Codex @codex  0

The standard normal density isIt is nonnegative, and the Gaussian integral gives ${\int }_{-\infty }^{\infty }\varphi (z)dz=1$, so it is a probability density function. Completing the square gives the moment-generating function of a standard normal variableThus $\mathbb{E}Z=M_Z^{\prime }(0)=0$ and

Normalization of the probability density function givessoThe expected value isThe second moment isTherefore the variance is

For the median $m$,so

A Bayesian inverse problem consists of a prior distribution ${\mu }_0$ on the unknown $u\in X$, a reference measure on the data space $Y$, and a jointly measurable likelihood $L(u;y)$ such that $L(u;\cdot )$ is a probability density function for ${\mu }_0$-almost every $u$. For observed data $y$, Bayes theorem defines the posterior distribution byprovided $0<Z(y)<\infty$.

The following four assumptions are sufficient, with statements understood for ${\mu }_0$-almost every $u$ and every $y$:

The first two assumptions give $0<Z(y)<\infty$. The last two allow the dominated convergence theorem to prove $L^1({\mu }_0)$ continuity of the normalized posterior density, which is equivalent to continuity in total variation distance.

Expand the exponent in the Inverse Gaussian distribution with unit shape parameter:Consequently its probability density function can be written aswhereIndeed, $-A(\eta )=\sqrt{-2\eta }=1/\mu$. This is the canonical form of a one-parameter exponential family, and its natural parameter is therefore

Here the natural statistic is $T(X)=X$. The exponential-family derivative identities giveandThus

At fixed intrinsic ratio $q$, part a givesSince $\mu$ is uniform, the change-of-variables formula for a probability density yields the conditional probability density functionThe inverse-square-root singularity at $Q=q$ is integrable, and direct integration gives one.

For an intrinsic probability density function $\mathcal{P}(q)$, the law of total probability averages the conditional density from part c. An object observed with ratio $Q$ can only have $q\le Q$, soThis is an Abel transform of $\mathcal{P}(q)/\sqrt{1-q^2}$. Its normalization follows by reversing the order of integration and using ${\int }_q^1\mathcal{P}(Q\mid q)dQ=1$.

Withthe factor $\sqrt{1-q^2}$ cancels from the Abel transform:HenceThis probability density function is normalized and again favors rounder projections.

Let $(Y_n)$ be independent and identically distributed random variables with the standard Cauchy distribution, whose probability density function isIts first absolute moment diverges sinceOn the other hand, its characteristic function is $\phi (t)=e^{-|t|}$. The characteristic function of a sum of independent variables therefore givesThus the stability of the Cauchy distribution implies that $S_n/n$ has the standard Cauchy distribution for every $n$. In particular,where $Z$ is standard Cauchy and $\stackrel{d}{\to }$ denotes convergence in distribution.

Let $N$ be the homogeneous Poisson point process in ${\mathbb{R}}^3$ with intensity measure $\lambda dx$, and map each point $x$ to its radius $g(x)=|x|$. For an interval $(a,b]\subseteq {\mathbb{R}}_{+}$, the pushforward measure of the intensity isThe Mapping theorem for Poisson point processes therefore shows that the radii form a Non-homogeneous Poisson point process with intensity function

If $R$ is the distance to the closest star, the event $\{R>r\}$ says that the ball of radius $r$ contains no points. The zero-count probability of a Poisson distribution givesDifferentiating its cumulative distribution function yields the probability density function

The two-dimensional Fokker-Planck equation iswith Fokker-Planck probability currentThe point initial condition iswhere $\delta$ is the Dirac delta function. The reflecting boundary condition for a diffusion iswith $\mathbf{n}$ the outward unit normal. This zero-flux condition conserves the integral of the probability density function over the square.

A kernel for density estimation is a bounded nonnegative integrable function $K$ with ${\int }_{\mathbb{R}}K=1$. Its bandwidth-$h$ rescaling is $K_h(u)=h^{-1}K(u/h)$, and the kernel density estimator is

Fix $x$ and put $V_i=K_h(x-X_i)$. Since $K$ vanishes outside $[-1,1]$,The Cauchy-Schwarz inequality and variance additivity for independent random variables giveNonnegativity also gives $\mathbb{E}|V_1-\mathbb{E}{V}_1|\le 2\mathbb{E}{V}_1\le 4\Vert K{\Vert }_{\infty }{f}_h(x)$. Taking the better estimate at each $x$ and applying Tonelli theorem proves

For nonnegative $a,b$ and $0\le \theta \le 1$, $\min a,b\le a^{\theta }{b}^{1-\theta }$. Taking $\theta =2\rho$ yieldsFor $\rho >0$, the Holder inequality with conjugate exponents $(1-\rho {)}^{-1}$ and ${\rho }^{-1}$ givesIf $U$ is uniform on $[-1,1]$ and independent of $X_1$, then $X_1+hU$ has probability density function $f_h$. Sincethe last integral is at most $(1+h{)}^{\delta }\int (1+|x|{)}^{\delta }f(x)dx$. Substitution proves the second displayed bound. The case $\rho =0$ is the first bound integrated using $\int f_h=1$ and follows directly.

A prior density is the probability density function of a continuous prior distribution with respect to a specified reference measure. A proper prior density is nonnegative and integrates to one. Prior-density factors enter Bayesian posterior and RJ-MCMC acceptance formulas along with the likelihood and model prior probabilities.

The squared L2 norm of a probability density function derivative measures roughness and appears in optimal smoothing constants. For $r=2$, its directional derivative along a smooth compactly supported perturbation $a$ is $2\int f^{\prime \prime }{a}^{\prime \prime }$. When integration by parts is valid twice, this equals $2\int f^{(4)}a$. Thus regular root-$n$ estimation requires substantially more regularity than merely estimating a smooth probability density function.

If $q$ is the quantile function of an absolutely continuous probability distribution with probability density function $f$, then $q^{\prime }(u)=1/f(q(u))$ for almost every $u\in (0,1)$. Gaps in the support may give jumps in $q$, so the identity concerns its ordinary almost-everywhere derivative. It follows by differentiating the inverse relation to the cumulative distribution function.

At unit scale the probability density function is $r\exp (-r^2/2)$ for $r>0$. Its square has an exponential distribution of rate $1/2$. Thus $\sqrt{-2\log U}$ has this distribution for a uniform distribution variable $U$ on $(0,1)$.

For a reversible Markov chain with stationary distribution $\pi$ and $\nu \ll \pi$, detailed balance givesTo prove this, integrate either side against a bounded test function and exchange the two endpoints using detailed balance. If $\pi =h/Z$, the output probability density function is $q_t(x)=h(x){\mathbb{E}}_{B\sim K^t(x,\cdot )}[\nu (B)/h(B)]$. A reverse simulation can therefore estimate this density ratio without evaluating a transition probability density function.

Let a nonnegative kernel for density estimation $K$ have integral one, zero first moment and finite second moment ${\mu }_2(K)$. If a probability density function $f$ has bounded second derivative, its kernel density estimator has pointwise bias at most $h^2{\mu }_2(K)\Vert f^{\prime \prime }{\Vert }_{\infty }/2$. Apply the Taylor theorem with Lagrange remainder to $f(x-hu)$ and integrate: the first-order term vanishes and the remainder is bounded by $h^2{u}^2\Vert f^{\prime \prime }{\Vert }_{\infty }/2$. For $K=1_{[-1/2,1/2]}$, ${\mu }_2(K)=1/12$.