# Does numeraire have to be a tradable asset

I thought we create replicating portfolios using underlying and the numeraire i.e. the numeraire has to be a tradable asset (assuming simple binomial model). But I have seen some examples which explicitly state that the numeraire does not have to be tradable.

• The numeraire in a sense is about units. It's quite natural to write the price of anything in units of dollars (which is tradable). In some abstract sense though, you could run an island economy where every price is in dollars but where no dollars circulate and instead cigarettes are used as money. This type of thinking may start you down the rabbit hole of "what's money," "what determines the value of money,", and other difficult questions where economics has something to say but does not provide clearly settled, definitive answers. – Matthew Gunn Jun 4 '18 at 19:40

This is an interesting question that I have asked myself. Below is my take.

Let us consider an economy $(\Omega,\mathcal{F},P)$ equipped with a filtration $(\mathcal{F})_{t \geq 0}$ consisting on a traded asset $S_t$ and a numéraire $N_t$ specified by the following stochastic differential equations: \begin{align} \text{d}S_t&=\alpha(t,S_t)\text{d}t+\beta(t,S_t)\text{d}W_t \\[3pt] \text{d}N_t&=a(t,N_t)\text{d}t+b(t,N_t)\text{d}\tilde{W}_t \end{align} Our economy has a derivative contract written on the asset $S_t$ with payoff function $h(\cdot)$ at maturity $T$. By derivative pricing theory, the price $V_t$ of the derivative is given by the following expectation under the measure $P^N$ associated to the numéraire $N_t$, conditional on the available information: $$\tag{1}V_t=N_tE^N\left(\frac{h(S_T)}{N_T}\bigg|\mathcal{F}_t\right)$$ Defining the function $g(\cdot)$ for $(s,n) \in \mathbb{R}_+^2$: $$g(s,n)=\frac{h(s)}{n}$$ By the Markov Property $-$ see e.g. theorem 6.3.1. in Stochastic Calculus for Finance II by Shreve $-$ we have for $0\leq t\leq T$: $$\tag{2} V_t=v(t,S_t,N_t)$$ Thus by Itô's lemma: \begin{align} \tag{3}\text{d}V_t=& \ \frac{\partial v}{\partial t}\text{d}t+\left(\frac{\partial v}{\partial S}\text{d}S_t+\frac{1}{2}\frac{\partial^2 v}{\partial S^2}(\text{d}S_t)^2\right)+\left(\frac{\partial v}{\partial N}\text{d}N_t+\frac{1}{2}\frac{\partial^2 v}{\partial N^2}(\text{d}N_t)^2\right) \\ &+\left(\frac{\partial^2v}{\partial S\partial N}\text{d}S_t\text{d}N_t\right) \end{align} We note two things:

• Observability: by equation $(2)$ the value today of a derivative depends upon the value today of the underlying asset and the numéraire $N_t$, therefore the numéraire needs to be at least observable, i.e. it cannot be some latent state variable. If the numéraire is unobservable we cannot compute the price.
• Tradability: most importantly, by equation $(3)$ we observe the variations in value of the derivative also depends upon the variations of the value of the underlying asset and the numéraire. If we are to set up a hedging portfolio, we need to be able to trade the numéraire $N_t$ in order to offset the fluctuations in the value of the derivative due to fluctuations of the numéraire.

References

Shreve, S. (2004). Stochastic Calculus for Finance II, Springer.

@AFK (2016). "Feynman Kac and choice of measure", Quant Stack Exchange.

@Quantuple (2016) "Other numeraire choices when applying Feynman Kac", Quant Stack Exchange.

• Actually in your analysis you are writing an option on two securities $S_t$ and $N_t$ with general payoff $g(S_t,N_t)$ which is why you need to hedge against and hence trade $N_t$. If you choose $N_t$ as the time $T$ zero coupon so that the payoff is $h(S_t/N_t)$ and work with $V_t/N_t$ then you do not need to trade $N_t$ (all accounting is done in units of $N_t$). – Antoine Conze Jun 5 '18 at 19:05
• I agree with you, however you are restricting yourself to a specific numéraire. If we want to encompass any numéraire then we need to work from the pricing equation $(1)$ above to answer the question "Are numéraires required to be traded?". – Daneel Olivaw Jun 5 '18 at 19:17
• You are right, but my point is that the numeraire needs to be traded only to the extent that the payoff in numeraire cannot be written as a function of the underlying. For instance when hedging a swaption you do not need to trade the annuity, you only need to enter into swaps. – Antoine Conze Jun 5 '18 at 20:25
• Fully agree @Antoine Conze. The whole question is rather theoretical and in practice irrelevant, since we tend to choose numéraires that simplify the payoff expression. – Daneel Olivaw Jun 5 '18 at 20:50

An obvious example is using the maturity $T$ zero coupon as numeraire, and a European option with premium paid at time $T$ hedged with maturity $T$ forward contracts. You do not need to trade the zero coupon, in fact you don't even really need to know its value in terms of \$prior to$T$, because all settlements will occur on$T$. As stated by @Matthew Gunn the numeraire's role is to act as a (self-financed) unit, and as long as you are able to do all your accounting in terms of the numeraire you don't need to trade it and you don't need to know its value (its "exchange rate") in terms of \$. Of course, being able to do all accounting in terms of the numeraire requires that the option payoff itself can be expressed as a function of the underlying in numeraire, which is why the zero coupon numeraire is nice to work with since it is always worth 1 at maturity.

• Your analysis is restricted to a specific numéraire, the zero-coupon bond, however OP is concerned with the broadly found statement that "numéraires are required to be traded assets", which you normally find in any paper discussing numéraires. To answer in all generality we need to start from the pricing equation. Of course, the whole question is rather theoretical and in practice this is not an issue. – Daneel Olivaw Jun 5 '18 at 19:33