# 随机微积分作业代写stochastic calculus代考| INTEREST RATE DERIVATIVES

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• 随机偏微分方程
• 随机控制
• Ito积分
• black-Scholes-Merton option pricing formula
• Fokker–Planck equation
• 布朗运动 Brownian motion

## 微积分作业代写calclulus代考|Swap rate models with prescribed volatilities

5. Fix a tenor structure $0=$ $T_{0}<T_{1}<\ldots<T_{n}$ and set $\delta_{i}=T_{i+1}-T_{i}, i=0,1, \ldots, n-1$. Let $\nu_{i}(t), i=$ $0,1, \ldots, n-1$, be any bounded, progressively measurable processes and let us turn to the construction of a deflatable market $B=\left(B_{0}, B_{1}, \ldots, B_{n}\right)$ such that the associated swap rate processes $S_{i}$ are Ito processes with proportional, $R^{d}$-valued, volatilities $\nu_{i}(t)$ and fit a given initial term structure $\left(S_{0}(0), S_{1}(0), \ldots, S_{n-1}(0)\right)$. Since the process $B_{i}$ should be thought of as the zero coupon bond $B\left(t, T_{i}\right)$, we also want to satisfy the zero coupon bond constraint $B_{i}\left(T_{i}\right)=1$.
As in the Libor case the idea is to solve the swap rate dynamics 5.h.eq. (8) recursively starting with $i=n-1$ and then to derive the zero coupon bonds $B_{i}$ from the forward swap rates $S_{i}$. In detail, let $W_{t}^{n}$ be an $R^{d}$-valued Brownian motion on $\left(\Omega, \mathcal{F},\left(\mathcal{F}{t}\right){t \in\left[0, T^{*}\right]}, Q\right)$ (the measure $Q$ plays the role of the forward martingale measure $P_{n}$ of $\left.5 . \mathrm{h}\right)$ and define $S_{n-1}$ as the solution of $d S_{n-1}=S_{n-1} \nu_{n-1} \cdot d W_{t}^{n}$, that is,
$$S_{n-1}(t)=S_{n-1}(0) \mathcal{E}{t}\left(\nu{n-1} \cdot W^{n}\right),$$
and let, for $0 \leq i \leq n-2, S_{i}$ be the solution of
\begin{aligned} d S_{i} &=-\sum_{j=i+1}^{n-1} \frac{\delta_{j-1} s_{i j} S_{i} S_{j}}{\left(1+\delta_{j-1} S_{j}\right) s_{i}} \nu_{i} \cdot \nu_{j} d t+S_{i}(t) \nu_{i}(t) \cdot W_{t}^{n} \ &=S_{i} \mu_{i} d t+S_{i} \nu_{i} \cdot d W_{t}^{n}, \end{aligned}
where
$$\begin{gathered} \mu_{i}=-\sum_{j=i+1}^{n-1} \frac{\delta_{j-1} s_{i j} S_{j}}{\left(1+\delta_{j-1} S_{j}\right) s_{i}} \nu_{i} \cdot \nu_{j} \quad \text { and } \ s_{i j}=\sum_{k=j}^{n-1} \delta_{k} \prod_{u=i+1}^{k}\left(1+\delta_{u-1} S_{u}\right), \quad s_{i}=s_{i i}, \quad 1 \leq i \leq j \leq n-1 . \end{gathered}$$
As in 5.h this implies the backward recursion
\begin{aligned} &s_{n-1}=\delta_{n-1} \quad \text { and } \ &s_{i-1}=\delta_{i-1}+\left(1+\delta_{i-1} S_{i}\right) s_{i}, \quad 1 \leq i \leq n-1 . \end{aligned}

Explicitly $S_{i}$ is given by $\quad S_{i}(t)=S_{i}(0) \exp \left(\int_{0}^{t} \mu_{i}(r) d r\right) \mathcal{E}{t}\left(\nu{i} \cdot W^{n}\right)$.
This defines the processes $S_{i}, s_{i j}, s_{i}$, for all $0 \leq i \leq j \leq n-1$ and we now turn to the definition of the zero coupon bonds $B_{i}$. Let $B_{n}$ be any continuous $Q$-semimartingale satisfying
$$B_{n}\left(T_{i}\right)=\frac{1}{1+S_{i}\left(T_{i}\right) s_{i}\left(T_{i}\right)}$$
and set $B_{i}=\left(1+S_{i} s_{i}\right) B_{n}$ in accordance with equation (2) of $5 . h$. Then (3) ensures that $B_{i}$ satisfies the zero coupon bond constraint $B_{i}\left(T_{i}\right)=1$. An easy backward induction using (2) shows that $S_{i}=\left(B_{i}-B_{n}\right) / B_{i, n}$ ensuring that $S_{i}$ is the intended forward swap rate associated with the zero coupon bonds $B_{i}$. To see that the market $B=\left(B_{0}, B_{1}, \ldots, B_{n}\right)$ is arbitrage free, it will suffice to show that the ratios
$$B_{i} / B_{n}=1+S_{i} s_{i}$$
are all $Q$-martingales. Indeed, from the dynamics (1),
$$s_{i} d S_{i}=-\sum_{j=i+1}^{n-1} \frac{\delta_{j-1} s_{i j} S_{i} S_{j}}{1+\delta_{j-1} S_{j}} \nu_{i} \cdot \nu_{j} d t+s_{i} S_{i} \nu_{i} \cdot d W_{t}^{n}$$
As in $5 . \mathrm{h}$ (equation (5)) the definition of $s_{i}$ implies
$$d\left\langle S_{i}, s_{i}\right\rangle=\sum_{j=i+1}^{n-1} \frac{\delta_{j-1} s_{i j}}{1+\delta_{j-1} S_{j}} d\left\langle S_{i}, S_{j}\right\rangle=\sum_{j=i+1}^{n-1} \frac{\delta_{j-1} s_{i j} S_{i} S_{j}}{1+\delta_{j-1} S_{j}} \nu_{i} \cdot \nu_{j} d t$$
and so, by addition, $\quad s_{i} d S_{i}+d\left\langle S_{i}, s_{i}\right\rangle=S_{i} s_{i} \nu_{i} \cdot d W_{t}^{n}$.
Now we claim that
$$d s_{i}=\gamma_{i} \cdot d W_{t}^{n}$$
with processes $\gamma_{i} \in L(W)$ satisfying the recursion
$$\gamma_{n-1}=0 \quad \text { and } \quad \gamma_{i-1}=\left(1+\delta_{i-1} S_{i}\right) \gamma_{i}+\delta_{i-1} S_{i} s_{i} \nu_{i} .$$
Indeed, proceeding by backward induction, (5), (6) are true for $i=n-1$ since $s_{n-1}=\delta_{n-1}$. Assume now that $d s_{i}=\gamma_{i} \cdot d W_{t}^{n}$ where $\gamma_{i} \in L\left(W^{n}\right)$. Then, using the recursion (2) and the stochastic product rule,
\begin{aligned} d s_{i-1} &=\left(1+\delta_{i-1} S_{i}\right) d s_{i}+\delta_{i-1} s_{i} d S_{i}+\delta_{i-1} d\left\langle S_{i}, s_{i}\right\rangle \ &=\left(1+\delta_{i-1} S_{i}\right) \gamma_{i} \cdot d W_{t}^{n}+\delta_{i-1}\left[s_{i} d S_{i}+d\left\langle S_{i}, s_{i}\right\rangle\right] \ &=\left(1+\delta_{i-1} S_{i}\right) \gamma_{i} \cdot d W_{t}^{n}+\delta_{i-1} S_{i} s_{i} \nu_{i} \cdot d W_{t}^{n} \ &=\gamma_{i-1} \cdot d W_{t}^{n} \end{aligned}
where $\gamma_{i-1}=\left(1+\delta_{i-1} S_{i}\right) \gamma_{i}+\delta_{i-1} S_{i} s_{i} \nu_{i} \in L\left(W^{n}\right)$, as desired. Let us now write this relation as
$$d s_{i}=s_{i} \sigma_{i} \cdot d W_{t}^{n}, \quad \text { where } \quad \sigma_{i}=\gamma_{i} / s_{i}$$

## 微积分作业代写calclulus代考|Valuation of swaptions in the log-Gaussian swap rate model

5.Continuing with the terminology of $5 . h$ and assuming that the volatilities $\nu_{i}(t)$ are nonstochastic, let $t \leq \hat{T} \leq T=T_{0}$. The payoff $F S_{\hat{T}}(\kappa)^{+}$of the forward payer swaption $P S(\hat{T}, \kappa)$ at time $\hat{T}$ can be written as
$$F S_{\hat{T}}(\kappa)^{+}=\sum_{j=0}^{n-1} \delta_{j} B\left(\hat{T}, T_{j+1}\right)(\kappa(\hat{T}, T, n)-\kappa)^{+}=B_{0, n}(\hat{T})\left(S_{0}(\hat{T})-\kappa\right)^{+}$$
Using the symmetric numeraire change formula 5.d.1.(d) and $B(\hat{T}, \hat{T})=1$ the arbitrage price $P S_{t}(\hat{T}, \kappa)$ of this swaption at time $t \leq \hat{T}$ is given by
\begin{aligned} P S_{t}(\hat{T}, \kappa) &=B(t, \hat{T}) E_{P_{\hat{T}}}\left[B_{0, n}(\hat{T})\left(S_{0}(\hat{T})-\kappa\right)^{+} \mid \mathcal{F}{t}\right] \ &=B{0, n}(t) E_{P_{0, n}}\left[\left(S_{0}(\hat{T})-\kappa\right)^{+} \mid \mathcal{F}{t}\right] \end{aligned} Set $Z(t)=\log \left(S{0}(t)\right)$. From the dynamics $d S_{0}(t)=S_{0}(t) \nu_{0}(t) \cdot d W_{t}^{0, n}$ it follows that
$$d Z(t)=-\frac{1}{2}\left|\nu_{0}(t)\right|^{2} d t+\nu_{0}(t) \cdot d W_{t}^{0, n}$$
Thus, using III.6.c.3,
$$E_{P_{\mathrm{D}^{, n}}}\left[\left(S_{0}(\hat{T})-\kappa\right)^{+} \mid \mathcal{F}{t}\right]=S{0}(t) N\left(d_{1}\right)-\kappa N\left(d_{2}\right)$$
where $\quad d_{1,2}=\frac{\log \left(S_{0}(t) / \kappa\right) \pm \frac{1}{2} \Sigma^{2}(t, \hat{T})}{\Sigma(t, \hat{T})}$ and $\Sigma(t, \hat{T})=\left(\int_{t}^{\hat{T}}\left|\nu_{0}(t)\right|^{2} d t\right)^{1 / 2}$
Consequently $P S_{t}(\hat{T}, \kappa)=B_{0, n}(t)\left(S_{0}(t) N\left(d_{1}\right)-\kappa N\left(d_{2}\right)\right)$ with $d_{1,2}$ as above. Recalling that
$$B_{0, n}(t)=\sum_{i=1}^{n} \delta_{j-1} B_{j}(t)=\sum_{i=1}^{n} \delta_{j-1} B\left(t, T_{j}\right) \quad \text { and } \quad S_{0}(t)=\kappa(t, T, n)$$
we can summarize these findings as follows:

## 微积分作业代写calclulus代考|Swap rate models with prescribed volatilities

5.修复男高音结构0= 吨0<吨1<…<吨n并设置d一世=吨一世+1−吨一世,一世=0,1,…,n−1. 让ν一世(吨),一世= 0,1,…,n−1，是任何有界的、可逐步衡量的过程，让我们转向建设一个可通货紧缩的市场乙=(乙0,乙1,…,乙n)这样相关的掉期利率过程小号一世是具有比例的 Ito 过程，Rd-估值，波动率ν一世(吨)并拟合给定的初始期限结构(小号0(0),小号1(0),…,小号n−1(0)). 由于过程乙一世应该被认为是零息债券乙(吨,吨一世)，我们也想满足零息债券约束乙一世(吨一世)=1.

S_{n-1}(t)=S_{n-1}(0) \mathcal{E} {t}\left(\nu {n-1} \cdot W^{n}\right),

d小号一世=−∑j=一世+1n−1dj−1s一世j小号一世小号j(1+dj−1小号j)s一世ν一世⋅νjd吨+小号一世(吨)ν一世(吨)⋅在吨n =小号一世μ一世d吨+小号一世ν一世⋅d在吨n,

μ一世=−∑j=一世+1n−1dj−1s一世j小号j(1+dj−1小号j)s一世ν一世⋅νj 和  s一世j=∑到=jn−1d到∏你=一世+1到(1+d你−1小号你),s一世=s一世一世,1≤一世≤j≤n−1.

sn−1=dn−1 和  s一世−1=d一世−1+(1+d一世−1小号一世)s一世,1≤一世≤n−1.
$$明确地小号一世由 \quad S_{i}(t)=S_{i}(0) \exp \left(\int_{0}^{t} \mu_{i}(r) dr\right) \mathcal{给出E} {t}\left(\nu {i} \cdot W^{n}\right).吨H一世sd和F一世n和s吨H和pr○C和ss和sS_{i}, s_{ij}, s_{i},F○r一种一世一世0 \ leq i \ leq j \ leq n-1一种nd在和n○在吨你rn吨○吨H和d和F一世n一世吨一世○n○F吨H和和和r○C○你p○nb○nds双}.一世和吨B_{n}b和一种n和C○n吨一世n你○你s问−s和米一世米一种r吨一世nG一种一世和s一种吨一世sF和一世nG乙n(吨一世)=11+小号一世(吨一世)s一世(吨一世)一种nds和吨B_{i}=\left(1+S_{i} s_{i}\right) B_{n}一世n一种CC○rd一种nC和在一世吨H和q你一种吨一世○n(2)○F5. H.吨H和n(3)和ns你r和s吨H一种吨双}s一种吨一世sF一世和s吨H和和和r○C○你p○nb○ndC○ns吨r一种一世n吨B_{i}\left(T_{i}\right)=1.一种n和一种s和b一种C到在一种rd一世nd你C吨一世○n你s一世nG(2)sH○在s吨H一种吨S_{i}=\left(B_{i}-B_{n}\right) / B_{i, n}和ns你r一世nG吨H一种吨S_{i}一世s吨H和一世n吨和nd和dF○r在一种rds在一种pr一种吨和一种ss○C一世一种吨和d在一世吨H吨H和和和r○C○你p○nb○nds双}.吨○s和和吨H一种吨吨H和米一种r到和吨B=\left(B_{0}, B_{1}, \ldots, B_{n}\right)一世s一种rb一世吨r一种G和Fr和和,一世吨在一世一世一世s你FF一世C和吨○sH○在吨H一种吨吨H和r一种吨一世○s乙一世/乙n=1+小号一世s一世一种r和一种一世一世问−米一种r吨一世nG一种一世和s.一世nd和和d,Fr○米吨H和d和n一种米一世Cs(1),s一世d小号一世=−∑j=一世+1n−1dj−1s一世j小号一世小号j1+dj−1小号jν一世⋅νjd吨+s一世小号一世ν一世⋅d在吨n一种s一世n5. \mathrm{h}(和q你一种吨一世○n(5))吨H和d和F一世n一世吨一世○n○Fs_{i}一世米p一世一世和sd⟨小号一世,s一世⟩=∑j=一世+1n−1dj−1s一世j1+dj−1小号jd⟨小号一世,小号j⟩=∑j=一世+1n−1dj−1s一世j小号一世小号j1+dj−1小号jν一世⋅νjd吨一种nds○,b和一种dd一世吨一世○n,\quad s_{i} d S_{i}+d\left\langle S_{i}, s_{i}\right\rangle=S_{i} s_{i} \nu_{i} \cdot d W_{t }^{n}.ñ○在在和C一世一种一世米吨H一种吨ds一世=C一世⋅d在吨n在一世吨Hpr○C和ss和s\gamma_{i} \in L(W)s一种吨一世sF和一世nG吨H和r和C你rs一世○nCn−1=0 和 C一世−1=(1+d一世−1小号一世)C一世+d一世−1小号一世s一世ν一世.一世nd和和d,pr○C和和d一世nGb和b一种C到在一种rd一世nd你C吨一世○n,(5),(6)一种r和吨r你和F○r我=n-1s一世nC和s_ {n-1} = \ delta_ {n-1}.一种ss你米和n○在吨H一种吨d s_{i}=\gamma_{i} \cdot d W_{t}^{n}在H和r和\gamma_{i} \in L\left(W^{n}\right).吨H和n,你s一世nG吨H和r和C你rs一世○n(2)一种nd吨H和s吨○CH一种s吨一世Cpr○d你C吨r你一世和,ds一世−1=(1+d一世−1小号一世)ds一世+d一世−1s一世d小号一世+d一世−1d⟨小号一世,s一世⟩ =(1+d一世−1小号一世)C一世⋅d在吨n+d一世−1[s一世d小号一世+d⟨小号一世,s一世⟩] =(1+d一世−1小号一世)C一世⋅d在吨n+d一世−1小号一世s一世ν一世⋅d在吨n =C一世−1⋅d在吨n在H和r和\gamma_{i-1}=\left(1+\delta_{i-1} S_{i}\right) \gamma_{i}+\delta_{i-1} S_{i} s_{i} \nu_ {i} \in L\left(W^{n}\right),一种sd和s一世r和d.一世和吨你sn○在在r一世吨和吨H一世sr和一世一种吨一世○n一种sds一世=s一世σ一世⋅d在吨n, 在哪里 σ一世=C一世/s一世 ## 微积分作业代写calclulus代考|Valuation of swaptions in the log-Gaussian swap rate model 5.继续术语5.H并假设波动率ν一世(吨)是非随机的，让吨≤吨^≤吨=吨0. 回报F小号吨^(ķ)+远期付款人掉期磷小号(吨^,ķ)有时吨^可以写成 F小号吨^(ķ)+=∑j=0n−1dj乙(吨^,吨j+1)(ķ(吨^,吨,n)−ķ)+=乙0,n(吨^)(小号0(吨^)−ķ)+ 使用对称计价变化公式 5.d.1.(d) 和乙(吨^,吨^)=1套利价格磷小号吨(吨^,ķ)这次交换的时间吨≤吨^由$$
\begin{aligned}
P S_{t}(\hat{T}, \kappa) &=B(t, \hat{T}) E_{P_{\hat{T}}}\left 给出[B_{0, n}(\hat{T})\left(S_{0}(\hat{T})-\kappa\right)^{+} \mid \mathcal{F} {t}\right ] \ &=B {0, n}(t) E_{P_{0, n}}\left[\left(S_{0}(\hat{T})-\kappa\right)^{+} \ mid \mathcal{F} {t}\right] \end{aligned} $$设置 Z(t)=\log \left(S {0}(t)\right).Fr○米吨H和d和n一种米一世Csd S_{0}(t)=S_{0}(t) \nu_{0}(t) \cdot d W_{t}^{0, n}一世吨F○一世一世○在s吨H一种吨d和(吨)=−12|ν0(吨)|2d吨+ν0(吨)⋅d在吨0,n吨H你s,你s一世nG一世一世一世.6.C.3, E_{P_{\mathrm{D}^{, n}}}\left[\left(S_{0}(\hat{T})-\kappa\right)^{+} \mid \mathcal{F } {t}\right]=S {0}(t) N\left(d_{1}\right)-\kappa N\left(d_{2}\right) 在H和r和d1,2=日志⁡(小号0(吨)/ķ)±12Σ2(吨,吨^)Σ(吨,吨^)一种ndΣ(吨,吨^)=(∫吨吨^|ν0(吨)|2d吨)1/2C○ns和q你和n吨一世和磷小号吨(吨^,ķ)=乙0,n(吨)(小号0(吨)ñ(d1)−ķñ(d2))在一世吨Hd1,2一种s一种b○v和.R和C一种一世一世一世nG吨H一种吨 B_{0, n}(t)=\sum_{i=1}^{n} \delta_{j-1} B_{j}(t)=\sum_{i=1}^{n} \delta_{ j-1} B\left(t, T_{j}\right) \quad \text { and } \quad S_{0}(t)=\kappa(t, T, n)$$