Volatility and drift of the instantaneous forward rate under risk neutral measure using the zero coupon bond

Question

I have question about this problem. I believe I have derived $f(t,T)$ correctly using the zero-coupon bond. But I am unsure about how to go forward with the question and how to use the second part.

Question

Note that $$ f(t, T)=-\frac{\partial}{\partial T} \log P(t, T) $$ Let us assume the HJM Framework. Assume that the $T$-bond price data from the market implies that $$ [f(\cdot, T), f(\cdot, T)]_{t}=\sigma e^{-\lambda(T-t)} $$ (a) What is the volatility of the forward rate? (b) If we assume there is no arbitrage, what is the drift of $f(t, T)$ under ELMM $\mathbb{Q}$ ?

My attempt:

First I derived $ f(t, T)=-\frac{\partial}{\partial T} \log P(t, T) $ as below:

Definition of instantaneous forward rate:

$$f(t, T)=f(0, T)+\int_{5}^{t} \alpha(s, T) d s+\int_{0}^{t} \sigma_{f}(s, T) d w(s)$$

$$d f\left(t, T\right)=\alpha(t, T) d t+\sigma^2_{f}(t, T) dw_{t}$$

Definition of zero-coupon bond price:

$$d P(t, T)=r_{t} P(t, T) d t+\sigma_{p}\left(t, T\right) P(t, T) d w_{t}$$

Taking differential of both sides of: $f(t, T)=-\frac{\partial}{\partial T} \log P(t, T) $ we get:

$$d f(t, T)=-\frac{d}{d T} d Log (P(t, T))$$

Using Ito's lemma on $d Log (P(t, T))$ we get:

$$d \ln P(t, T)=r_{t} d t+\sigma_{p}\left(t, T\right) d W_{t}-\frac{1}{2} \sigma_{p}(t, T)^{2} d t$$

hence:

$$d f(t, T)=\sigma_{p}(t, T) \frac{d}{d T} \sigma_{p}\left(t, T\right) d t-\frac{d}{d T} \sigma_{p}(t, T) d w_{t}$$

Now this is the part where I am confused. We have two definitions of $d f(t, T)$ as below:

$$d f\left(t, T\right)=\alpha(t, T) d t+\sigma^2_{f}(t, T) dw_{t}$$

$$d f(t, T)=\sigma_{p}(t, T) \frac{d}{d T} \sigma_{p}\left(t, T\right) d t-\frac{d}{d T} \sigma_{p}(t, T) d w_{t}$$

Hence, I thought the drift and volatility terms must equal as such:

$$\begin{aligned} &-\frac{d}{d T} \sigma_{p}(t, T) d \omega_{t}=\sigma_{f}(t, T) d w_{t} \\ &\int_{t}^{T} \sigma_{f}(t, u) d u+c=-\sigma_{p}(t, T) \end{aligned}$$

C = 0 by the definition of the zero-coupon bond. Now for the volatility part:

$$\sigma^2(t, T) d t=\sigma_{p}(t, T) \frac{d}{d T} \sigma_{p}(t, T) d t$$

$$\sigma(t, T) d t=\sqrt{\sigma_{f}(t, T) \int_{t}^{T} \sigma_{f}(t, u) d u}$$

Hence the dynamics of the instantaneous forward rate under the risk neutral measure is:

$$d f(t, t)=\sqrt{\left(\sigma_{f}(t, T) \int_{t}^{\top} \sigma_{f}(t, u) d u\right) d t}+\sigma_{f}(t, t) d w_{t}$$

I feel like I am wrong and I didn't even use the second part of the question where $[f(\cdot, T), f(\cdot, T)]_{t}=\sigma e^{-\lambda(T-t)}$. Help please.

Answer 1

You have numerous typos the most serious of which is hat in your $df$ equation the $\sigma_f(t,T)$ should not be squared.

The basic HJM equations are \begin{align} f(t,T)&=f(0,T)+\int_0^t\alpha(s,T)\,ds+\int_0^t\sigma(s,T)\,dw_s\,,\\ P(t,T)&=e^{-\int_t^Tf(t,u)\,du}\,. \end{align} Clearly, \begin{align} \int_t^Tf(t,u)\,du&=\int_t^Tf(0,u)\,du+\int_t^T\int_0^t\alpha(s,u)\,ds\,du+\int_t^T\int_0^t\sigma(s,u)\,dw_s\,du\\ &\stackrel{\text{Fubini}}{=}\int_t^Tf(0,u)\,du+\int_t^T\int_0^t\alpha(s,u)\,ds\,du+\int_0^t\int_t^T\sigma(s,u)\,du\,dw_s\,. \end{align} Therefore, \begin{align} d\left(\int_t^Tf(t,u)\,du\right)&=\underbrace{-f(0,t)\,dt-\left(\int_0^t\alpha(s,t)\,ds\right)\,dt-\left(\int_0^t\sigma(s,t)\,dw_s\right)\,dt}_{-f(t,t)\,dt\,=-r(t)\,dt}\\ &+\left(\int_t^T\alpha(t,u)\,du\right)\,dt+\left(\int_t^T\sigma(t,u)\,du\right)\,dw_t \,. \end{align} It follows that \begin{align} dP(t,T)&=P(t,T)\Bigg\{r(t)\,dt -\left(\int_t^T\alpha(t,u)\,du\right)\,dt- \left(\int_t^T\sigma(t,u)\,du\right)\,dw_t\\ &+\frac{1}{2}\left(\int_t^T\sigma(t,u)\,du\right)^2\,dt\Bigg\}\,. \end{align} This is the SDE for the zero bond $P(t,T)\,.$ Obviously, the relationship between the volatility function $\sigma(t,T)=\sigma_f(t,T)$ of the instantaneous forward rate $f(t,T)$ and that of the zero bond is $$ \boxed{\sigma_P(t,T)=-\int_t^T\sigma_f(t,u)\,du\,.} $$ You are asking what is the volatiliy of the forward rate?

Answer: it is the function $\sigma_f(t,T)$. In other words: each forward rate $f(t,T)$ has the time dependent volatility $\sigma_f(t,T)\,.$

The term $[f(.,T),f(.,T)]_t=\sigma e^{-\lambda(T-t)}$ is the quadratic variation. By the first HJM equation we know that this is $$ \int_0^t\sigma_f^2(s,T)\,ds\,. $$ Thus, $$ \sigma_f^2(t,T)=\frac{d}{dt}\sigma e^{-\lambda(T-t)}=\sigma\lambda e^{-\lambda(T-t)}\,. $$ This looks a bit odd but is possible. Are you sure your problem statement was not $$ [f(.,T),f(.,T)]_t=\frac{\sigma^2}{2\lambda}e^{-2\lambda(T-t)}\,? $$ This quadratic variaton leads to the more familiar $$ \sigma_f(t,T)=\sigma e^{-\lambda(T-t)} $$ that embeds the Vasicek model into the HJM framework.