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Let $q\in (0,1]$ be the branching extinction probability of a Galton-Watson process with offspring probability generating function $f$. Conditional on extinction, the probability of $j$ children is multiplied by $q^j/q$: every child's descendants must also become extinct. Hence the new probability generating function is $f_q$ and its mean is $f^{\prime }(q)$. If $f^{\prime }(q)<1$, the total progeny of a branching process has conditional expected value $1/(1-f^{\prime }(q))$, and unconditionally $\mathbb{E}[T;T<\infty ]=q/(1-f^{\prime }(q))$. The Markov inequality gives a finite-total-progeny tail bound without discarding the survival mass.

For a finite-variance random variable, shift its centred version by $b>0$ and apply Markov inequality to the square. Minimizing $({\sigma }^2+b^2)/(a+b{)}^2$ at $b={\sigma }^2/a$ proves the bound. A two-point distribution attains equality. If the variance is zero, the centred variable vanishes almost surely.

For every $\lambda >0$, Markov inequality applied to $e^{\lambda X}$ givesOptimizing over $\lambda$ gives the Chernoff bound in terms of the moment-generating function of $X$.

A coalescing coupling of probability distributions of two Markov chains keeps them identical after a meeting time $T$. The coupling inequality for total variation gives $\Vert P^t(x,\cdot )-P^t(y,\cdot ){\Vert }_{\mathrm{TV}}\le \mathbb{P}(T>t)$. A uniform bound on $\mathbb{E}T$, combined with Markov inequality, bounds the mixing time of a Markov chain.

For $t>0$, the number $B_t$ of blocks is finite almost surely. On a finite restriction, the expected time to reach $m$ blocks is at most ${\sum }_{j>m}2/[j(j-1)]=2/m$. Markov inequality, followed by increasing the restriction, gives $\mathbb{P}(B_t>m)\le 2/(mt)$. Let $m\to \infty$, then use rational times and monotonicity to obtain the assertion simultaneously for every $t>0$.

Convergence in $L^2$ means $\mathbb{E}[(X_n-X{)}^2]\to 0$. It implies convergence in probability by Markov inequality.

For independent uniform couplings on $[0,g_{\ast }]$ and equatorial initial bath spins,At fixed finite $N$, the mean tends to zero and the mean square tends to $2^{-N}$ as time grows. At any fixed nonzero time, the mean-square bracket is strictly below one; the Markov inequality proves coherence tends to zero in probability as $N$ grows. Individual finite realizations still exhibit finite spin-bath coherence recurrence.

For $Z$ with a standard normal distribution and $a\ge 0$, split the Gaussian integral at zero and complete the square to obtain $\mathbb{E}{e}^{a|Z|}=2e^{a^2/2}\Phi (a)\le 2e^{a^2/2}$, where $\Phi$ is the standard normal distribution function. This converts a bound on an absolute Gaussian random variable into a two-sided Gaussian tail bound using Markov inequality.

For a nonnegative integer-valued random variable $N$, Markov inequality givesConsequently, $\mathbb{E}N<1$ proves that some outcome has $N=0$, while $\mathbb{E}N\to 0$ shows that $N=0$ with probability tending to one.

Fix $k\ge 1$, ${\gamma }_0>1/(k+1)$, and ${\gamma }_0\le \gamma \le 1-\omega (n)/\log n$ with $\omega (n)\to \infty$. A binomial random graph has one giant component of the displayed order and all other graph components are tree components of orders at most $k$, with high probability. The expected value and variance of the isolated vertex count concentrate it around $n^{1-\gamma }\to \infty$. A Cayley formula upper bound for connected sets excludes component sizes $k+1$ through a small fixed fraction of $n$; the empty-graph cut bound excludes the remaining sizes up to $n/2$. An extra-edge count excludes cyclic small graph components. The expected value of the number of vertices in small nonisolated tree components is $O(nd{e}^{-2d})=o(n{e}^{-d})$, where $d=np$. The Markov inequality completes the count of vertices outside the unique large graph component.

If two parameters are separated by at least $2r$ in a metric, any estimator gives a test by deciding whether its distance from the first parameter is below $r$. For $0<\eta <1$, the event $|Z-1|\le \eta$ and Markov inequality give the displayed minimax risk bound, where $Z$ is the product likelihood ratio. The unrestricted claim for every positive $\eta$ is false when both factors on the right are negative.