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From what I am reading arrears swap are paid on the same day(actually, +2 business days for JPY and USD) as the reset date. To me then, a week before the reset date the floating rate is not known. Which means it is like predicting the rate a week ahead using nothing but ${\sigma}$ - volatility. The arrears swap rates are swap rates + convexity adjustment. Surely, there is no prediction here, just correcting yield to account for non linear price to yield relationship. Apparently price is linear with time but yield has to go through convexity correction. I hope someone will explain. I have read Hull's chapters several times, read literature on the internet etc, I have not found a clear description. I can add more details if it is not clear. Thanks.

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  • $\begingroup$ This question is for 1. libor in-arrears swap 2. why paying a floating amount just announced is not estimated using a stochastic process? 3. would be nice if one can add explanation as to how convexity adjustment corrects the arrears-swap valuation. $\endgroup$ – user12348 Nov 3 '15 at 2:48
  • $\begingroup$ I'm confused...why would there be a convexity adjustment on a LIBOR swap? $\endgroup$ – user9403 Jan 11 '16 at 11:01
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We consider a single Libor rate. The application to a swap is straightforward.

Consider the Libor calculation period $[T_1, \, T_2]$ and the Libor payment made at $T_1$. We denote by $\Delta = T_2-T_1$ the length in years of the calculation period. Here, we ignore the two-day payment delay as its impact to pricing is immaterial. We assume that, under the $T_2$-forward measure $P_{T_2}$, the Libor rate process $\{L(t, T_1, T_2) \mid 0 \le t \le T_1\}$, where \begin{align*} L(t, T_1, T_2) = \frac{1}{\Delta} \left(\frac{P(t, T_1)}{P(t, T_2)}-1\right), \end{align*} is a martingale and satisfies an SDE of the form \begin{align*} dL(t, T_1, T_2) = \sigma L(t, T_1, T_2) d W_t, \end{align*} where $\{W_t \mid t \ge 0\}$ is a standard Brownian motion. Then, for $0 \le t \le T \le T_1$, \begin{align*} L(T, T_1, T_2) = L(t, T_1, T_2) e^{-\frac{1}{2}\sigma^2 (T-t) + \sigma \int_{t}^{T} dW_s}. \end{align*}

Let $B_t$ be the money -market account value at time $t$. Then, for $ 0 \le t \le T_2$, \begin{align*} \frac{dP}{dP_{T_2}} \big|t = \frac{B_t P(0, T_2)}{P(t, T_2)} \equiv \eta_t. \end{align*} Moreover, let $E$ and $E_{T_2}$ be the respective expectation operators under the risk-neutral measure and the $T_2$-forward measures.

Then the value, at time $t\le T_1$, of the Libor rate $L(T_1, T_1, T_2)$, both set and paid at $T_1$, is given by \begin{align*} B_t E\left(\frac{L(T_1, T_1, T_2)}{B_{T_1}}\mid \mathcal{F}_t \right) &= B_t E_{T_2}\left(\frac{\eta_{T_1}}{\eta_t}\frac{L(T_1, T_1, T_2)}{B_{T_1}}\mid \mathcal{F}_t \right)\\ &=P(t, T_2) E_{T_2}\left(\frac{1}{P(T_1, T_2)}L(T_1, T_1, T_2)\mid \mathcal{F}_t \right)\\ &= P(t, T_2) E_{T_2}\left(\left(\Delta L(T_1, T_1, T_2) + 1 \right)L(T_1, T_1, T_2)\mid \mathcal{F}_t \right)\\ &= P(t, T_2)E_{T_2}\left(L(T_1, T_1, T_2) + \Delta L(T_1, T_1, T_2)^2\mid \mathcal{F}_t \right)\\ &= P(t, T_2) \left(L(t, T_1, T_2) + \Delta L(t, T_1, T_2)^2 e^{\sigma^2 (T_1-t)}\right)\\ &=P(t, T_1) \frac{L(t, T_1, T_2) + \Delta L(t, T_1, T_2)^2 e^{\sigma^2 (T_1-t)}}{\Delta L(t, T_1, T_2) + 1} \\ &= P(t, T_1)\left(c_t + L(t, T_1, T_2) \right), \end{align*} where \begin{align*} c_t = \frac{\Delta L(t, T_1, T_2)^2}{\Delta L(t, T_1, T_2) + 1}\big(e^{\sigma^2 (T_1-t)} -1 \big) \end{align*} is the convexity adjustment. Note that, there is no approximation needed, as long as we can estimate the volatility.

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lets take it as simple that all payment are at the same time. When you want to calculate the floating, you always look at the previous period. But to calculate the first payment, we do not have forward rate at time t=0. But when you want to calculate between two periods, you have forward rates and you can always fix the rate for next period (floating one). So you can always calculate how much will be the floating one,one period before it.

When you are discounting and you have a floating at time T and you have another t , if you want to discount and see how much does it dost today, you can either discount it back with D(T) or you discount first to time t with D(t,T) and then D(t). you know the rate D(T) and D(t ) but you don’t know the rate D(t,T). but if we discount in either way, we should get the same result, then we should have

$$D(T)=D(t).D(t,T)$$

Think that you want to calculate the Forward rate between $[T_{i-1}, T_i]$ which will be :

$$ D(T_i)=D(T_{i-1}).D(T_{i-1},T_i)$$ Since $D(T_{i-1},T_i)$ is less than a year, we can put simple compounding formula and get: $$D(T_i)= D(T_{i-1}).D(T_{i-1},T_i)=\frac{D(T_{i-1})}{1+\Delta_i.F_i}$$

and now u can solve for $F_i$: $$F_i=\frac{D(T_{i-1})-D(T_i)}{\Delta_i .D(T_i)}$$

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  • $\begingroup$ I really want to thank you for your effort. But it does not answer my question. please see the comment to my question. $\endgroup$ – user12348 Nov 3 '15 at 2:44

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