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

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.

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.
• Thank you for helping me understand my typos and errors. I still a confused about this part of this part of the question: $[f(\cdot, T), f(\cdot, T)]_{t}=\sigma e^{-\lambda(T-t)} \text { (a) What is the volatility of the forward rate? }$ So according to your answer, the volatility of the forward rate is just the function $\sigma_{P}(t, T)$ of the zero coupon bond? Aug 28, 2021 at 1:36
• I check again, the problem states $[f(\cdot, T), f(\cdot, T)]_{t}=\sigma e^{-\lambda(T-t)}$. I'm still somewhat confused, wouldn't this be the drift term $\sigma_{f}^{2}(t, T)=\frac{d}{d t} \sigma e^{-\lambda(T-t)}=\sigma \lambda e^{-\lambda(T-t)}$ as it is the $dt$ term in the HJM model? Aug 29, 2021 at 11:13
• We know that in the HJM model the drift and volatility term are closely related under the risk-neutral measure. There, the drift of $f(t,T)$ is $\alpha(t,T)=-\sigma_f(t,T)\int_t^T\sigma_f(t,u)\,du\,.$ Aug 29, 2021 at 13:09