative random variable and $a>0$, equality in the uncapped Markov's inequality holds exactly when $Y$ takes only $0$ and $a$ almost surely. Indeed the nonnegative difference $Y-a{1}_{\{Y\ge a\}}$ has zero expected value exactly when it vanishes almost surely. Giving $a$ probability $p$ yields $\mathbb{E}Y=pa$ and $\mathbb{P}(Y\ge a)=p$. This shows the bound cannot generally be reduced using only the nonnegative random variable's expected value.

Applying Markov's inequality to the positive random variable $e^{tX}$ givesHere $M_X(t)=\mathbb{E}[e^{tX}]$ is the moment-generating function, and the useful bound requires it to be finite at the chosen $t$. Optimizing over $t>0$ gives the Chernoff bound method: an exponential function weights large values more strongly before applying the same average-budget argument.

For example, let $N$ have the Poisson distribution with parameter one. Its expected value and variance are both one, and its moment-generating function is $M_N(t)=\exp (e^t-1)$. At threshold four, choosing $t=\log 4$ givesThis $t$ minimizes the exponent $e^t-1-4t$ because its derivative is $e^t-4$, with positive second derivative. The direct Markov's inequality bound is $1/4$, and the two-sided Chebyshev's inequality bound, applied to a deviation of at least three from the mean one, is $1/9$. The exact Poisson distribution tail probability is $1-e^{-1}(1+1+1/2+1/6)\approx 0.0189882$. These compare three valid bounds obtained from progressively different information; none is an exact tail probability formula for general random variables.

Expurgation discards messages with unusually high conditional decoding error. If a codebook with $2M$ equally likely messages has average error at most $ϵ$, at least $M$ messages have individual error at most $2ϵ$, by Markov's inequality. Retaining those messages and restricting the decoder's candidates gives a codebook with maximal error at most $2ϵ$, provided removal of candidates cannot spoil a previously successful decoding. This is the case for bounded-distance decoding with errors on ambiguity, and for a decoder that retains its former decisions whenever the decoded message remains.

For a sequence of block lengths $n$, halving a book changes its logarithmic rate by $1/n$, which tends to zero. Thus a vanishing average-error bound from a random coding construction can be converted into a vanishing maximal decoding error probability at the same limiting rate. The probabilistic construction first supplies a deterministic book with sufficiently small average error; expurgation is a subsequent deterministic step.

The first moment method uses the expected value of a count to control whether any counted occurrence exists. For a random variable $N$ taking nonnegative integer values, Markov's inequality givesConsequently, $\mathbb{E}N<1$ implies $\mathbb{P}(N=0)\ge 1-\mathbb{E}N>0$, proving that an outcome with no counted occurrence exists. For a sequence of such counts, $\mathbb{E}N\to 0$ shows that $N=0$ with probability tending to one. The first moment method uses information about the average count, while the complementary second moment method also uses its variance.

For $\alpha >0$, Markov's inequality gives the displayed bound, so $\mathbb{E}|Y_n{|}^{\alpha }\to 0$ implies convergence in probability to zero even when the first absolute moment does not tend to zero. For products $Y_n={\prod }_{j=1}^n{U}_j$ of independent and identically distributed factors, it suffices that $\mathbb{E}|U_1{|}^{\alpha }<1$. With the uniform distribution on $(0,b)$, the product of independent uniform random variables formula makes this $\alpha \log b<\log (1+\alpha )$. Since $\log (1+\alpha )/\alpha \to 1$ as $\alpha \downarrow 0$, such an $\alpha$ exists whenever $b<e$.

Suppose $(X_n)$ are identically distributed independent random variables with $X_n\ge 0$, $\mathbb{E}{X}_n=1$, and $X_1$ not constant almost surely. For each $0<r<1$, strict concavity of $t^r$ givesThe moment is finite because $t^r\le 1+t$ on $[0,\infty )$. Let $Y_n={\prod }_{k=1}^n{X}_k$. Independence and Markov's inequality giveThe series of the latter probabilities is finite. The First Borel-Cantelli lemma, applied for the countable choices $\epsilon =1/m$, proves $Y_n\to 0$ almost surely. This proof requires neither existence nor integrability of $\log X_1$.

Although this nonnegative martingale has $\mathbb{E}{Y}_n=1$ at every time, its terminal expected value is zero. It is therefore not uniformly integrable. The contrast is useful: bounded expected values ensure almost sure convergence here, but do not allow preservation of the expected value through the limit.

For Brownian motion started at zero, condition at time ${\lambda }^2$. If the stopping time exceeds ${\beta }^2{\lambda }^2$ while the running absolute maximum stays below $\delta \lambda$, the path must remain in a fixed interval for another $({\beta }^2-1){\lambda }^2$ time units. Its expected exit time from this interval is at most ${\delta }^2{\lambda }^2$. Markov's inequality yields the displayed bound. Integrating it against $d\Phi$ for a function of moderate growth, then absorbing the small tail term, proves $\mathbb{E}\Phi (\sqrt{\tau })\le C_{\Phi }\mathbb{E}\Phi (B_{\tau }^{\ast })$.

The Hoeffding inequality turns finite interval bounds and independence into exponentially small tail probabilities for a sum. It requires neither a normal distribution nor identical probability distributions. Let $X_1,\dots ,X_n$ be independent essentially bounded real random variables, with deterministic finite bounds $a_i\le X_i\le b_i$ almost surely. Define the centered sum and its squared-width budget byFor $V>0$ and $t>0$,The union bound therefore gives $\mathbb{P}(|S_n|\ge t)\le \min \{1,2e^{-2t^2/V}\}$. If $V=0$, every $X_i$ is constant almost surely, so $S_n=0$ and both positive-threshold tail probabilities vanish. The equal-width version is stated in the authored Stanford notes.^[ref]

The quantity $V$ is a sum of squared interval widths, not the variance of $S_n$. Centering changes the endpoints to $a_i-\mathbb{E}{X}_i$ and $b_i-\mathbb{E}{X}_i$, leaving each width unchanged. For example, independent random variables in $[-2,1]$ and $[4,5]$ have $V=3^2+1^2=10$, whatever their expected values or probability distributions. A deviation at least three above their total expected value has probability at most $e^{-18/10}\approx 0.165299$. If all widths equal $b-a>0$, the sample mean ${\overline{X}}_n=n^{-1}{\sum }_i{X}_i$ satisfies

Independence is essential to this general guarantee. Take one Bernoulli random variable $B$ with success probability $1/2$ and set every $X_i=B$. Each lies in $[0,1]$, but for every $n$ the sample mean is either zero or one. Thus $\mathbb{P}(|{\overline{X}}_n-1/2|\ge 1/4)=1$, whereas incorrectly inserting these dependent observations into the formula would give $2e^{-n/8}$, less than one for $n\ge 6$. Some special dependent models have their own valid concentration inequalities; arbitrary statistical dependence cannot be ignored.

For the short proof, apply Markov's inequality to $e^{\lambda S_n}$ with $\lambda >0$, factor the expectation using independence, and apply the Hoeffding lemma to each factor:Minimizing this quadratic polynomial at $\lambda =4t/V$ gives the upper tail probability; applying it to $-S_n$ gives the lower one. This is the Chernoff bound method. The original 1963 article is Probability Inequalities for Sums of Bounded Random Variables by Wassily Hoeffding; its publisher record identifies the article, but its full text was unavailable during this source check. The argument above is a direct derivation rather than a claim to have inspected that article's proof.

Let $Z$ be a nonnegative integer-valued random variable with $\mathbb{E}Z=A/q$, where $A\ge 0$ and $q\ge 1$ are integers. On $Z>A/q$, the integer $qZ$ is at least $A+1$. Markov's inequality gives $\mathrm{Pr}(Z>A/q)\le A/(A+1)$. Independent restarts therefore need at most $A+1$ attempts in expected value to achieve $Z\le \mathbb{E}Z$. The denominator bound on the expected value matters for this runtime guarantee.

Arrange a uniform random permutation of $n$ distinct ordered elements around a circle, and let $L$ be the greatest number of consecutive entries that appear in increasing order. Put $k_0=\min \{k:k!\ge n\}$. Then $\mathbb{P}(k_0-2\le L\le k_0)\to 1$ as $n\to \infty$.

Let $X_k$ count increasing consecutive blocks of length $k$. Each of the $n$ starting positions succeeds with probability $1/k!$, so $\mathbb{E}{X}_k=n/k!$. For $k\le n/2$, disjoint blocks have independent relative orders; blocks overlapping with a start displacement $s<k$ are both increasing precisely when their length-$k+s$ union is increasing. ThusConsequently Markov's inequality gives $\mathbb{P}(L\ge k_0+1)\le 1/(k_0+1)$, while Chebyshev's inequality gives $\mathbb{P}(L<k_0-2)\le 2/(k_0-1)$ for sufficiently large $n$. The latter uses $n/(k_0-2)!>k_0-1$. Both bounds vanish. These are consecutive cyclic runs; allowing arbitrary skipped entries is a different permutation statistic.

For a nonnegative random variable $Y$ and threshold $a>0$, Markov's inequality givesThe useful case has finite expected value; an infinite right side gives no finite information. The bound also holds if $Y$ is nonnegative almost surely. Combining it with the elementary bound one gives $\mathbb{P}(Y\ge a)\le \min \{1,\mathbb{E}Y/a\}$. Nonnegativity is essential: negative values could otherwise cancel the contribution of large positive values to the expected value.

The idea is an average budget. Every outcome where $Y\ge a$ contributes at least $a$ times its probability to $\mathbb{E}Y$, and the other outcomes contribute nonnegative amounts. Formally, the indicator function gives $Y\ge a{1}_{\{Y\ge a\}}$; taking the expected value proves the bound. ^[ref].

For example, let $Y$ take $0$, $1$, $2$, $5$ with respective probabilities $0.5$, $0.2$, $0.2$, $0.1$. Its expected value is $1.1$. At $a=2$, the tail probability is $0.3$, while the bound is $1.1/2=0.55$. At $a=5$, the tail probability is $0.1$, while the bound is $0.22$. A different probability distribution with the same expected value, putting masses $0.78$ and $0.22$ at $0$ and $5$, attains the bound $0.22$ exactly at $a=5$.

Markov's inequality is useful when a moment is easy to compute but a full probability distribution is difficult to determine. A nonnegative count with expected value less than one must have some positive probability of being zero, giving the first moment method. Applying the same bound to a squared centered random variable gives Chebyshev's inequality. Applying it to an exponential function of a random variable gives an exponential Markov bound, which can exploit more than a single moment.

A radius-$r$ Hamming decoder outputs the unique codeword within distance $r$ of the received word, if there is one. Ambiguous cases and absent candidates can be declared errors, or assigned arbitrarily for an error bound. The average error is the mean of the conditional error probabilities over the equally likely transmitted messages.

Let $V_n(r)={\sum }_{j=0}^r\left(\genfrac{}{}{0ex}{}{n}{j}\right)$ be the size of a Hamming ball. For any transmitted random codeword, the channel noise weight is $U\sim \mathrm{Bin}(n,p)$. Decoding can fail only if $U>r$ or another codeword lies within distance $r$ of the output. Conditional on the output, each independently sampled competing codeword is still uniform on the whole word space, so has probability $2^{-n}{V}_n(r)$ of lying in that ball. A union bound yieldsThis is an average over the random code construction as well as over messages and noise. It implies existence of at least one deterministic codebook with average error no greater than this bound, rather than asserting the same inequality for every randomly sampled book.

Choose a target rate $R<1-h(p)$ and a small $\delta >0$ such that $p+\delta <1/2$ and $R<1-h(p+\delta )$. Let $M_n=\lfloor 2^{nR}\rfloor$ and $r=\lfloor n(p+\delta )\rfloor$. The standard Hamming-volume estimate is $V_n(r)\le 2^{nh(r/n)}$: for $s=r/n\le 1/2$, each binomial probability term $s^j(1-s{)}^{n-j}$ with $j\le r$ is at least $s^r(1-s{)}^{n-r}$, and summing proves this bound. Thus the union-bound term decays exponentially, while $P(U>r)\to 0$ by the law of large numbers or a Chernoff bound.

There exists a $2M_n$-message book with average error ${\epsilon }_n\to 0$. At least half its messages have individual error at most $2{\epsilon }_n$, by Markov's inequality on the uniformly chosen message. Keep $M_n$ of these and discard the rest. Restricting the decoder's candidate set cannot spoil a previously unique successful decoding. The retained book therefore has maximum error at most $2{\epsilon }_n\to 0$. For large $n$ its retained codewords must be distinct: identical words could not both have conditional error below one half. This expurgation of a codebook proves reliable communication at every rate below $1-h(p)$, while converting an average-error estimate into the requested maximum-error conclusion.

Chebyshev's inequality assumes a real random variable $X$ with finite expectation $\mu$ and finite variance ${\sigma }^2$. For every real $a>0$,No independence assumption is needed. Applying Markov's inequality to the nonnegative random variable $(X-\mu {)}^2$ proves the formula, including the zero-variance case.

Set $N=\left(\genfrac{}{}{0ex}{}{n}{4}\right)$. Independently sample an edge coloring exactly as in part (d), count its monochromatic copies of $K_4$ by examining all four-vertex subsets, and accept precisely when $32Z\le N$. Otherwise start over with a fresh edge coloring.

To prove an expected polynomial time bound, the existence argument alone is insufficient: we need a lower bound on the acceptance probability at this exact threshold. Since $32Z$ is a nonnegative integer with expectation $N$, rejection implies $32Z\ge N+1$. Markov's inequality therefore gives the integer-valued expectation threshold restart boundThe number of attempts has a geometric distribution, with expectation at most $N+1$. Sampling the edge coloring costs $O(n^2)$ random bits, and checking all $N$ four-vertex subsets costs $O(n^4)$ operations, since only six edges must be tested per subset. HenceThis is a randomized algorithm that always returns an edge coloring meeting the original threshold and terminates with probability one. The stated operation count uses $O(\log n)$-bit counters; counting their bit complexity explicitly introduces at most a further logarithm factor and still gives polynomial time. For $n<4$, return any edge coloring immediately.

For a variable $Z$ of finite variance and $t>0$, Chebyshev's inequality isOn that event, $(Z-EZ{)}^2\ge t^2$. Thus $t^{2}P(|Z-EZ|\ge t)\le E[(Z-EZ{)}^2]$, proving the assertion. Equivalently apply Markov's inequality to the squared deviation.

Let $X$ count the four-vertex sets spanning a clique. Its expectation is $\mu =\left(\genfrac{}{}{0ex}{}{n}{4}\right)p^6$. If $p\ll n^{-2/3}$, then $\mu \to 0$ and Markov's inequality gives $P(X>0)\to 0$.

Now suppose $p\gg n^{-2/3}$. Then $\mu \to \infty$. Two distinct clique indicators are independent unless their vertex sets share at least two vertices. An overlap of two vertices gives $O(n^6)$ ordered pairs requiring eleven distinct edges; an overlap of three gives $O(n^5)$ pairs requiring nine edges. ConsequentlyThe preceding second-moment bound therefore gives $P(X=0)\to 0$. Thusis a threshold for the appearance of a four-clique. Sharing just one vertex creates no dependence because the two edge sets are disjoint.