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4

There is no possibility to convert any two of your mentioned variables into the remaining one. For the compound and arithmetic return you can derive an inequality, but that's the best you can do. The definitions for your statements are: $$r_{\mathrm{compound}}= \prod_{t=0}^{n}{\left( 1+r_t \right)}$$ $$r_{\mathrm{arithmetic}}=\frac{1}{n} \sum_{t=0}^n{r_t}... 4 They are different because there are different conventions in different places. Whilst it would make the maths more consistent to use discount (or accrual) factors to describe interest, the foremost concern historically has been absolute clarity on amounts of money, so having a simple way to calculate the coupon payment on a given bond. So a UK Gilt paying 5%... 3 I grappled with different compounding conventions for a while when I first joined the world of finance. After some time, I came to the conclusion that the best analogy is that of different units of speed in different countries: i.e. in the UK, the unit of speed is miles, whilst in continental Europe, it is kilometers. A car travelling at 100mph is travelling ... 3 Let us start from your last equation, and focus specifically on the expectation. Assuming that the end date of each period is the start period of the next, the idea is to simplify it using conditional expectations. Since t < t_{n-2}, we can write using the tower property of conditional expectations:$$ \begin{aligned} \Bbb{E}_{t}^{Q^{t_n}} \left[\prod_{...

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Continuous-time formulation is much easier for some basic asset pricing theories. In continuous-time you will have to deal with integrals rather than sums which makes your life much easier. And for those, you will need continuous discounting. Here's an excerpt from John Cochrane's Asset Pricing: The choice of discrete vs. continuous time is one of ...

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You add 1 to every monthly return of a given quarter, take the product of those returns, and then subtract 1. In R (without any package): Suppose r are the monthly returns, and dt are the timestamps. r <- rep(0.01, 12) dt <- seq(from = as.Date("2020-1-1"), to = as.Date("2020-12-1"), by = "1 month") tapply(r, paste(as....

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Hint: Let $$z = \mathrm{e}^{-y}$$ That way you get a quadratic equation in $z$ (note that $z$ is positive) and then you can get back to $y$ using: $$y = -\ln (z)$$

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Look this is just a geometric sum: Assume interest is paid monthly at rate $r = 0.08/12$ (you can use the exact monthly equivalent if you want) and let $x_n =$total after $n$ months (including that month's interest and deposit). So $x_0= 100$ and $x_{n+1} = x_n(1+r) + d$, where $d = 5$ is your deposit amount (added at the end of the month). Applying the ...

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Often the choice of discount curves are dependent upon the context of the assets/liabilities being valued. Banks, for example who have active and to-the-second accurate valuations of these due to active trading desks reporting daily mark-to-market should have more accurate curves, so that all of their operations are consistently valued and do not lead to ...

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By no arbitrage, market participants need to agree on the values of the discount factor, even if they are using different conventions (day count, compounding period) to convert the discount factor into a rate. For example, consider two discount factors computed using continuous compounding, where one is computed using the 30/360 day count (year fraction $t_{... 2 Convention for most currencies is flat compounding. This includes the most liquid G4 basis swaps: EUR 3m vs 6m, 3m vs 12m, 1m vs 3m USD 1m vs 3m, 3m vs 6m GBP 3m vs 6m, 1m vs 3m JPY 3m vs 6m, 1m vs 6m, 1m vs 3m Typically the spread is quoted in terms of the leg with the shorter tenor. Two notable exceptions are EONIA v 3m and the (relatively new) SOFR ... 2 Depends on which OIS you are referring to. For EUR OIS Swaps, the EONIA Swap rate is calculated via the usual compounding formula (notice that in the example below, the rate$r_i$is updated every night): Example is shown here: For USD OIS Swaps, the link to Investopedia that you shared is correct: it is pretty much the same formula as for the EUR swap ... 2 In continuous compounding, a nominal (or an index value) in time$t$is given by formula $$N_t = N_0\mathrm{e}^{rt},$$ where$r$is return (or interest) rate per annum. Based on the equation above, the$r$can be calculated as $$r = \frac{1}{t}\ln\frac{N_t}{N_0}.$$ So, for$t = 1we have the annualized return: $$r_{t=1} = \frac{1}{1}\ln\frac{4086}{... 2 Sometimes it is easier to work with continuos compounding in some models, especialy when you compound interest daily. Moreover, it can come from history when it was more difficult to calculate higher powers than tabulated exponential function. Take for example annual interest rate i = 5 \%. Then effective annual interes rate based on daily compounding is ... 2 So, you: 1- take your daily return series. I've used the SPY ETF including divis 2- take a log return series, ln(1) 3- add ln(1.015)/261 to 2, given 261 trading days on average each year 4- do a running sum series of 2 and 3 5- exp(4) to give you a price Gives you: The ratio between the two is 1.35 = 1.015^20, ie your 150bps compounded over the ... 2 Let's add a time variable to extend to non-annual periods$$1 + r_d t = e^{r_c t}The taylor expansion of exponential is \begin{align} e^{r_c t} &= \sum_{n=0}^\infty {\frac {(r_c t)^n} {n!}}\\ &= 1 + r_c t + {\frac 1 2}(r_c t)^2 + \cdots \end{align} so by equating the two equations, we see thatr_d = r_c + {\frac 1 2}r_c^2 t + O(t^2)$$Two things ... 1 They're not equivalent, but you can use log identities to derive something similar after applying a log. Eg, ln\left(\left(\frac{1+E_2}{1+E_1}\right)^{\frac{d}{360}}\right) \frac{d}{360}*\left(1+E_2-\left(1+E_1\right)\right) \left(E_2-E_1\right)*\frac{d}{360} 1 Let's assume that the collateral rate on cash equals the overnight rate, that we have a schematic (lined/tiled up accrual periods and payments dates) strip of dates/times T_0<T_1<\ldots <T_n, accrual factor \tau_t := \tau(t-1,t), and c_t collateral rate at t (overnight t-1 to t). The floating coupon is then:$$ \prod_{s=T_{i-1}}^{T_i}\... 1 Well let's just do the math. 10% p.a. with semi annual means 10%/2 for 6 months, so you get1.05*1.05=1.1025$That is, for 1 dollar you'll have 1.1025 in 1 year, i.e. 10.25% p.a. if it was annualy compounded. What should be the rate for continuous compounding (annual)? Well:$e^{r\times1}=1.1025$gives$r=0.0975803$or 9.758% as stated in your book. 1 This would be specified in the ISDA or term sheet. There are four alternative methods: No compounding: Meaning neither the rate nor the spread get compounded. Compounding: Meaning both the spread and the libor rate get compounded. Spread exclusive compounding:Meaning the libor get compounded but the spread does not. This was not covered by ISDA 2006 but ... 1 Let$\Delta P = P_t - P_{t-1}\$ and expand the continuously compounded return in a Taylor series $$r = \log\left(\frac{P_t}{P_{t-1}}\right) = \log\left(\frac{P_{t-1}+\Delta P}{P_{t-1}}\right) = \log\left(1+\frac{\Delta P}{P_{t-1}}\right) \approx \frac{\Delta P}{P_{t-1}} - \mathcal{O}\left(\left(\frac{\Delta P}{P_{t-1}}\right)^2\right) = \frac{P_t - P_{t-1}}{... 1 Since the simple interest r_{s} and the continuous compounded interest r_{c} are connected by$$(1 + r_{s} \cdot (T_{2}-T_{1})) = e^{r_{c} \cdot (T_{2}-T_{1})}$$it follows for the continuous compounded interest:$$r_{c} = \frac{1}{T_{2}-T_{1}} \cdot \ln{(1+r_{s} \cdot (T_{1}-T{2}))}$$your convexity formula becomes than:$$ \frac{1}{T_{2}-T_{1}} \cdot \...

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