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

2022年10月14日

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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代写

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