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Put $\nu =(T^{†}{)}_{\#}\mu$ and use the two Kantorovich potentialsFor every $x,y\in {\mathbb{R}}^d$,Thus the pair is dual feasible, with equality precisely on $y=\lambda x$. Finite second moments make both potentials integrable; compactness of $X$ is more than sufficient. For any admissible transport map $T$, integrating the inequality and using its pushforward measure givesThe admissible map $T^{†}(x)=\lambda x$ attains equality. ConsequentlyThe same certificate proves optimality of its graph transport plan among all transport plans. Notice that $u$ need not be a convex function when $\lambda >1$: it is a cost dual potential. The associated convex gradient potential is instead $\Phi (x)=\lambda |x{|}^2/2$.

Write $L=W_p(\mu ,\nu )$. The assumed equality of the Monge optimal transport problem and Kantorovich optimal transport problem values givesEach interpolated pushforward measure ${\mu }_t=(P_t{)}_{\#}\mu$ has a finite $p$th absolute moment, since $|P_t(x){|}^p\le 2^{p-1}(|x{|}^p+|T^{†}(x){|}^p)$ and $(T^{†}{)}_{\#}\mu =\nu$.

For any $s,t\in [0,1]$, the common-source transport planprovides the upper boundFor the reverse bound assume $0\le s\le t\le 1$. Since ${\mu }_0=\mu$ and ${\mu }_1=\nu$, the triangle inequality for the p-Wasserstein distance and the upper bounds already established giveHence $W_p({\mu }_s,{\mu }_t)\ge (t-s)L$. Combining the bounds and using symmetry provesThis is the constant speed property of displacement interpolation. The given invertibility of $P_s$ also lets one realize the competitor as the map $P_t\circ P_s^{-1}$, assuming its inverse is measurable, but the transport plan argument proves the result without invertibility.

Saying that $U$ has law $\mu$ means that $\mu$ is the pushforward measureFor a real random variable, $\mu$ is a probability measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$.

The pair $U,V$ consists of independent random variables whenfor every pair of Borel sets $A,B$; equivalently, their joint law is the product of their marginal laws.

SetThese are bounded random variables with values in $(0,1]$, and for every positive integer $n$,Part c shows that $U$ and $V$ have the same law. Since $x\mapsto e^{-x}$ is injective on $[0,\infty )$ with inverse $u\mapsto -\log u$, taking the pushforward measure by this inverse shows that $X$ and $Y$ have the same law. This proves the uniqueness theorem for Laplace transforms of nonnegative random variables.

The mapping theorem for Poisson point processes applies provided $f$ is a measurable function and its pushforward measureis locally finite, meaningfor every bounded Borel set $B\subseteq {\mathbb{R}}^s$. If “Poisson process” means a simple point process, one also requires $f_{\ast }\Lambda$ to be atomless, equivalentlyUnder these conditions, $f(\Pi )$ is a Poisson point process with mean measure

The pushforward measure of $\mu$ under $g$ isThe Lebesgue decomposition theorem says that for sigma-finite measures $P,Q$ there are unique measures $P_{\mathrm{ac}}$ and $P_{\mathrm{s}}$ such that

For a convex lower-semicontinuous function $f$ with $f(1)=0$, choose any measure $\lambda$ dominating probability measures $P,Q$, put $p=dP/d\lambda$ and $q=dQ/d\lambda$, and define the f-divergenceusing the lower-semicontinuous perspective value when $q=0$. This definition includes the singular part and is independent of $\lambda$.

To prove the data processing inequality for f-divergences, let $\mathcal{G}=\sigma (g)$ and use $\lambda =P+Q$. The densities of $g_{\ast }P$ and $g_{\ast }Q$, pulled back to $\mathcal{G}$, are ${\mathbb{E}}_{\lambda }(p\mid \mathcal{G})$ and ${\mathbb{E}}_{\lambda }(q\mid \mathcal{G})$. Since the perspectiveis jointly convex, conditional Jensen inequality givesIntegration proves

The chi-squared divergence iswhen $P\ll Q$, and is infinite otherwise.

Fix a probability measure $Q$. Let $J$ be uniform on $\{1,\dots ,M\}$ and, conditionally on $J=j$, draw $X$ from $P_j$. Compare this joint law with the reference law under which $J$ is uniform and independent of $X\sim Q$. Forthe target expectation is $M^{-1}{\sum }_j{P}_j(A_j)$, while its reference expectation isbecause the $A_j$ form a partition. Its reference variance is $M^{-1}(1-M^{-1})$. The Cauchy-Schwarz inequality applied to the likelihood ratio givesThe last divergence separates over $J$:Taking the infimum over all probability measures $Q$ proves the required inequality.

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 map $F$ is a nonsingular transformation with respect to $\omega$ whenEquivalently, the pushforward measure $F_{\ast }\omega (B)=\omega (F^{-1}(B))$ satisfies $F_{\ast }\omega \ll \omega$.

For an essentially bounded observable $g$, define the Koopman operatorNonsingularity makes this well defined on almost-everywhere equivalence classes. By the Radon-Nikodym theorem,Consequently the bounded Koopman operator criterion is

The pushforward measure of $\mu$ under $g$ is $g_{\ast }\mu$, defined for $B\in \mathcal{B}$ byThe Lebesgue decomposition theorem says that if $\nu$ and $\mu$ are sigma-finite measures on the same measurable space, then uniquely

Let $f:[0,\infty )\to \mathbb{R}\cup +\infty$ be convex with $f(1)=0$. If $\lambda$ dominates the probability distributions $P,Q$, with densities $p,q$, their f-divergence isusing the lower-semicontinuous perspective convention where $q=0$. This definition is independent of the dominating measure.

The data processing inequality for f-divergences statesTo prove it, take $\lambda =P+Q$ and let $\mathcal{G}=g^{-1}(\mathcal{B})$. If $p=dP/d\lambda$ and $q=dQ/d\lambda$, then the pullbacks of the densities of $g_{\ast }P$ and $g_{\ast }Q$ with respect to $g_{\ast }\lambda$ are respectively ${\mathbb{E}}_{\lambda }[p\mid \mathcal{G}]$ and ${\mathbb{E}}_{\lambda }[q\mid \mathcal{G}]$. The perspective $F(a,b)=bf(a/b)$ of a convex function is jointly convex. Conditional Jensen inequality therefore givesIntegration proves the claim.

The Squared Hellinger distance isIf $P,Q$ have densities $p,q$ with respect to a sigma-finite measure $\mu$, this becomes

Fix any probability distribution $Q$ and set $p_j=P_j(A_j)$, $q_j=Q(A_j)$, and $t_j=H^2(P_j,Q)$. Applying the data processing inequality to the indicator function of $A_j$ gives the Bernoulli Hellinger boundThe hinted inequality impliesThe function $t\mapsto \sqrt{t(1-t/4)}$ is concave on $[0,2]$. Since the $A_j$ form a set partition, ${\sum }_j{q}_j=1$, and Jensen inequality givesTaking the infimum over $Q$ proves the stated inequality.

If $\mu$ has Poincaré constant $C$ and $\varphi$ is $L$-Lipschitz, then the pushforward measure ${\varphi }_{\ast }\mu$ has Poincaré constant at most $C{L}^2$. Apply the original inequality to $g\circ \varphi$ and use the chain rule.

Applying a measurable per-claim retention $g$ to independent claim sizes preserves the compound Poisson distribution form. The transformed severity is the pushforward measure of the original severity law. With count parameter $\lambda$, the retained aggregate has expected value $\lambda \mathbb{E}g(X)$ and variance $\lambda \mathbb{E}[g(X{)}^2]$. Zero retained marks may be removed by Poisson thinning.