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If $X_n\stackrel{d}{\to }X$ and a measurable function $g$ is continuous at $X$ almost surely, then $g(X_n)\stackrel{d}{\to }g(X)$. The same principle holds for random elements of metric spaces.

The Einstein tensor isThe contracted Bianchi identity and metric compatibility give
${\nabla }^{\nu }{G}_{\mu \nu }=0$, ensuring compatibility of the Einstein equations with covariant stress-energy conservation.

An adapted process $M$ is a local martingale when there are stopping times $T_n\uparrow \infty$ almost surely such that each stopped process $M_{t\wedge T_n}$ is a martingale. Every martingale is local, while a positive local martingale is a supermartingale.

A birth-death process moves only between neighbouring nonnegative integer states, with state-dependent birth and death rates.

A batch-birth linear-death process has transitionsWriting $P_n(t)=\mathbb{P}(N_t=n)$ and taking $P_j=0$ for $j<0$, its master equation is

For a finite-state chain with transition semigroup $P(t)=e^{tQ}$, a probability row vector $\pi$ is invariant when $\pi P(t)=\pi$ for every $t\ge 0$. Equivalently,

Explosion occurs when infinitely many jumps take place in finite time.

Explosion is determined by whether the successive holding times of the continuous-time Markov chain have a finite sum.

The jump chain records the successive states visited by a continuous-time Markov chain while discarding its holding times.

For $h>0$, the fixed-time skeleton has matrix $P(h)$. If $q_i$ is the holding rate, then for $nh\le t\le (n+1)h$,Thus ${\int }_0^{\infty }{p}_{ii}(t)dt$ diverges exactly when ${\sum }_n{p}_{ii}(nh)$ does, proving recurrence equivalence for an irreducible chain.

Conditioned on a finite jump path $i_0,\dots ,i_n$, the holding times are independent exponentials with the corresponding rates. Multiplying the probability of occupying $i_n$ at time $t$ by $q_{i_n}$ turns its simplex density into a product symmetric under reversal of the time portions and the state sequence.

For transition rates $q_{nm}$, the generator acts on a function $f$ byWhenever the expectations are finite,Choosing polynomial functions $f$ gives differential equations for the moments.

A Markov reward model attaches a reward rate to each state and integrates it over the time spent there.

For a time-homogeneous continuous-time Markov chain, $P_{ij}(t)={\mathbb{P}}_i(X_t=j)$ and the matrices satisfy $P(s+t)=P(s)P(t)$, the Chapman-Kolmogorov equation. On a finite state space with Q-matrix $Q$, the matrix exponential gives $P(t)=e^{tQ}$.

The distinction from a discrete-time transition matrix is the elapsed-time parameter: $P(t)$ describes where the process is after $t$ time units, allowing any number of intervening jumps. Each $P(t)$ is a stochastic matrix, with $P(0)=I$. A Q-matrix instead records instantaneous rates and has row sums zero. Sampling at a fixed spacing $\Delta$ yields a Discrete-time Markov chain with transition matrix $P(\Delta )$. See Hao Wu's MIT lecture 21, transition semigroups and lecture 22, infinitesimal rates.

A finite-state process has generator $Q$ exactly when, for every function $f$ on the state space,is a martingale with respect to the process filtration.

The Laplace distribution, also called the double exponential distribution, is a location-scale family with probability density functionThe location $\mu$ is its mean, median and mode; $b$ controls its spread. In the centred rate convention, the same law has $\mu =0$ and $\theta =1/b>0$, giving $f(x)=\theta e^{-\theta |x|}/2$. Keeping track of whether a parameter is a scale or its reciprocal prevents reversing the effect of increasing it. ^[ref]

The two halves of the probability density function are reflected exponential decay curves, each with total probability $1/2$. Their meeting point has a sharp corner: the left and right derivatives are $1/(2b^2)$ and $-1/(2b^2)$. This finite peak is not a positive probability mass at $\mu$. Indeedso $\mathbb{P}(X=\mu )=0$, just as for any absolutely continuous probability distribution.

The figure compares a Laplace distribution with $\mu =0$, $b=1$ against a normal distribution with mean zero and standard deviation $\sigma =\sqrt{2}$: both have variance 2. The right panel uses a logarithmic scale for the two-sided tail probability $\mathbb{P}(|X|>r)$. Its straight Laplace distribution curve comes from $e^{-r}$; the normal distribution tail falls faster. Thus values far from the center can be more plausible under a Laplace distribution even after matching the variance. These are theoretical curves, not measured frequencies. The tails still have exponential decay; this is not a heavy-tailed distribution in the convention requiring every positive exponential moment to diverge.

Integrating either half of the probability density function gives the cumulative distribution functionIn particular $F(\mu )=1/2$. The basic moments areThus the scale is the expected value of the absolute distance from the center, and is not the standard deviation. ^[ref] A short way to obtain the moments is to note that $R=|X-\mu |$ has an exponential distribution with rate $1/b$: $\mathbb{P}(R>r)=e^{-r/b}$ for $r\ge 0$. Integrating gives $\mathbb{E}R=b$ and $\mathbb{E}{R}^2=2b^2$; the symmetric distribution gives the centered first moment zero.

The exponential distribution with rate $c>0$ has densityIt is the gamma distribution with shape one.

The gamma distribution with shape $a>0$ and rate $r>0$ has density

A probability density function represents probabilities as areas under a nonnegative measurable function. For a real random variable with an absolutely continuous probability distribution, it satisfiesfor every Borel set $A$. Its height is probability per unit of the outcome variable, not the probability of that exact outcome. Thus a probability density function may exceed one, while the probability of a singleton is zero. NIST describes the interval-probability interpretation through integration. ^[ref].

For the example $f_Y(y)=2y$ on $[0,1]$, $f_Y(0.8)=1.6$, but $\mathbb{P}(Y=0.8)=0$ and $\mathbb{P}(0.7<Y<0.9)=0.9^2-0.7^2=0.32$. The cumulative distribution function is $F_Y(y)=y^2$ within that interval, connecting density height to the local accumulation rate.

Changing the scale changes the density height. If $Z=100Y$, then $f_Z(z)=f_Y(z/100)/100=z/5000$ on $[0,100]$. At the corresponding value $z=80$, its height is $0.016$, while $\mathbb{P}(70<Z<90)=0.32$ is unchanged. The factor $1/100$ compensates for the wider integration variable. This is a change of variables, not a change in the underlying probabilities.

A probability density function is determined only almost everywhere: changing its values on a Lebesgue measure zero set changes no probabilities. The Cantor distribution has no such representation, and a mixed probability distribution can have both a point mass and a density component, so differentiating its cumulative distribution function does not recover all its mass.

An isomorphism of Banach spaces is a bijective bounded linear map whose inverse is bounded. The bounded inverse theorem makes boundedness of the inverse automatic.

Two distinct reduced fractions with denominators below $N$ differ by more than $1/N^2$, so an interval of radius $1/(2N^2)$ contains at most one.