0
$\begingroup$

Given that the price of market risk (or market price of interest rate risk) is $\lambda(r_t, t)=0$ and that we have the following dynamics of the interest rate (under the physical measure $P$.

$$dr_t = \sigma dW_t^P \quad , \quad \sigma \in \mathbb{R}, \; W_t \text{ is a Wiener Process}.$$

If we furhter more have the relation $dW_t^Q=dW_t^P-\lambda(r_t,t)$ then we also have $$dr_t = \sigma dW_t^Q,$$ where $Q$ denotes the risk neutral measure.

I want to find

  1. The price of a Zero Coupon Bond and
  2. The yield of a Zero Coupon Bond.

I think that I have most of the calculations right, but I am missing a few pieces. My work goes as follows:

For the price, $p(t,T)$, for at ZCB at time $t$ with maturity $T$, I want to find the the pricing function $F(t,r_t;T)=p(t,T)$ satisfying the Term Structure Equation $$F_t^T+\frac{1}{2}\sigma^2F_{rr}^T-rF^T=0$$ where subscripts denote differantials and we also have the boundary condition $F^T(T,r_t;T)=p(T,T)=1.$

To do this I apply the Feynmann-Kac theorem to get the price of a ZCB as $$p(t,T)=F(t,r_t;T)=e^{-r_t(T-t)}E^Q_t[1]=e^{-r_t(T-t)}$$.

However as the (continuosly compounded) Zero Coupon Yield is given by $$y(t,T)=-\frac{\log p(t,T)}{T-t}$$

then by insereting my result for the price I would get $y(t,T)=r_t$.

I think I've done something wrong as the last result does not make much sence to me. E.g. I would not be able to make a yield curve from this a. Also what would $r_0$ be?

$\endgroup$
2
  • $\begingroup$ You should have $p(t,T)=\mathbb{E}_t^Q[e^{-\int_t^{T}r_sds}]$ $\endgroup$
    – fes
    Commented Mar 12, 2022 at 17:53
  • $\begingroup$ Okay, I can get to that expression but how am I gonna continue from there? $\endgroup$
    – Landscape
    Commented Mar 12, 2022 at 18:38

1 Answer 1

1
$\begingroup$

The standard pricing formula applies:

$$p(t,T)=\mathbb{E}_t^Q[e^{-\int_t^{T}r_sds}]$$

From $dr_t=\sigma dW_t$ you can solve:

$$r_t=r_0+\sigma W_t$$

Note (see: Integral of Brownian motion w.r.t. time)

$$ \int_t^{T}W_sds \sim N(0,\frac{1}{3}(T-t)^3)$$

Hence

$$-\int_t^{T}r_sds \sim N(-r_0(T-t),\frac{1}{3}(T-t)^3\sigma^2)$$

Using the formula for the mean of a log-normal variable:

$$p(t,T)=\exp(-r_0(T-t)+\frac{1}{6}(T-t)^3\sigma^2)$$

Hence

$$y(t,T)=r_0-\frac{1}{6}(T-t)^2\sigma^2$$

In this model all yields equal the current short rate minus (a typically small) convexity adjustment.

$\endgroup$

Your Answer

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

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