# 金融代写|金融模型代写Modelling in finance代考|BFW3540

#### Doug I. Jones

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
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## 金融代写|金融模型代写Modelling in finance代考|Approximation of the arithmetic average

When computing the price of compounded rate with pseudo-discount factors implementation, if we use the simplifying assumption of deterministic spread (which is trivially satisfied when pricing OISs with OIS discounting), we do not need to compute every overnight rate and actually compound them. It is possible to use the direct formula of the ratio of start and end discount factors
$$P^D\left(0, t_n\right) \mathrm{E}^n\left[\left(\prod_{i=1}^n \frac{P^O\left(t_{i-1}, t_{i-1}\right)}{P^O\left(t_{i-1}, t_i\right)}\right)-1\right]=P^D\left(0, t_n\right)\left(\frac{P^O\left(0, t_0\right)}{P^O\left(0, t_n\right)}-1\right) .$$
This result is described in Section 2.7.
This type of direct computation result, which speeds-up the computation dramatically, is not available for arithmetic average. Nevertheless some approximations are available and we describe one here. It was initially introduced in Takada (2011).
Let $A_c$ be the amount paid for the compounded rates, that is,
$$A_c=\left(\prod_{i=1}^n\left(1+\delta_i I^O\left(t_{i-1}\right)\right)\right)-1 .$$
To the first order,
$$1+\delta_i I^O\left(t_{i-1}\right) \simeq \exp \left(\delta_i I^O\left(t_{i-1}\right)\right) .$$
As $\delta_i$ is small, the approximation is relatively good. Let $A_a$ denote the amount paid for the arithmetic average. We have
\begin{aligned} \Lambda_a &=\sum_{i=1}^n \delta_i I^O\left(t_{i-1}\right) \ & \simeq \ln \left(\prod_{i=1}^n\left(1+\delta_i I^O\left(t_{i-1}\right)\right)\right)=\ln \left(1+A_c\right) . \end{aligned}
As $A_c$ can be computed efficiently, if the approximation is good enough and the adjustment due to the non-linearity of the function is not too large, we have an efficient approximate way to compute the arithmetic average coupon.

In Table $6.2$ we analyse the quality of the approximation for several levels of rates. The curves we use are constant (on zero-coupon continuously compounded rates) with level $1 \%, 5 \%$ and $10 \%$. We have computed the difference between the exact arithmetic average and the one provided by the above approximation. The coupons are three months long with the dates computed using market conventions. We run the test with a samplc of 36 different starting dates to cover different month lengths and weekend effects. Depending on the level of rates, the error is from below $0.01$ basis points to $0.25$ basis points. The first error is certainly small enough; it is that of a $1 \%$ curve level, which is roughly the rate level in the main currencies at the time of writing. The error for a rate level of $10 \%$ is probably beyond the level of precision one would like for market making.

## 金融代写|金融模型代写Modelling in finance代考|Federal Funds swaps

Now that we have established that the approximation is good enough in the current rate environment, we have to analyse if it is useful. Is the computation time of the approximation lower enough to justify the use of the approximation? The answer to that question is summarised in Table 6.3. We have analysed the performance in three ways: the calculation itself for present value and curve sensitivity (bucketed delta) and the performance of the instrument construction and the present value. The last measure requires probably a little bit of explanation. For simple instruments, the time consuming part is often to construct the dates associated to the instrument following the different conventions, not to compute the relatively simple present value. This is certainly the case for overnight-related products, where a lot of dates are calculated. If the actual value computation time was dwarfed by the instrument construction time, the approximation exercise would be relatively meaningless. Note also that the curve sensitivity is computed by algorithmic differentiation and the results are (almost) independent of the number of points used to calibrate the curve and the number of node points impacting the instrument.

The conclusion of the table is that for the computation of present value and curve sensitivity, the approximation formula is significantly faster (between 20 and 40 times). This is not surprising as we use only two dates (start and end) to compute the result instead of all the daily points (around 60 dates for three months). If we add the instrument construction time, the impact is lesser but still relatively important. The time for construction and present value is divided by two. Which of those two numbers is relevant in practice: 2 or 20? A little bit in between! If you want only to compute the present value of a new instrument, the ratio 2 is the realistic one. But if you build curves with those instruments, the instrument description is computed once and the present value and curve sensitivities are computed numerous times. Suppose that to calibrate the curves you need for each instrument one instrument construction, ten present valucs, and two sensitivitics, the ratio in that case would be around eight. The small approximation error is probably a price you would accept to divide the curve construction time by eight.

At this stage, the analysis covers only the static approximation, that is the computation of the arithmetic average when the rates are known. A second level of approximation is required as the rates in the arithmetic average are not paid at the right time; they are paid at the end of the accrual period and not at the end of their reference period.

# 金融模型代写

## 金融代写|金融模型代写金融建模代考|算术平均的近似

$$P^D\left(0, t_n\right) \mathrm{E}^n\left[\left(\prod_{i=1}^n \frac{P^O\left(t_{i-1}, t_{i-1}\right)}{P^O\left(t_{i-1}, t_i\right)}\right)-1\right]=P^D\left(0, t_n\right)\left(\frac{P^O\left(0, t_0\right)}{P^O\left(0, t_n\right)}-1\right) .$$

$$A_c=\left(\prod_{i=1}^n\left(1+\delta_i I^O\left(t_{i-1}\right)\right)\right)-1 .$$

$$1+\delta_i I^O\left(t_{i-1}\right) \simeq \exp \left(\delta_i I^O\left(t_{i-1}\right)\right) .$$

\begin{aligned} \Lambda_a &=\sum_{i=1}^n \delta_i I^O\left(t_{i-1}\right) \ & \simeq \ln \left(\prod_{i=1}^n\left(1+\delta_i I^O\left(t_{i-1}\right)\right)\right)=\ln \left(1+A_c\right) . \end{aligned}

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MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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