When a Numeraire Can Be Used Without Trading It

# Does numeraire have to be a tradable asset


# 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.

Could someone please help clear this for me, thanks!

## Answer by Daneel Olivaw (score 14, accepted)

https://quant.stackexchange.com/a/40126

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: $$\tag{1}V_t=N_tE^N\left(\left.\frac{h(S_T)}{N_T}\right|\mathcal{F}_t\right)$$ 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 $-$ there exists a function $v$ such that 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:





To discuss further on tradability, note that $V_T=h(S_T)$ and let us define the price processes normalized by the numéraire as $V^*_t:=V_t/N_t$ and $S^*_t:=S_t/N_t$. Equation $(1)$ can be rewritten: $$\tag{4}V^*_t=E^N\left(\left.V^*_T\right|\mathcal{F}_t\right)$$ As observed by @AntoineConze, a clever choice in practice is to choose $N$ such that the following homogeneity condition is satisfied: $$\tag{5}\frac{h(S_T)}{N_T}=h\left(\frac{S_T}{N_T}\right)$$ We can then focus on normalized processes and rewrite equation $(4)$ as: $$\tag{6}V^*_t=E^N\left(\left.h(S^*_T)\right|\mathcal{F}_t\right)$$ In this case we can use normalized prices for both the derivative and the asset, thus neglecting trading on 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.

## Answer by Antoine Conze (score 6)

https://quant.stackexchange.com/a/40171

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.