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If $Y$ is a real square-integrable random variable and $X=\mathbb{E}[Y\mid \mathcal{G}]$, then $\mathbb{E}[XY]=\mathbb{E}{X}^2$ and the displayed identity follows by expansion. Thus equality of the second moments forces $X=Y$ as an almost sure equality. This is the equality case of the conditional Jensen inequality for the strictly convex function $z\mapsto z^2$.

Let $Y$ be an integrable random variable, let $H$ be a finite $\mathcal{G}$-measurable random variable, and suppose $HY$ is an integrable random variable. Then a known factor can be taken outside the conditional expectation:For a bounded $H$, first verify the defining integral identity when $H$ is the indicator function of a measurable set in $\mathcal{G}$, extend by linearity to simple functions, and then use bounded approximation and the dominated convergence theorem. For nonnegative $H,Y$, truncating $H$ and using the conditional monotone convergence theorem proves the identity with extended nonnegative conditional expectations. In particular,For general $H,Y$ under the stated integrability assumptions, the conditional Jensen inequality bounds $|H\mathbb{E}[Y\mid \mathcal{G}]|$ by that integrable random variable. Apply the bounded-factor result to truncations of $H$, then use the dominated convergence theorem on the products and the L1 contraction of conditional expectation. This proves the identity without assuming $H$ itself is an integrable random variable.

The product-integrability condition is essential to the ordinary finite conditional expectation statement. An unbounded factor cannot be moved through an expected value simply because its partner is an integrable random variable.^[ref]

For a centered square-integrable martingale with $v=\mathbb{E}{M}_n^2$ and $\lambda >0$, apply the Doob maximal inequality for a nonnegative submartingale to $(M_k+c{)}^2$ for $c\ge 0$. Conditional Jensen inequality gives the submartingale property, and crossing $\lambda$ forces its square above $(\lambda +c{)}^2$. The bound $(v+c^2)/(\lambda +c{)}^2$ is minimized by $c=v/\lambda$. This extends the Cantelli inequality from one random variable to a martingale maximum, without requiring independent increments.

Write $Y=\mathbb{E}[X\mid \mathcal{G}]$. The conditional expectation is an $L^2$ contraction by the conditional Jensen inequality, so $X,Y$ are square-integrable. The defining integral identity for conditional expectation, first with bounded measurable truncations of $Y$ and then with their $L^2$ limit, givesFor clarity, if $Y^{(K)}=\max (-K,\min (Y,K))$, then $\mathbb{E}[X{Y}^{(K)}]=\mathbb{E}[Y{Y}^{(K)}]$, and the Cauchy-Schwarz inequality permits the limit $K\to \infty$. Expanding the square now proves the equality case for conditional second moments:A nonnegative random variable of zero expected value vanishes almost surely. Therefore $\boxed{X=Y\text{ almost surely}.}$

Let $(Z_k,{\mathcal{F}}_k{)}_{0\le k\le n}$ be an integrable real submartingale, and put $Z_n^{\ast }={\max }_{0\le k\le n}{Z}_k$. The refined Doob maximal inequality isIn particular, for a nonnegative submartingale the right-hand side is at most $\mathbb{E}{Z}_n$, giving the usual weak maximal bound.

For the proof, partition $A=\{Z_n^{\ast }\ge \lambda \}$ by its first crossing time: $A_k=\{Z_j<\lambda (j<k),Z_k\ge \lambda \}\in {\mathcal{F}}_k$. The submartingale property gives $\mathbb{E}[Z_n\mid {\mathcal{F}}_k]\ge Z_k$, soSumming over the disjoint crossing events proves the refined inequality, including a possible crossing at time zero. No independence assumption is needed.

The strong form is the Doob Lp maximal inequality. For $p>1$ and a nonnegative submartingale with $Z_n\in L^p$,Indeed, integrate the refined inequality against $p{\lambda }^{p-2}$ up to $K$. The tail integral formula for moments and the Tonelli theorem giveApply the Holder inequality, divide by $\Vert Z_n^{\ast }\wedge K{\Vert }_p^{p-1}$ when it is nonzero, and let $K\to \infty$ by the monotone convergence theorem. For a martingale $M$, the process $|M_k|$ is a submartingale by the conditional Jensen inequality, giving the corresponding bound for $\max |M_k|$. On a bounded continuous-time interval the nonnegative, right-continuous version follows by applying these inequalities on refining finite time grids and taking limits.

Put $V=\mathbb{E}{X}_n^2$. For $c\ge 0$, the process $(X_m+c{)}^2$ is a nonnegative submartingale, by the conditional Jensen inequality. A crossing of $x$ by $X$ implies a crossing of $(x+c{)}^2$ by this squared process. Since $\mathbb{E}{X}_n=\mathbb{E}{X}_0=0$, the Doob maximal inequality givesFor $V>0$, differentiation of the last expression gives $2(cx-V)/(x+c{)}^3$, so its minimum over $c\ge 0$ occurs at $c=V/x$. Substitution proves the one-sided martingale maximal inequalityIf $V=0$, then $X_n=0$ almost surely and $X_m=\mathbb{E}[X_n\mid {\mathcal{F}}_m]=0$ for every $m\le n$, so the same conclusion holds. The constant is sharp: at one step take $X_1=x$ with probability $V/(V+x^2)$ and $X_1=-V/x$ otherwise. This has mean zero, variance $V$, and achieves equality.

Put $C={\sup }_p\mathbb{E}|X_p|<\infty$. Since $x\mapsto x^{+}$ is an increasing convex function, the conditional Jensen inequality and the submartingale property giveFor fixed $n$ and $p\ge n$, the tower property of conditional expectation therefore impliesChoose versions for this countable family so that all these inequalities hold outside one null set. Its nonnegative increasing limit $M_n$ is ${\mathcal{F}}_n$-measurable. The monotone convergence theorem shows thatIn particular the limit is finite almost surely. Thus the increasing conditional means converge to an integrable nonnegative $M_n$, rather than merely to a possibly infinite extended value. This is the construction of the positive martingale majorant of an L1-bounded submartingale.

A family $(Y_i{)}_{i\in I}$ of integrable random variables is uniformly integrable ifThis controls the contributions of large values uniformly over the family, rather than merely bounding its L1 norms.

Put $X_n=\mathbb{E}[Z\mid {\mathcal{F}}_n]$ and $A_n(K)=\{|X_n|>K\}\in {\mathcal{F}}_n$. The conditional Jensen inequality gives $|X_n|\le \mathbb{E}[|Z|\mid {\mathcal{F}}_n]$, hence $\mathbb{E}|X_n|\le \mathbb{E}|Z|$. The Markov inequality therefore givesUsing the measurability of $A_n(K)$ in the defining integral identity for conditional expectation, for every $R>0$ we getThe first term tends to zero as $R\to \infty$ because $Z$ is integrable, by the dominated convergence theorem. Given $\epsilon >0$, first choose $R$ to make it less than $\epsilon /2$, then choose $K$ to make the second term less than $\epsilon /2$. These choices work for every $n$. This proves uniform integrability of conditional expectations:In fact the argument works for an arbitrary family of sigma-algebras; nesting is not needed for this conclusion.

The square root is concave on $[0,\infty )$, and $\sqrt{S_T}$ is integrable since $\mathbb{E}\sqrt{S_T}\le \sqrt{\mathbb{E}{S}_T}$. For $t\le T_1\le T_2$, the conditional Jensen inequality and the martingale property giveTaking conditional expectations with respect to ${\mathcal{F}}_t$ and using the tower property of conditional expectation yieldsThus the maturity curve for a square-root stock claim is nonincreasing; it need not be strictly decreasing, as a constant stock gives equality. The inequality holds for every ordered pair of maturities. Under the usual right-continuous versions of the market processes, it gives the corresponding nonincreasing version of the maturity curve.

Put $v=\mathbb{E}{X}_n^2$. For $c\ge 0$, conditional Jensen inequality applied to the convex function $x\mapsto (x+c{)}^2$ shows thatis a nonnegative submartingale. Its integrability follows from the square integrability of $X_k$. If $X_k\ge \lambda$, then $Z_k\ge (\lambda +c{)}^2$, since $c\ge 0$. The Doob maximal inequality for a nonnegative submartingale yieldswhere zero mean removes the cross term. For completeness, the maximal inequality follows by stopping at the first crossing: on the event of a crossing at $k$, the submartingale property gives $\mathbb{E}[Z_n{1}_{\{T=k\}}]\ge \mathbb{E}[Z_k{1}_{\{T=k\}}]$. Sum over $k\le n$, and use nonnegativity on the event of no crossing.

The derivative of the last ratio isFor $v>0$, the minimum over $c\ge 0$ is attained at $c=v/\lambda$. Substitution gives the one-sided maximal inequality for a centered square-integrable martingale:If $v=0$, $X_n=0$ almost surely and $X_k=\mathbb{E}(X_n\mid {\mathcal{F}}_k)=0$ almost surely for every $k\le n$, so the bound also holds. The optimization is the same one underlying the Cantelli inequality, but the submartingale argument controls the entire finite-time maximum.

The Rao-Blackwell theorem says that an unbiased finite-variance estimator $\delta (X)$ of a parameter function $\tau (\theta )$ can be improved using a sufficient statistic $T$. DefineBecause the conditional distribution given $T$ is parameter-independent, this is a statistic computable without knowing $\theta$. The tower property of conditional expectation gives ${\mathbb{E}}_{\theta }{\delta }^{\ast }={\mathbb{E}}_{\theta }\delta =\tau (\theta )$, so it remains unbiased. The law of total variance givesHence the improved estimator has no larger variance or mean squared error:Equality holds precisely when $\delta ={\delta }^{\ast }(T)$ almost surely under the relevant parameter. More generally, even without unbiasedness, conditional Jensen inequality shows that conditioning on a sufficient statistic cannot increase risk for any convex loss in the estimate, provided the expectations exist. The squared-loss result above is the usual variance form of the theorem.

Under the usual no-arbitrage interpretation from part (a), let $Q^N$ be an equivalent martingale measure for the strictly positive numéraire $N$. Then $M_t=S_t/N_t$ is a martingale. Define the discounted European call option payoff $Y_t=(S_t-K{)}^{+}/N_t=(M_t-K/N_t{)}^{+}$. Because $N_{t+1}\ge N_t$ and $K>0$, we have $K/N_{t+1}\le K/N_t$. The conditional Jensen inequality therefore givesFor the middle step, the strike $K/N_t$ is fixed conditionally on ${\mathcal{F}}_t$; alternatively use $\mathbb{E}[X^{+}\mid {\mathcal{F}}_t]\ge \max \{\mathbb{E}[X\mid {\mathcal{F}}_t],0\}$. Thus $Y$ is a submartingale. Risk-neutral valuation in units of $N$ gives $C(T,K)=N_0{\mathbb{E}}^{Q^N}{Y}_T$, with the usual fixed initial prices. ConsequentlyThe source's word “increasing” means nondecreasing: a European call option can have zero payoff at successive maturities, so strict increase is not guaranteed. Integrability is understood in the claim class for which the stated replication prices exist.

The Rao-Blackwell theorem applies to an estimator $Y$ with finite second moment and a sufficient statistic $T$. The conditional estimator $Y^{\ast }={\mathbb{E}}_{\theta }[Y\mid T]$ can be chosen as a function of the observed $T$ that does not depend on the unknown parameter: this uses the parameter-independent conditional data distribution in the definition of sufficient statistic. The tower property of conditional expectation gives ${\mathbb{E}}_{\theta }{Y}^{\ast }={\mathbb{E}}_{\theta }Y$, so their bias agrees. The law of total variance givesFor any target $a(\theta )$ it follows thatEquality holds exactly when $Y=Y^{\ast }$ almost surely under that parameter value. This proves variance reduction and reduction of mean squared error, whether or not the original estimator is an unbiased estimator. More generally, conditional Jensen inequality proves the corresponding inequality for any convex loss in the estimate, whenever the expectations exist.

The function ${\varphi }_a(t)=t\wedge a$ is a concave function, and $X\wedge a$ is integrable since $|X\wedge a|\le |X|+a$. The concave form of the conditional Jensen inequality givesThe random variables $X$ and $Y$ have the same probability distribution, so $\mathbb{E}[X\wedge a]=\mathbb{E}[Y\wedge a]$. Using the law of total expectation, the nonnegative difference between the right and left sides of the inequality has zero expectation. It must vanish almost surely. ThereforeThe equality comes from the distributional assumption as well as the conditional Jensen inequality.

For a submartingale with ${\sup }_n\mathbb{E}|X_n|=C<\infty$, its positive part is a submartingale by the conditional Jensen inequality. Thus the displayed conditional expectations increase in $p$. Their limits are finite almost surely and have expectations at most $C$, by the monotone convergence theorem. Conditional monotone convergence and the tower property of conditional expectation give $\mathbb{E}[M_{n+1}\mid {\mathcal{F}}_n]=M_n$. Moreover $M_n\ge X_n^{+}$. Consequently $Y_n=M_n-X_n$ is a nonnegative supermartingale, ${\sup }_n\mathbb{E}{Y}_n\le 2C$, and $X=M-Y$. The resulting martingale need not have uniform integrability.