# Zero Coupon Bond - Price and Yield when interest rate is a diffusion process and 0 "price of market risk"

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?

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

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.