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Because $s>1$, the function $t\mapsto s^t$ is increasing and nonnegative. Thus $\{X\ge x\}=\{s^X\ge s^x\}$. Applying Markov's inequality to the nonnegative random variable $s^X$, whose expected value is finite by assumption, yieldsUnder the usual probability generating function convention $X$ takes nonnegative integer values. The argument itself works for any real random variable if $G_X(s)$ is understood as the finite expected value $\mathbb{E}[s^X]$.

For a rate-one Poisson process with Poisson arrival times $T_n$, suppose $t_n\to \infty$ and $t_n/n\to \lambda \in [0,\infty ]$. Thenwhere $1/\sqrt{\infty }=0$ and $\mathrm{BN}(\rho )$ denotes the bivariate normal distribution with standard normal marginal distributions, including the singular cases. In particular, equal scales give $\rho =-1$ and separated scales give $\rho =0$.

To prove this, fix real $x,y$ and put $s_n=n+y\sqrt{n}>0$ for large $n$. The event $(T_n-n)/\sqrt{n}\le y$ is exactly $N_{s_n}\ge n$. Counts at the two deterministic times $t_n,s_n$ are sums of a shared count on $[0,\min (t_n,s_n)]$ and an independent count on the remaining interval. The central limit theorem for a centered Poisson random variable of diverging parameter follows by writing a count at a real time $u$ as its count at $\lfloor u\rfloor$, a sum of independent unit increments, plus an independent remainder with expected value at most one. Markov's inequality makes that remainder negligible after division by $\sqrt{u}$.

Apply this central limit theorem to the two independent count blocks whenever their lengths diverge. A block with bounded length, or length negligible compared with the scale of its coordinate, contributes a term tending to zero in probability. The standardized pair of counts therefore has a joint normal limit whose correlation coefficient isThis includes $\lambda =1$: the difference block then has negligible variance after normalization. In the arrival event the standardized count at $s_n$ must be at least $(n-s_n)/\sqrt{s_n}\to -y$. Reversing this inequality changes the sign of the second limiting normal random variable, giving the asserted negative correlation coefficient. Continuity of the marginal distributions permits the moving thresholds and the weak inequalities, including at the singular limits.

Every $\rho \in [-1,0)$ is realized by $t_n=\lambda n$ with $\lambda ={\rho }^{-2}\ge 1$, and zero by $t_n=n^4$. Conversely, even when $t_n/n$ has no limit, it has a subsequence converging in the extended interval $[0,\infty ]$. If the original pair has a joint bivariate normal distribution limit, that subsequence has the same limit, forcing its correlation coefficient into $[-1,0]$.

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.

Starting from completeness $2/3$ and soundness $1/3$, run an odd number $r$ of parallel copies and take majority. The QMA parallel repetition with entangled witnesses argument proves soundness for arbitrary repeated witnesses. Exponential Markov's inequality gives majority error at most $(2\sqrt{2}/3{)}^r$. Hence $O(\log (1/ϵ))$ copies suffice for error $ϵ$, with polynomial overhead when the target error is inverse-polynomial or exponentially small in input length.

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.