# If any zero coupon bond $P(T)$ can be chosen as a numéraire, then why can the rolling bond for any time discretization be chosen as numéraire

Let us consider some finite time horizon $$[0,T]$$, and we assume that $$P(t)$$, the zero coupon bond maturing in $$t$$ for any $$t\in [0,T]$$ can be chosen as a numéraire, i.e. such that the numéraire-relative prices of traded assets are martingales.

Next, we define for a given time discretization of $$\{0=T_{0}<...< T_{n}=T \}$$, the rolling bond $$R$$ is defined as follows:

$$R(\{T_{0},...,T_{n} \};t):=P(T_{m(t)+1};t)\prod\limits_{i=0}^{m(t)}(1+L(T_{i},T_{i+1};T_{i})(T_{i+1}-T_{i}))$$

where $$L(T_{i},T_{i+1}):= \frac{1}{T_{i+1}-T_{i}}\frac{P(T_{i})-P(T_{i+1})}{P(T_{i+1})}$$, i.e. the forward rate and further

$$m(t):=\max\{i:T_{i}\leq t\}$$

Question: Why does it follow that indeed the rolling bond $$R$$ may also be chosen as a numéraire? That is we can associate a measure $$\mathbb Q^{R}$$ such that the $$R$$-relative prices of traded assets are martingales under that measure.

Edit:

I guess I should edit my question in the following way to be more precise:

Given that we know for every $$t \in [0,T]$$ the martingale measure $$\mathbb Q^{P(t,\cdot)}$$ exists, i.e. the dynamics of $$\frac{S}{P(t,\cdot)}$$ are driftless under $$\mathbb Q^{P(t,\cdot)}$$, where $$S$$ represents a traded asset, how do we know that $$R$$ necessarily has a martingale measure associated to it? In this particular case, the forwards $$(L_{i})$$ for the given time discretization are assumed to be the traded assets. In order for the rolling bond $$R$$ (for a given tenor discretization) to have an associated martingale measure $$\mathbb Q^{R}$$, we need to find the drift of the traded assets that render $$\frac{S}{R}$$ driftless under $$\mathbb Q^{R}$$.

• The discrete money-market numeraire starts at 1 and compounds at the money-market rate ($\tau$ is the annual fraction). $$\prod\limits_{i=0}^{n}(1+\tau L(T_{i},T_{i+1};T_{i}))$$ The rolling bond is about combining the zero-coupon bond numeraire and the discrete money-market numeraire. The spot zero coupon bond is taken and when it matures (at maturity, it is equal to 1), it basically turns into the classical discrete money-market numeraire that compounds using the formula you wrote. Jan 21, 2022 at 17:51
• Hi Jan, I think my initial question may have been misleading as this does not answer what I intended it to. Please see my edit for the updated question. Jan 21, 2022 at 21:23

The rolling bond $$R(t)$$ as defined in your question is a valid numéraire. Indeed, this bond can synthetized with the following iterative trading strategy in basic assets:

1. At any time $$T_i\in\{T_0,\dots,T_{m(t)-1}\}$$, invest your available wealth into the spot-starting zero-coupon bond expiring at $$T_{i+1}$$, with price $$P(T_i,T_{i+1})$$.
2. At $$T_{i+1}$$, you receive the proceeds from the investment made in 1, namely your initial wealth compounded by $$1+\tau_iL(T_i,T_{i+1})$$ where $$\tau_i:=T_{i+1}-T_i$$.
3. Repeat steps 1-2.

It is easy to see the above strategy is self-financing: once initiated at $$T_0$$ by investing your available wealth in $$P(T_0,T_1)$$, there is no need for any inflow or outflow of cash to carry on. It only requires buying basic assets, namely zero-coupon bonds, which are themselves valid numéraires with positive price processes. Therefore, the rolling bond is also a valid numéraire.

Alternatively, we can refer to the seminal paper by Geman et al. (1995) which established numéraire theory. Working on a probability space $$(\Omega,\mathscr{F},\mathbb{P})$$ over the interval $$[0,T]$$, a numéraire is defined as follows:

Definition 2. A numéraire is a price process $$X(t)$$ almost surely strictly positive for each $$t\in[0,T]$$.

The price process of any zero-coupon bond is strictly positive (note that the traded assets are the bonds $$P$$, not the forward rates $$L$$), hence the price process of the rolling bond is also positive $$-$$ thus $$R$$ is a valid numéraire. Moreover, per your edited question, you have assumed that a martingale measure exists for each of the zero-coupon bonds $$P(\cdot,T_i)$$ for $$i=1,\dots,n$$ therefore Assumption 1 of their paper is fulfilled:

Assumption 1. There exists a non-dividend-paying asset $$n(t)$$ and a probability $$\pi$$ equivalent to the initial probability $$P$$ such that for any basic security $$S_k$$ without intermediate payments, the price of $$S_k$$ relative to $$n$$, i.e. $$S_k(t)/n(t)$$, is a local martingale with respect to $$\pi$$.

You can then apply Theorem 1, which states that $$R$$ induces a new martingale measure $$\mathbb{Q}^R$$ equivalent to the other zero-coupon (i.e. forward) measures $$\mathbb{Q}^{P(\cdot,T_1)},\dots,\mathbb{Q}^{P(\cdot,T_n)}$$. You do not need to determine the drift of the assets under $$\mathbb{Q}^R$$ to determine whether such measure exists. Indeed, as long as you define the following Radon-Nikodym derivative to change between the zero-coupon(s) and the rolling measures: $$\left.\frac{\text{d}\mathbb{Q}^R}{\text{d}\mathbb{Q}^{P(\cdot,T_i)}}\right|_{\mathscr{F}_{T_i}}=\frac{R(T_i)P(0,T_i)}{R(0)P(T_i,T_i)}=\frac{R(T_i)P(0,T_i)}{R(0)}$$ You know that any asset price divided by the numéraire $$R$$ will be a martingale under measure $$\mathbb{Q}^R$$, per the aforementioned Theorem.

The measure associated to the rolling bond is usually known as spot measure or rolling measure, see for example Proposition 6.3.3 in Brigo and Mercurio's book on interest rates. This measure was originally introduced by Jamashidian (1997). Another reference is Antonov & Lee (2004).

Often, the rolling measure is used in practice under Monte Carlo based pricing engines where the theoretical pricing measure should be the risk-neutral one: to avoid interpolation problems, a change of measure is performed from the risk-neutral measure to the rolling measure where the anchor points $$T_1,\dots,T_n$$ are chosen to match the simulation grid.

References

Antonov & Lee (2004). "Interest Rate Modelling Framework in Discrete Rolling Spot Measure", available at SSRN.

Brigo and Mercurio (2006). Interest Rate Models — Theory and Practice, Springer.

Geman, El Karoui and Rochet (1995). "Changes of Numéraire, Changes of Probability Measure and Option Pricing", Journal of Applied Probability, Vol. 32, No. 2, pp. 443-458.

Jamshidian (1997). "Libor and Swap Market Models and Measures", Finance and Stochastics, Vol. 1, pp. 293-330.

• Thanks for the background information, but I cannot see how this answers my question: If $(\mathbb Q^{P(t;\cdot)})_{0 \leq t \leq T}$ are equivalent martingale measure then why is $\mathbb Q^{R}$ an equivalent martingale measure? Jan 21, 2022 at 18:24
• @user9078057 I misinterpreted your question due to that typo that you fixed. I added further details, hope it is clearer now. Jan 21, 2022 at 19:05
• ah thanks, I think I understood your original comment (please correct me if I am wrong): We can find a drift term for every tradeable asset such that its $R$-relative dynamics under $\mathbb Q^{R}$ is that of a martingale (i.e. drift is $0$). But it is not clear to me how this depends on the fact that for every $t \in [0,T]$ the $\mathbb Q^{P(t,\cdot)}$-dynamics of every tradeable asset divided by $P(t,\cdot)$ is such that the tradeable asset has zero drift under $\mathbb Q^{P(t,\cdot)}$ Jan 21, 2022 at 19:26
• @user9078057 added additional details based on the seminal paper by Geman, El Karoui and Rochet (1995) on numéraires and changes of measure. Jan 22, 2022 at 16:44
• @user9078057 Basically: if you already assume a certain martingale measure exists for an asset $X$, and you have some other asset $Y$ with no dividends and strictly positive price, then you can define a new probability measure induced by $Y$ under which $Y$-relative dynamics of the other assets will be martingales. Jan 22, 2022 at 16:50