Convexity Adjustment of Daily Compounded Swap under Hull-White Model

I am working on a problem that deals OIS daily compounded swap under Hull-White 1-factor model. I am struggling with pricing the floating leg, on a delayed payment date:

$$E^{T^p}_t[\prod_{i=0}^{n-1} (1 + \tau_i L_i)]$$

where $$L_i = L(t_i; t_{i+1}, t_{i+1})$$ is the (daily) forward rate, $$\tau_i$$ is the accrual factor between the consecutive business days, and $$T^p > t_{n}$$ is the delayed payment date.

My thought was to derive the distribution of $$\prod_{i=0}^{n-1} (1 + \tau_i L_i) = \prod_{i=0}^{n-1}\frac{1}{P(t_i, t_{i+1})}$$ under $$T^p$$-measure, but the computation was too nasty, could anyone give me some hints/reference to do this? Thank you very much.

You can approximate the daily compounding of O/N rate with continuous one of the short rate: $$\prod_{i=0}^{n-1} (1 + \tau L_i)= \prod_{i=0}^{n-1} \frac{1}{P(t_i,t_{i+1})} \approx \exp \left(\int_{t_0}^{t_n} r(u)du \right)$$

This is a reasonable approximation. If you need a justification or an assessment the error, you can start from the zero-coupon bond price expression under Hull-White of the form: $$P(t,t+\varepsilon) = \exp \left(A(t,t+\varepsilon) - r(t)B(t,t+\varepsilon) \right)$$ and look at what happens to $$A$$ and $$B$$ terms when: $$\varepsilon \rightarrow 0$$.

Next, you can derive the expression of $$\int_{t_0}^{t_n} r(u)du$$ under Hull-White model and under the $$T_p$$ measure. I think the easiest would be to work under the risk-neutral measure to get the expression of the integral of short rate from that of the short rate, and then use Girsanov to change measures to $$T_p$$.

Once you have this, it will be straightforward to get the desired expectation.

I didn't include any explicit derivations, rather hints as you asked. But let me know if you are stuck in one of the steps, and I might be able to help.

We know that $$1+\tau_iL_i=\frac{1}{P(t_i,t_{i+1})}$$ and, under the risk-neutral measure, $$P(t,T)=P(0,T)\exp\left(\int_0^tr(s)\,ds+\int_0^t\sigma(s,T)\,dW_s-\frac{1}{2}\int_0^t\sigma^2(s,T)\,ds\right)\,.\quad\quad\quad (1)$$ Using $$P(t_i,t_i)=1$$ this implies \begin{align} P(t_i,t_{i+1})&=\frac{P(0,t_{i+1})}{P(0,t_i)}\exp\Bigg(\int_0^{t_i}\sigma(s,t_{i+1})-\sigma(s,t_i)\,dW_s\\ &\quad\quad-\frac{1}{2}\int_0^{t_i}\sigma^2(s,t_{i+1})-\sigma^2(s,t_i)\,ds \Bigg)\,. \end{align} Therefore, \begin{align} \prod_{i=0}^{n-1}1+\tau_iL_i&=\frac{P(0,t_0)}{P(0,t_n)}\exp\Bigg(\sum_{i=0}^{n-1}\int_0^{t_i}\sigma(s,t_i)-\sigma(s,t_{i+1})\,dW_s\\ &\quad\quad-\sum_{i=0}^{n-1}\frac{1}{2}\int_0^{t_i}\sigma^2(s,t_i)-\sigma^2(s,t_{i+1})\,ds \Bigg)\,. \end{align} Further, (1) implies with $$P(T,T)=1$$ that $$\exp\left(-\int_0^{T}r(s)\,ds\right)=P(0,T)\exp\left(\int_0^T\sigma(s,T)\,dW_s-\frac{1}{2}\int_0^T\sigma^2(s,T)\,ds\right)\,.$$ So we have to calculate the expectation of \begin{align} \exp\left(-\int_0^{T}r(s)\,ds\right)\left(\prod_{i=0}^{n-1}1+\tau_iL_i\right) \end{align} which is $$\frac{P(0,t_0)P(0,T)}{P(0,t_n)}\exp\left(-\sum_{i=0}^{n-1}\frac{1}{2}\int_0^{t_i}\sigma^2(s,t_i)-\sigma^2(s,t_{i+1})\,ds-\frac{1}{2}\int_0^T\sigma^2(s,T)\,ds\right)$$ times the expectation of the lognormal variable $$e^Y:=\exp\left(\int_0^T\sigma(s,T)\,dW_s+\sum_{i=0}^{n-1}\int_0^{t_i}\sigma(s,t_i)-\sigma(s,t_{i+1})\,dW_s\right)\,.$$ To calculate that expectation you need to know only the variance of $$Y$$ which is a sum of integrals of the form \begin{align} &\int_0^{t_i}(\sigma(s,t_i)-\sigma(t_{i+1}))(\sigma(s,t_j)-\sigma(t_{j+1}))\,ds\quad\quad i\le j\\ &\int_0^{t_i}(\sigma(s,t_i)-\sigma(t_{i+1}))\sigma(s,T)\,ds\,. \end{align} This is as explicit as it gets. Even using the time dependent HW-form of $$\sigma(t,T)\,,$$ that is, $$\sigma(t,T)=\int_t^T\sigma(s)\exp\left(-\int_s^T\lambda(u)\,du\right)\,ds$$ does not help much, unless you assume constant mean reversion $$\lambda$$ and constant volatility $$\sigma$$ of the short rate.

• Hi, may I know how you derive formula (1) and $P(t_i, t_{i+1})$? Thanks! Sep 3 '21 at 17:17
• Formula (1) says that $P(t,T)\exp(-\int_0^tr(s)\,ds)$ is a martingale. This is required by no arbitrage. The equation for $P(t_i,t_{i+1})$ you get as follows: (1) and $P(t_i,t_i)=1$ give you $1=P(0,t_i)\exp(\int_0^{t_i}r(s)\,ds+\int_0^{t_i}\sigma(s,t_i)\,dW_s-\frac{1}{2}\int_0^{t_i}\sigma^2(s,t_i)\,ds)\,.$ Then divide the expression for $P(t_i,t_{i+1})$ by the RHS of that (which is 1). And gone is the $r(s)$-term ! I believe this trick is due to El Karoui. Sep 3 '21 at 17:32
• Could you put more details for deriving formula (1)? The martingale property only says that $P(0, T) = E^Q[e^{-\int_0^t r(s) ds} P(t, T)]$, how you get the explicit expression? Thank you very much. Sep 3 '21 at 18:39
• Clearly, by (1), the process $e^{-\int_0^tr(s)\,ds}P(t,T)$ is of the form $e^{M_t-\langle M\rangle_t/2}\,.$ Because $M_t=\int_0^t\sigma(s,T)\,dW_s$ is a martingale, the exponential is one as well. Sep 4 '21 at 5:36

Why have you defined this valuation under the T-forward measure? If you consider the Hull-White model you can simply simulate the process under the risk-neutral measure and discount it with the money-savings account, $$M(t)$$. Every Libor rate, $$L_i(t)$$, will be simply defined in the terms of the short rate, $$r(t)$$.

• Hi Lech, thanks for your reply, I need to derive an analytical formula for the swap. Given Hull-White model has an analytical ZCB formula, I thin define it under $T$-forward measure is a good way to do it. Aug 22 '21 at 15:04