### 统计代写|离散时间鞅理论代写martingale代考|Derivation of the Monge–Amp`ere equation

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• Statistical Inference 统计推断
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• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 统计代写|离散时间鞅理论代写martingale代考|Axiomatic construction of marginals: Stieltjes moment problem

We have explained previously that marginals $\mathbb{P}^{i}$ can be inferred from market values of $T$-Vanilla call/put options. However, in practice, only a finite number of strikes are quoted and therefore these liquid prices need to be interpolated and extrapolated in order to imply the marginals $\mathbb{P}^{i}$ (supported in $\mathbb{R}{+}$). We report here how this can be achieved. This problem can be framed as Stieltjes moment problem. By construction, our $T$-marginal should belong to the infinite-dimensional convex set $$\mathcal{M}=\left{\mathbb{P}: \mathbb{E}^{\mathbb{P}}\left[S{T}\right]=S_{0}, \quad \mathbb{E}^{\mathbb{P}}\left[\left(S_{T}-K_{i}\right)^{+}\right]=C\left(K_{i}\right) \equiv c_{i}, \quad i=1, \ldots n\right}$$
$\mathcal{M}$ is relatively compact from Prokhorov’s theorem (See Remark 1.4). For instance, we add the technical assumption that the elements in $\mathcal{M}$ should also be compactly supported in the interval $\left[0, S_{\max }\right]$ with $S_{\max }$ large in order to get that $\mathcal{M}$ is compact. This implies that from Krein-Milman’s theorem, this set can be reconstructed from its extremal points $\operatorname{Ext}(\mathcal{M})$ :
$$\mathcal{M}=\overline{\operatorname{Conv}(\operatorname{Ext}(\mathcal{M}))}$$
Furthermore, from Choquet’s theorem, one can show that all arbitrage-free prices $C(K)$ can be obtained by linearly combining extremal points. They are supported by a probability measure $\mu$ on $\operatorname{Ext}(\mathcal{M})$ (probability on probability space!) and for all $K$,
$$C(K)=\int_{\operatorname{Ext}(\mathcal{M})} \mathbb{E}^{\mathbb{P}}\left[\left(S_{T}-K\right)^{+}\right] d \mu(\mathbb{P})$$
Enumerating all the extremal points (and therefore elements in $\mathcal{M}$ ) is a difficult task. We follow a different route. A canonical point of $\mathcal{M}$ can be obtained by minimising a convex lower semi-continuous functional $\mathcal{F}$ :
$$I \equiv \inf {\mathbb{P} \in \mathcal{M}} \mathcal{F}(\mathbb{P})=\mathcal{F}\left(\mathbb{P}{c_{1}, \ldots, c_{n}}^{}\right), \quad \mathbb{P}{c{1}, \ldots, c_{n}}^{} \in \mathcal{M}$$

## 统计代写|离散时间鞅理论代写martingale代考|Spence–Mirrlees condition

The Spence-Mirrlees condition, i.e., $c_{12}>0$, required for the Fréchet-Hoeffding solution to hold, is very natural from a financial point of view. If we shift the payoff $c$ by some European payoffs $\Lambda_{1} \in L^{1}\left(\mathbb{P}^{1}\right), \Lambda_{2} \in L^{1}\left(\mathbb{P}^{2}\right)$ :
$$\bar{c}\left(s_{1}, s_{2}\right)=c\left(s_{1}, s_{2}\right)+\Lambda_{1}\left(s_{1}\right)+\Lambda_{2}\left(s_{2}\right)$$
then the Monge-Kantorovich bound for $\bar{c}$ should be
$$\mathrm{MK}{2}(\bar{c})=\mathrm{MK}{2}(c)+\mathbb{E}^{\mathbb{P}^{1}}\left[\Lambda_{1}\left(S_{1}\right)\right]+\mathbb{E}^{\mathbb{P}^{2}}\left[\Lambda_{2}\left(S_{2}\right)\right]$$
as the market price of $\Lambda_{i}\left(s_{i}\right)$ is fixed by $\mathbb{E}^{\mathbb{P}^{x}}\left[\Lambda_{i}\left(S_{i}\right)\right]$. The payoff $\bar{c}$ is precisely invariant under the Spence-Mirrlees condition : $\bar{c}{12}=c{12}$.

## 统计代写|离散时间鞅理论代写martingale代考|Mirror coupling: Co-monotone rearrangement map

Similarly, the upper bound under the condition $c_{12}<0$ is attained by the co-monotone rearrangement map $$T\left(s_{1}\right)=F_{2}^{-1} \circ\left(1-F_{1}\left(-s_{1}\right)\right)$$ This can be obtained by applying the parity transformation $\mathcal{P}\left(s_{1}, s_{2}\right)=$ $\left(-s_{1}, s_{2}\right)$. For each measure $\mathbb{P}$ matching the marginals $\mathbb{P}^{1}$ and $\mathbb{P}^{2}$, we can associate the measure $\mathcal{P}{} \mathbb{P}$ matching the marginals $\mathcal{P}{} \mathbb{P}^{1}$ and $\mathbb{P}^{2}$ with cumulative distributions $\bar{F}{1}\left(s{1}\right) \equiv 1-F_{1}\left(-s_{1}\right)$ and $F_{2}\left(s_{2}\right)$. We conclude as the Monge-Kantorovich bounds for $c$ and $\tilde{c}\left(s_{1}, s_{2}\right) \equiv c\left(-s_{1}, s_{2}\right)$ coincides as $\mathbb{E}^{\mathbb{P}}[c]=\mathbb{E}^{\mathcal{P}} \mathbb{P}[\bar{c}]$. Similarly, by replacing $c$ by $-c$, we obtain that the comonotone rearrangement map gives the lower bound under the condition $c_{12}>0$
Example $2.4$ Lower bound, $c\left(s_{1}, s_{2}\right)=\left(s_{1}-K_{1}\right)+1_{s_{2}>K_{2}}$
By applying Anti-Fréchet-Hoeffding solution, the lower bound is attained by
$$\mathrm{MK}{2}=\int{F_{2}\left(K_{2}\right)}^{\max \left(1-F_{1}\left(K_{1}\right), F_{2}\left(K_{2}\right)\right)}\left(F_{1}^{-1}(1-u)-K_{1}\right) d u$$
with
\begin{aligned} &\lambda_{2}(x)=\left(\bar{F}{1}^{-1} \circ F{2}\left(K_{2}\right)-K_{1}\right)^{+} 1_{x>K_{2}} \ &\lambda_{1}(x)=\left(x-K_{1}\right)^{+} 1_{F_{2}^{-1} \circ F_{1}(x)>K_{2}}-\left(\bar{F}{1}^{-1} \circ F{2}\left(K_{2}\right)-K_{1}\right)^{+} 1_{F_{2}^{-1} \circ F_{1}(x)>K_{2}} \end{aligned}

## 统计代写|离散时间鞅理论代写martingale代考|Axiomatic construction of marginals: Stieltjes moment problem

\mathcal{M}=\left{\mathbb{P}: \mathbb{E}^{\mathbb{P}}\left[S{T}\right]=S_{0}, \quad \mathbb{E }^{\mathbb{P}}\left[\left(S_{T}-K_{i}\right)^{+}\right]=C\left(K_{i}\right) \equiv c_{ i}, \quad i=1, \ldots n\right}\mathcal{M}=\left{\mathbb{P}: \mathbb{E}^{\mathbb{P}}\left[S{T}\right]=S_{0}, \quad \mathbb{E }^{\mathbb{P}}\left[\left(S_{T}-K_{i}\right)^{+}\right]=C\left(K_{i}\right) \equiv c_{ i}, \quad i=1, \ldots n\right}

C(ķ)=∫分机⁡(米)和磷[(小号吨−ķ)+]dμ(磷)

## 统计代写|离散时间鞅理论代写martingale代考|Spence–Mirrlees condition

Spence-Mirrlees 条件，即C12>0从财务的角度来看，Fréchet-Hoeffding 解决方案所需的 ，是非常自然的。如果我们改变收益C通过一些欧洲的回报Λ1∈大号1(磷1),Λ2∈大号1(磷2):

C¯(s1,s2)=C(s1,s2)+Λ1(s1)+Λ2(s2)

## 统计代写|离散时间鞅理论代写martingale代考|Mirror coupling: Co-monotone rearrangement map

$$\mathrm{MK} {2}=\int{F_{2}\left(K_{2}\right)}^{\max \left (1-F_{1}\left(K_{1}\right), F_{2}\left(K_{2}\right)\right)}\left(F_{1}^{-1}(1 -u)-K_{1}\right) 杜 在一世吨H λ2(X)=(F¯1−1∘F2(ķ2)−ķ1)+1X>ķ2 λ1(X)=(X−ķ1)+1F2−1∘F1(X)>ķ2−(F¯1−1∘F2(ķ2)−ķ1)+1F2−1∘F1(X)>ķ2$$

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