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A nonnegative regression kernel supplies weights $K((x_i-x)/h)$ in a local weighted least squares fit. Unlike a probability density function, its integral need not equal one, because rescaling every weight does not change the fitted coefficients.

Let $L_n$ be the probability density function of the absolutely continuous part of $Q_n$ relative to $P_n$. If $L_n$ converges in distribution under $P_n$ to $L>0$ almost surely with $\mathbb{E}L=1$, then $P_n,Q_n$ are mutually contiguous. Indeed, $\mathbb{E}{L}_n\le 1$ and convergence to a mean-one limit imply uniform integrability and vanishing singular mass, giving forward contiguity. For reverse contiguity, $P_n(A_n)\le P_n(L_n\le c)+Q_n(A_n)/c$, and one first takes $n\to \infty$ and then $c\downarrow 0$.

For a probability density function $f$ on $\mathbb{R}$ with Lipschitz bound $L>0$, the triangular lower envelope $f(x+u)\ge (f(x)-L|u|{)}_{+}$ gives $f(x{)}^2\le L$. If $f$ is strictly positive everywhere, the inequality is strict. Equality forces the entire probability density function to coincide with the triangular envelope, which vanishes outside a finite interval.

The location score is the score function for translating a probability density function $f$. The minus sign comes from differentiating $f(y-\theta )$ in $\theta$. Under regular tail conditions, integration by parts gives $E_f\rho =0$ and $E_f(\epsilon \rho )=1$. These identities determine projections of the score function onto the span of the centered error.

A marginal density is a probability density function for one component of a joint probability distribution. Integrating the joint probability density over the other coordinates gives it; for two components, $f_X(x)=\int f_{X,Y}(x,y)dy$.

For estimating the expected value over all continuous probability density functions on $[0,1]$, compare $f=1$ and $g_n(u)=1+(u-1/2)/\sqrt{n}$. Their means differ by $1/(12\sqrt{n})$, and ${\chi }^2(P_{g_n}\Vert P_f)=1/(12n)$. The chi-squared divergence of product measures bounds joint total variation distance by $1/2$. The metric squared-loss two-point bound then gives the displayed uniform minimax risk lower bound.

Mean-field variational inference restricts the trial probability density function to $q(x)={\prod }_i{q}_i(x_i)$. With the other factors fixed, $q_j^{\ast }(x_j)\propto \exp ({\mathbb{E}}_{q_{-j}}\log \pi (x_j,X_{-j}))$, provided this expression has a finite positive normalization constant. Subtracting the objective at $q_j^{\ast }$ leaves $\mathrm{KL}(q_j\Vert q_j^{\ast })\ge 0$, proving the update is optimal when the terms are well defined.

For any estimator $T$ in a metric parameter space, $d(T,f{)}^2+d(T,g{)}^2\ge d(f,g{)}^2/2$ by the triangle inequality. Average the two risk functions and replace their probability density functions by their minimum. This proves $\max (R_f,R_g)\ge d(f,g{)}^2\int \min (p_f,p_g)/4$. The overlap equals one minus total variation distance. The same proof applies to squared error for a real-valued statistical functional of the parameter, without requiring that functional to be injective.

For a proper convex function with sequential lower semicontinuity $U=-\log \mu$, form $U_{\lambda }(x)={\inf }_y\{U(y)+\lambda \Vert x-y{\Vert }^2\}$ with $\lambda >0$. The Moreau envelope theorem gives $\nabla U_{\lambda }(x)=2\lambda (x-{\mathrm{prox}}_{U/(2\lambda )}(x))$. If $e^{-U_{\lambda }}$ is integrable, it defines a smooth surrogate probability density function. Infimizing the logarithm of a probability density function with a positive quadratic penalty has the opposite sign and does not generally smooth it. For the Laplace distribution, that incorrect operation leaves $-\log 2-|x|-1/(4\lambda )$, which is not differentiable at zero.

The standard normal distribution has probability density function $\varphi (z)=(2\pi {)}^{-1/2}{e}^{-z^2/2}$. Its rapid decay explains both concentration at one location and the logarithmic growth of a Gaussian maximum over many locations.

A normalizing constant makes a nonnegative integrable function into a probability density function: if $0<Z=\int h<\infty$, then $h/Z$ is a probability density function.