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Put $\Delta B_h=B_{t+h}-B_t$ and $R_h={\int }_t^{t+h}(H_s-H_t)d{B}_s$. The constant-in-time coefficient $H_t$ on $(t,t+h]$ is previsible, so the ratio minus $H_t$ is $R_h/\Delta B_h$. Since $|H|\le C$, continuity and the dominated convergence theorem implyThe Itô isometry gives $\mathbb{E}{R}_h^2\le h{\eta }_h$. For $\epsilon ,\delta >0$, use the Chebyshev inequality and the normal distribution of $\Delta B_h/\sqrt{h}$ to obtainFirst let $h\downarrow 0$, then $\delta \downarrow 0$; the normal distribution has no atom at zero. This provesNo independence between $R_h$ and the denominator is needed.

The first estimate suggested by the hint is valid: Brownian scaling and the Gamma function give $\mathbb{E}|\Delta B_h{|}^{-1/2}=h^{-1/4}{2}^{-1/4}\Gamma (1/4)/\sqrt{\pi }$. The second hint requires care about the placement of its expectation: with $V_h={\int }_t^{t+h}(H_s-H_t{)}^{2}ds$, the interpretation $\mathbb{E}|R_h{|}^{1/2}\le (\mathbb{E}{V}_h{)}^{1/4}$ is correct by the Jensen inequality and the Itô isometry, whereas the interpretation $\mathbb{E}|R_h{|}^{1/2}\le \mathbb{E}(V_h^{1/4})$ is false with constant $1$. For the latter interpretation, the Burkholder-Davis-Gundy inequality supplies a constant $C_{1/2}$ instead. One can see the obstruction with a Brownian motion stopped on exiting $(-1,1)$: if $\tau$ is that Brownian exit time, then $|B_{\tau }|=1$, $\mathbb{E}\tau =1$, and strict Jensen inequality gives $\mathbb{E}{\tau }^{1/4}<1=\mathbb{E}|B_{\tau }{|}^{1/2}$.

This obstruction also survives the continuity condition on $H$. On a long finite interval approximate $1_{\{s<\tau \}}$ by a continuous adapted process equal to $1$ until $\tau$ and decreasing linearly to $0$ during the next $ϵ$ units of time, and take its initial value to be $0$ by inserting a short continuous ramp before starting the exit experiment. The Itô isometry controls these approximations, and letting the interval length tend to infinity recovers the strict inequality above; hence some bounded continuous adapted example also violates the latter interpretation with constant $1$. The direct proof avoids relying on this ambiguous hint.

Fix the observed $Y$, let $L(\theta )=p_{\theta }(Y)$, and let $q_0(z)=p_{{\theta }_0}(z\mid Y)$. The E-step formsFor finite expected log-likelihoods and $L({\theta }_0)>0$, the Jensen inequality givesThe first inequality allows additional mass outside the old complete-data support; it is equality when the relevant supports coincide. Ratios are taken on the support of $q_0$. If a candidate assigns zero density on positive $q_0$-mass, its $Q$ is minus infinity and it cannot be an improving finite M-step.

An M-step of the expectation-maximization algorithm chooses ${\theta }_1$ with $Q({\theta }_1\mid {\theta }_0)\ge Q({\theta }_0\mid {\theta }_0)$, soThis EM likelihood monotonicity also holds for a generalized M-step that merely increases $Q$, rather than maximizing it exactly. It is a statement about the observed-data likelihood function.

Apply the Jensen inequality to the convex exponential function under the trial equilibrium measure:The given identity therefore implies $e^{-F}\ge e^{-F_0-⟨H-H_0{⟩}_0}$. Taking minus the logarithm reverses the inequality and proves the Feynman-Bogoliubov inequalityThe units $\beta =1$ make $H$ and $F$ dimensionless here. Equality holds when $H-H_0$ is constant on the support of the trial measure. Optimizing the trial parameters tightens an upper bound, which is the basis of the Gaussian variational approximation.

For every admissible transport map $T$, the pushforward measure condition gives $T(x)\in [1,2]$ for $\mu$-almost every $x\in [0,1]$. Hence $|T(x)-x|=T(x)-x$, andApply the Jensen inequality to the convex function $h$ and the probability measure $\mu$:Translation by one sends the source uniform distribution to the target uniform distribution, so $T^{†}(x)=x+1$ is admissible. Its displacement is constantly one, givingNo monotonicity assumption on $h$ is needed, because all displacements have the same sign. The same argument with a transport plan also gives the identical minimum for the Kantorovich optimal transport problem.

As in part (c), every admissible transport map has nonnegative displacement $D(x)=T(x)-x$ with ${\int }_0^{1}D(x)dx=1$. The square root is a strictly concave function, so the reversed Jensen inequality givesThe map $T^{†}(x)=x+1$ is admissible and attains one. ThereforeThus “worst” means largest cost among admissible transport maps. Equality in the Jensen inequality for a strictly concave function forces $D(x)$ to be constant almost everywhere, so this maximizer is unique up to a $\mu$-null set. For comparison, the admissible reflection $T(x)=2-x$ has smaller costThe upper bound also holds for every transport plan.

The stochastic gradients obeyThe nonexpansiveness of metric projection and the subgradient inequality imply, after conditioning on ${\beta }_s$,Summing telescopes. Since both ${\beta }_1$ and $\hat{\beta }$ lie in the radius-$R$ set $C$, their distance is at most $2R$. By the Jensen inequality and convexity of $f$,ChooseThe two terms are then equal, and

With varying positive step sizes, the same one-step inequality isSum from $s=1$ to $k$, use $\Vert {\beta }_1-\hat{\beta }{\Vert }_2\le 2R$, and divide by $2A_1$. The Jensen inequality for the weighted average $\tilde{\beta }=A_1^{-1}{\sum }_s{\eta }_s{\beta }_s$ giveswhere $A_1={\sum }_s{\eta }_s$ and $A_2={\sum }_s{\eta }_s^2$.

A function $f$ on an interval is a convex function ifJensen inequality states that for an integrable random variable $X$, when all terms are defined,

The function $\varphi (x)=x\log x$ is convex on $(0,\infty )$ becauseApplying Jensen's inequality to the uniform distribution on the positive numbers $x_1,\dots ,x_n$ givesWriting $S={\sum }_i{x}_i$ and multiplying by $n/S$ yields

SetUnder the probability measure with density $Z$, the Jensen inequality for the concave logarithm givesThe sub-Gaussian assumption with variance parameter $\nu /4$ gives $M(2\lambda )\le e^{\nu {\lambda }^2/2}$. A second application of Jensen inequality gives $M(\lambda )\ge e^{\lambda \mathbb{E}X}=1$. Consequentlyas required.

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.

A set $R\subseteq V(G)$ is $(s,k)$-rich when every $s$-element subset of $R$ has at least $k$ common neighbours.

Choose $t$ vertices $x_1,\dots ,x_t$ independently and uniformly from $V(G)$, allowing repetitions, and letBy Jensen inequality applied to the convex function $z\mapsto z^t$,

Let $Y$ count the $s$-subsets $S\subseteq X$ having fewer than $k$ common neighbours in $G$. For each such $S$, the probability that $S\subseteq X$ issoThe assumed inequality gives $\mathbb{E}(|X|-Y)\ge r$, so some choice has $|X|-Y\ge r$. Delete one vertex from each bad $s$-subset of $X$. The remaining set $R$ has size at least $|X|-Y\ge r$ and contains no bad $s$-subset, so it is $(s,k)$-rich. This is the basic dependent random choice argument.

Let $X_1,\dots ,X_N$ be independent with laws $p_1,\dots ,p_N$. We prove the claim by induction. Split the law of total variance at the last coordinate:The first term is at most $c_N^2\mathbb{E}|{\partial }_{N}f(X){|}^2$. Apply the induction hypothesis to $g(x_1,\dots ,x_{N-1})={\mathbb{E}}_{X_N}f(x_1,\dots ,x_{N-1},X_N)$. Differentiation under the expectation and Jensen inequality giveWriting $c={\max }_i{c}_i$ and combining the terms yieldsThis is the Tensorization of a Poincaré inequality.

For $0\le t\le 1$, concavity of $U$ impliesso $G$ is concave.

Since $\mathbb{E}Z=0$, Jensen inequality givesIf $0\le s<t$, concavity and $s=(1-s/t)0+(s/t)t$ yieldHence $G$ is decreasing on $[0,\infty )$; this is scaled centered risk under concave utility.

Condition on the information available before time $n$. The current wealth is then fixed, while ${\xi }_n\sim N(b,\Sigma )$ is independent of that information. PutFor any portfolio $\theta$, use the inner product induced by the positive-definite matrix $\Sigma$ to writeThe linear image of a multivariate normal vector shows that the two portfolio returns are jointly normal, andThus independence of uncorrelated jointly normal variables makes the residual return ${\eta }^T{\xi }_n$ independent of the aligned return; it also has expected value zero.

The continuation value $V(n,\cdot )$ is concave by part (c). Conditional Jensen inequality therefore shows that adding the independent centered residual cannot improve the objective:where $m=(1+r)X_{n-1}$. Since the optimal portfolio is unique, its residual must be zero.

It remains to determine the sign. If $\lambda <0$, the portfolios $\lambda {\Sigma }^{-1}b$ and $-\lambda {\Sigma }^{-1}b$ have the same return variance, namely ${\lambda }^{2}B$, while their means are $\lambda B$ and $-\lambda B$. The latter return has the distribution of the former plus the positive constant $-2\lambda B$. Since $V(n,\cdot )$ is increasing, replacing $\lambda$ by $-\lambda$ cannot reduce the objective, contradicting uniqueness. Hence $\lambda \ge 0$.

Applying this conditional argument at every time gives nonnegative, past-measurable random variables ${\lambda }_n$ such thatThis is the Gaussian one-fund theorem.

The function $U(s)=V_0{e}^{-2s}$ is strictly convex. Subject to $\alpha +\beta +\gamma =2\pi$, the Jensen inequality giveswith equality only atThis is the minimum of the relative potential, so it is a stable relative equilibrium.

Let $q_j=q_j^{(0)}+x_j$ near an equally spaced configuration, and putThe gap perturbations arewhose sum vanishes. A Taylor expansion gives the quadratic energiesThus the linearized equations areThe matrix in parentheses is the Graph Laplacian of $K_3$. Its constant eigenvector has eigenvalue zero, while every vector whose components sum to zero has eigenvalue three. One orthonormal set of three normal modes and their angular frequencies is thereforeThe zero mode is rigid rotation. The positive and equal squared frequencies on the two-dimensional relative subspace prove stability and agree with the general normal modes of three equal masses with a symmetric gap potential.

Let $w=N(U)/N(R)$, so $1-w=N(V)/N(R)$. Since class counts add,The function $g(p)=p(1-p)$ is concave function on $[0,1]$. Jensen inequality therefore giveswhich is precisely $Q\le 0$. Thus an axis-aligned split cannot increase the weighted empirical Gini impurity.

Use the distribution $R$ associated with the code in part b. Since $2^{L(x)}=(KR(x){)}^{-1}$,For $\alpha =1/(1+\rho )$, Holder inequality, equivalently the indicated Jensen inequality, givesTherefore