### 统计代写|离散时间鞅理论代写martingale代考|OT versus MOT: A summary

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## 统计代写|离散时间鞅理论代写martingale代考|Martingale Brenier’s solution

The enormous development of OT in the last decades was initiated by Brenier’s celebrated theorem, briefly reviewed in Theorem 2.3. Hence a most natural question is to obtain similar results also for the martingale version of the transport problem. The literature on this topic includes [82, 83]. This seems a potentially very interesting problem for mathematicians working in OT to tackle this problem, particularly in $\mathbb{R}^{d}$.

We briefly state below MOT in $\mathbb{R}{+}^{d}$. We denote $\mathbb{P}^{1}$ and $\mathbb{P}^{2}$ the marginals of $S{1}$ and $S_{2}$ in $\mathbb{R}{+}^{d}$ and $S{1}^{i}$ the $i$-component of $S_{1}$. The knowledge of marginals $\mathbb{P}^{1}$ and $\mathbb{P}^{2}$ is not very common in finance as the (known) marginals are usually one-dimensional (e.g. Vanillas), see however our discussion in Section 2.1.3. A notable exception arises in fixed income and foreign exchange markets (see Example 2.1) where Vanillas on spread swap rates, i.e., $\left(S_{2}-K S_{1}\right)^{+}$, are quoted on the market.
$$\widetilde{\mathrm{MK}}{2}=\inf {\lambda_{1} \in \mathrm{L}^{1}\left(\mathbb{P}^{1}\right), \lambda_{2} \in \mathrm{L}^{1}\left(\mathbb{P}^{2}\right),\left(H^{i}(-)\right){1 \leq i \leq d}} \mathbb{E}^{\mathbb{P}^{1}}\left[\lambda{1}\left(S_{1}\right)\right]+\mathbb{E}^{\mathbb{P}^{2}}\left[\lambda_{2}\left(S_{2}\right)\right]$$
such that $\lambda_{1}\left(s_{1}\right)+\lambda_{2}\left(s_{2}\right)+\sum_{i=1}^{d} H^{i}\left(s_{1}\right)\left(s_{2}^{i}-s_{1}^{i}\right) \geq c\left(s_{1}, s_{2}\right), \quad \forall\left(s_{1}, s_{2}\right) \in$ $\left(\mathbb{R}{+}^{d}\right)^{2}$. Taking for granted that the primal is attained (the dual is attained by weak compactness), the (strong) duality result implies as before that $$\lambda{1}\left(s_{1}\right)+\lambda_{2}\left(s_{2}\right)+\sum_{i=1}^{d} H^{i}\left(s_{1}\right)\left(s_{2}^{i}-s_{1}^{i}\right)=c\left(s_{1}, s_{2}\right), \quad \mathbb{P}^{}-\text { a.s. }$$ We have $d+2$ unknown functions $\left(\lambda_{1}, \lambda_{2},\left(H^{i}(\cdot)\right){1 \leq i \leq d}\right.$ ) (defined on (a subset of) $\mathbb{R}{+}^{d}$ ) and it is tempting to guess that the optimal martingale measure $\mathbb{P}^{}$ is localized on some maps $\left(T^{\alpha}\right){\alpha=1, \ldots, N}$. For each map – denoted schematically by $T$ with components $\left(T{1}, \ldots, T_{d}\right)$ – we should have: $\forall s_{1} \in \mathbb{R}^{d}$,
\begin{aligned} &\lambda_{1}\left(s_{1}\right)+\lambda_{2}\left(T\left(s_{1}\right)\right)+\sum_{i=1}^{d} H^{i}\left(s_{1}\right)\left(T_{i}\left(s_{1}\right)-s_{1}^{i}\right)=c\left(s_{1}, T\left(s_{1}\right)\right) \ &\partial_{s_{2}^{i}} \lambda_{2}\left(T\left(s_{1}\right)\right)+H^{i}\left(s_{1}\right)=\partial_{s_{2}^{i}} c\left(s_{1}, T\left(s_{1}\right)\right), \quad \forall i=1, \ldots, d \end{aligned}
On the dual side, we should have :
$$\mathbb{P}^{*}\left(d s_{1}, s_{2}\right)=\sum_{\alpha=1}^{N} q_{\alpha}\left(s_{1}\right) \delta_{T a\left(s_{1}\right)}\left(d s_{2}\right) \mathbb{P}^{1}\left(d s_{1}\right)$$
where the functions $\left(q_{\alpha}\right){\alpha=1, \ldots, N}$ are constrained by the algebraic equations: $$\sum{\alpha=1}^{N} q_{\alpha}\left(s_{1}\right)=1, \quad \sum_{\alpha=1}^{N} q_{\alpha}\left(s_{1}\right)\left(T^{\alpha}\left(s_{1}\right)-s_{1}\right)=0$$

## 统计代写|离散时间鞅理论代写martingale代考|Mirror coupling: The right-monotone martingale transport plan

Suppose that $c_{s_{1} s_{2} s_{2}}<0$. Then, the upper bound $\widetilde{\mathrm{MK}}{2}$ is attained by the right-monotone martingale transport map $$\begin{array}{r} \mathbb{P}{*}\left(d s_{1}, d s_{2}\right)=\mathbb{P}^{1}\left(d s_{1}\right)\left(q\left(s_{1}\right) \delta_{\bar{T}{u}\left(s{1}\right)}\left(d s_{2}\right)+\left(1-q\left(s_{1}\right)\right) \delta_{\bar{T}{d}\left(s{1}\right)}\left(d s_{2}\right)\right) \ q(x)=\frac{x-\bar{T}{d}(x)}{\bar{T}{u}(x)-\bar{T}{d}(x)} \end{array}$$ where $\left(\bar{T}{d}, \bar{T}{u}\right)$ is defined as in $(2.31,2.32)$ with the pair of probability measures $\left(\overline{\mathrm{P}}^{1}, \overline{\mathrm{P}}^{2}\right):$ $$\bar{F}^{1}\left(s{1}\right) \equiv 1-F^{1}\left(-s_{1}\right), \text { and } \bar{F}^{2}\left(s_{2}\right) \equiv 1-F^{2}\left(-s_{2}\right) .$$ To see this, we rewrite the OT problem equivalently with modified inputs: \begin{aligned} \bar{c}\left(s_{1}, s_{2}\right) \equiv c\left(-s_{1},-s_{2}\right), & \overline{\mathbb{P}}^{1}\left(\left(-\infty, s_{1}\right]\right) \equiv \mathbb{P}^{1}\left(\left[-s_{1}, \infty\right)\right) \ \overline{\mathbb{P}}^{2}\left(\left(-\infty, s_{2}\right]\right) \equiv \mathbb{P}^{2}\left(\left[-s_{2}, \infty\right)\right) \end{aligned} so that $\bar{c}{s{1} s_{2} s_{2}}>0$, as required in Theorem 2.8. Note that the martingale constraint is preserved by the map $\left(s_{1}, s_{2}\right) \mapsto\left(-s_{1},-s_{2}\right)$ (and not by our parity transformation $\left(s_{1}, s_{2}\right) \mapsto\left(s_{1},-s_{2}\right)$ in OT $)$.

Suppose that $c_{s_{1} s_{2} s_{2}}>0$. Then, the lower bound problem is explicitly solved by the right-monotone martingale transport plan. Indeed, it follows from the first part of the present remark that:
\begin{aligned} \inf {\mathbb{P} \in \mathcal{M}\left(\mathbb{P}^{1}, \mathbb{P}^{2}\right)} \mathbb{E}^{\mathbb{P}}\left[c\left(S{1}, S_{2}\right)\right] &=-\sup {\mathbb{P} \in \mathcal{M}\left(\mathbb{P}^{1}, \mathbb{P}^{2}\right)} \mathbb{E}^{\mathbb{P}}\left[-c\left(S{1}, S_{2}\right)\right] \ &=-\sup {\mathbb{P} \in \mathcal{M}\left(\mathbb{P}^{1}, \mathbb{R}^{2}\right)} \mathbb{E}^{\mathbb{P}}\left[-\bar{c}\left(-S{1},-S_{2}\right)\right] \ &=-\sup {\mathbb{P} \in \mathcal{M}\left(\mathbb{P}^{1}, \mathbb{R}^{2}\right)} \mathbb{E}^{\mathbb{P}}\left[-\bar{c}\left(S{1}, S_{2}\right)\right] \ &=\mathbb{E}^{\mathbb{P}}\left[c\left(S_{1}, S_{2}\right)\right] \end{aligned}

## 统计代写|离散时间鞅理论代写martingale代考|Change of num´eraire

We define the involution $\mathcal{S}[34]$ (i.e., $\mathcal{S}^{2}=\mathrm{Id}$ ) on a payoff function $c$ by
$$(\mathcal{S c})\left(s_{1}, s_{2}\right) \equiv s_{2} c\left(\frac{1}{s_{1}}, \frac{1}{s_{2}}\right)$$
We have
\begin{aligned} \sup {\mathbb{P} \in \mathcal{M}\left(\mathbb{P}^{1}, \mathbb{P}^{2}\right)} \mathbb{E}^{\mathbb{P}}\left[(\mathcal{S} c)\left(S{1}, S_{2}\right)\right]=& \sup {\mathbb{P} \in \mathcal{M}\left(\mathbb{P}^{1}, \mathbb{P}^{2}\right)} \mathbb{E}^{\mathbb{P}}\left[S{2} c\left(\frac{1}{S_{1}}, \frac{1}{S_{2}}\right)\right] \ =S_{0} & \sup {\left.\mathbb{Q} \in \mathcal{M}\left(\mathcal{S}^{1}\right), \mathcal{P}\left(\mathbb{(}^{2}\right)\right)} \mathbb{E}^{\mathbb{Q}}\left[c\left(\bar{S}{t_{1}}, \bar{S}{t{2}}\right)\right] \end{aligned}
where $\mathcal{S}\left(\mathbb{P}^{i}\right), i=1,2$ has a density $\left(\mathcal{S} f^{i}\right)(s)=\frac{1}{S_{0} s^{3}} f^{i}\left(\frac{1}{s}\right)$ where $f^{i}$ the density of $\mathbb{P}^{i}$. We have used that by working in the numéraire associated to the discrete martingale $S_{t}$ :
$$\mathbb{E}^{\mathrm{P}}\left[S_{2} c\left(\frac{1}{S_{1}}, \frac{1}{S_{2}}\right)\right]=S_{0} \mathbb{E}^{\mathbb{Q}}\left[c\left(\frac{1}{S_{1}}, \frac{1}{S_{2}}\right)\right]$$
with $\left.\frac{d \mathbb{Q}}{}\right|{\mathcal{F}{t_{i}}}=\frac{S_{i}}{S_{0}}$. Under $\mathbb{Q}, \frac{1}{S_{i}}$ is a discrete martingale: $\mathbb{E}^{\mathbb{Q}}\left[\frac{1}{S_{2}} \mid \frac{1}{S_{1}}\right]=\frac{1}{S_{1}}$. This involution $\mathcal{S}$ satisfies
$$(\mathcal{S c}){122}\left(s{1}, s_{2}\right)=-\frac{1}{s_{1}^{2} s_{2}^{3}} c_{122}\left(\frac{1}{s_{1}}, \frac{1}{s_{2}}\right)$$
and exchanges therefore the left and right-monotone martingale transport plan where the marginals have support in $\mathbb{R}_{+}$.

## 统计代写|离散时间鞅理论代写martingale代考|Martingale Brenier’s solution

MOT 读取

λ1(s1)+λ2(s2)+∑一世=1dH一世(s1)(s2一世−s1一世)=C(s1,s2),磷− 作为 我们有d+2未知功能(λ1,λ2,(H一世(⋅))1≤一世≤d) (定义在 (的一个子集)R+d) 并且很容易猜测最优鞅测度磷在某些地图上进行了本地化(吨一个)一个=1,…,ñ. 对于每张地图——示意性地表示为吨带组件(吨1,…,吨d)- 我们本应该：∀s1∈Rd,

λ1(s1)+λ2(吨(s1))+∑一世=1dH一世(s1)(吨一世(s1)−s1一世)=C(s1,吨(s1)) ∂s2一世λ2(吨(s1))+H一世(s1)=∂s2一世C(s1,吨(s1)),∀一世=1,…,d

∑一个=1ñq一个(s1)=1,∑一个=1ñq一个(s1)(吨一个(s1)−s1)=0

## 统计代写|离散时间鞅理论代写martingale代考|Mirror coupling: The right-monotone martingale transport plan

F¯1(s1)≡1−F1(−s1), 和 F¯2(s2)≡1−F2(−s2).为了看到这一点，我们用修改后的输入等效地重写了 OT 问题：

C¯(s1,s2)≡C(−s1,−s2),磷¯1((−∞,s1])≡磷1([−s1,∞)) 磷¯2((−∞,s2])≡磷2([−s2,∞))以便C¯s1s2s2>0，如定理 2.8 所要求的。请注意，鞅约束由地图保留(s1,s2)↦(−s1,−s2)（而不是通过我们的平价变换(s1,s2)↦(s1,−s2)我不).

## 统计代写|离散时间鞅理论代写martingale代考|Change of num´eraire

(小号C)(s1,s2)≡s2C(1s1,1s2)

(小号C)122(s1,s2)=−1s12s23C122(1s1,1s2)

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