Testing Whether a Direct Zero-Coupon Bond Model Is Arbitrage-Free
Summary
The document asks whether a term-structure model can be built by specifying discounted zero-coupon bond prices as positive martingales, then defining a numeraire from the bond maturing at the current time. It explores whether that numeraire can be replicated by a self-financing portfolio of other zero-coupon bonds, which would support the model’s no-arbitrage claim.
For a one-dimensional Brownian driver, the proposed discrete-time argument chooses holdings to cancel the shared random shock while matching the target portfolio value. The author is uncertain whether this carries over to continuous time. The document offers no resolution or proof: in particular, it does not establish continuous-time replication, admissibility, or the required trading strategy. It is therefore useful as a formulation of the modeling question and a warning that martingale assumptions alone do not demonstrate tradability or arbitrage freedom.
Key ideas
- The proposed model takes discounted zero-coupon bond prices to be positive martingales.
- The candidate numeraire is defined using the discounted bond that matures at the current time.
- The author’s discrete-time replication argument cancels a shared shock while matching portfolio value.
- The document leaves continuous-time replication and the no-arbitrage conclusion unresolved.
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Full text
# Is this term structure model valid? (Modeling the Zerobonds directly)
# Is this term structure model valid? (Modeling the Zerobonds directly)
Let us define the dynamics of the discounted Zerobonds as
$$ \tilde{P}(t,T) = \int \sigma(t,T) dW_t + \tilde{P}(0,T)$$ Lets assume $\sigma(t,T)$ is s.t. $\tilde{P}(t,T) $ is a martingale and positive (and nonzero) $\mathbb{P} \ a.s.$ . Lets define a numeraire as follows:
$$ B(t) = \frac{1}{\tilde{P}(t,t)} $$ And finally define the dynamics of the actual Zerobonds by
$$P(t,T) = \tilde{P}(t,T) B(t)$$
Now by construction $P(t,t) =1 \forall t$ and there exists a numeraire (namely $B(t)$) s.t. the discounted Zerobonds are martingales.
I think $B(t)$ is a valid numeraire. It is by assumption strictly positive and I could show in the discrete case that it is an tradable asset (it can be replicated by trading in zerobonds). For the continuous case I have the following starting point
Lets assume the brownian motion is one dimensional. We need a selffinancing trading strategy $\phi$ in Zerobonds s.t.
$$ \sum\limits_{i=1}^{n} \phi_{i,t} P(t,t_{i}) = B(t)$$
which is equivalent to
$$ \sum\limits_{i=1}^{n} \phi_{i,t} \tilde{P}(t,t_{i}) = 1$$ This is where I am not sure in the continuous case. in the discrete case $\tilde{P}(t,t_{i}) $ results from $\tilde{P}(t-\Delta,t_{i}) + \sigma_{t,T}\varepsilon$ with $\varepsilon\sim \mathcal{N}(0,\Delta)$. Since $\varepsilon$ is equal for all Zerobonds we can find $\phi_{1t}$ and $\phi_{2t}$ s.t. on one hand $$ \phi_{1t}\sigma_{t,t_1}\varepsilon + \phi_{2t} \sigma_{t,t_2}\varepsilon=0 $$ (where I assumed that $\sigma_{t,T}$ is predictable) and on the other hand
$$\phi_{1t}\tilde{P}(t,t_{1}) + \phi_{2t}\tilde{P}(t,t_{2})=1 $$ These equation guarantee that the value of the portfolio matches the value of the numeraire and furthermore that the cost of setting up this portfolio at $t-\Delta $ equals:
$$\phi_{1t}P(t-\Delta,t_{1}) + \phi_{2t}P(t-\Delta,t_{2}) =B(t-\Delta)(\phi_{1t}\tilde{P}(t,t_{1}) + \phi_{2t}\tilde{P}(t,t_{2})) =B(t-\Delta) $$
Thus we have a selffinancing portfolio with initial cost $B(0)$ that replicates the $B(t)$ which makes it a numeraire.
So all in all this implies that this model is arbitrage-free.
Is this reasoning valid? If not where got I wrong? If yes, why haven't I seen this anywhere? Is it not practical to work directly with zerobonds?Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.