2
$\begingroup$

Let $R(t)$ be an interest rate process and $D(t)=e^{-\int_0^t R(s)ds}$ and $B(t,T)$ the price at time t of a zero coupon bond maturing at T.

Can anyone explain why this formula holds? $D(t)B(t,T) = \widetilde{E}[D(T)|F(t)]$?

The following is my understanding. Please correct me if I am incorrect. To apply to argument of risk neutral pricing, one needs to first construct a portfolio of assets whose payoff matches the payoff of the derivative security under dynamic hedge. For example, the replicating portfolio of a European call is its underlying stock and risk free bond.

Similar, to price a zero coupon bond, one needs to find a replicating portfolio as well. If this is right, What is the replicating portfolio then?

$\endgroup$

3 Answers 3

2
$\begingroup$

You are not completely right. Risk-neutral pricing means that the price you'll quote for a given instrument at a given point in time will be the same as the capital you'd require at the same point in time to build a self-financing strategy which will deliver exactly the same future cash flows. The replication mechanism need not be dynamic, it may be static (see for instance static replication of a forward contract by "cash & carry" argument).

This concept is quite intuitive and easy to grasp. However, assuming the existence of a risk-free investment vehicle, best known as the risk-free asset, along with a complete market model defined under a probability measure $\mathbb{P}$, noting that the above claim is strictly equivalent to saying that "any self-financing strategy should emerge as a martingale under a probability measure $\tilde{\mathbb{P}}$ equivalent to $\mathbb{P}$ and associated to the risk-free asset numéraire" is far less intuitive.

There is a lot of theory behind this last statement (see fundamental theorems of asset pricing). I invite you to investigate on your own as this would require a book to answer. But if you want to be convinced, just go back to the binomial model set up and observe how by using a replication argument, one can get rid of historical probabilities and define a new set of probabilities under which discounted asset prices are martingales.

At the end of the day, if you have a complete market model, by absence of arbitrage, there should be a unique risk-neutral measure $\tilde{\mathbb{P}}$ under which: $$ \frac{V_t}{B_t} \text{ is a } \tilde{\mathbb{P}}\text{-martingale} \tag{1} $$ for any self-financing strategy $V_t$ and $B_t = \exp(\int_0^t r(s) ds)$ the $t$-value of a risk-free cash account in which $1$ unit of currency has been invested at $t=0$.

Applying this in your case:

  • Investing in the ZC bond is a self-financing strategy: $V_t \to B(t,T)$
  • What you have defined as $D_t$ is the inverse of what we just defined as $B_t$: $B_t \to 1/D(t)$
  • Claiming $(1)$ by definition means that, $\forall t$ $$ \frac{V_t}{B_t} = \mathbb{E}^{\tilde{\mathbb{P}}}\left[ \frac{V_T}{B_T} \vert \mathcal{F}_t \right] $$ which using the above notations writes: $$ B(t,T) D(t) = \mathbb{E}^{\tilde{\mathbb{P}}}\left[ \underbrace{B(T,T)}_{=1} D(T) \vert \mathcal{F}_t \right] $$ hence $$ B(t,T) = \mathbb{E}^{\tilde{\mathbb{P}}}\left[ 1 e^{-\int_t^T r(s) ds} \right] $$ where the first line is exactly the equality you gave and the second finds the following interpretation: the value of a contingent claim is the expectation of the discounted value of the future cash flows it shall pay, where the discount is performed at the risk-free rate.
$\endgroup$
1
$\begingroup$

To replicate a zero-coupon bond,

  • either the zero-coupon bond is a tradeable asset and you buy the zero coupon bond,

  • or you believe in a model and using other rates instruments, you set up a delta-hedging strategy.

$\endgroup$
0
$\begingroup$

Since rate is not a tradable directly, zero-coupon bond is an underlying itself. So you cannot replicate a zero-coupon because it is not a derivative

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.