# HJM framework problem - showing that HJM drift condition implies that $b(z)=b+βz$ and $(ρ)^2=α$

Hi I am looking for some general clarification to Heath–Jarrow–Morton framework. I am analyzing a problem where the forward rate is modeled as $$f(t,T)=e^{\beta(T-t)} Z_t+h(T-t) \tag{1}$$ for some constant $\beta$ and some smooth function $h:R \rightarrow \ R \ with \quad h(0)=0$ and some scalar diffusion $$dZt=b(Z_t)dt+\rho dW_t^* \tag{2}$$ with continuous drift and functions $b(z)$ and $\rho(z)$.

To show that HJM drift condition implies that $\ \mathbf{b(z)=b+\beta z} \$ and $\ \mathbf{(\rho)^2=\alpha} \$ for some constant b and $\alpha$

Q1
Can anybody explain in simple terms what this above statement mean? what needs to be showed/proved? I am confused here what is the relation between the variables given in the equations (1) and (2) with those $\alpha$ and b.

There is the analyzes I have done on this problem. I understand the most of the transformations, but some points are unclear.

Taking the derivative of $f(t,T)$ in order to get $df(t,T)$ and substituting for $dz_t$ I get $$df(t,T)=\big{(} -h'(T-t)+e^{\beta(T-t)}(b(z_t)-\beta z) \big{)} \ dt+e^{\beta(T-t)} \rho(z_t) dW_t^* \tag{3}$$ the Q-dynamics of the forward rates in HJM framework are of the form $$f(t,T)=f(0,T) + \int_0^t \big{(} \sigma(s,T) \int_s^T \sigma(s,u) \ du \big{)} \ ds + \int_0^t \sigma(s,T) dW_t^* \tag{4}$$ or equivalently $$df(t,T)= \sigma(t,T) \int_t^T \sigma(t,u) \ du + \sigma(s,T) dW_t^*\tag{5}$$ therefore the HJM drift equals $$e^{\beta(T-t)} \rho(z_t) \int_t^T e^{\beta(u-t)} \rho(z_t) \ du =\rho(z_t)^2 e^{\beta(T-t)} \frac{ e^{-\beta t}}{\beta} (e^{\beta T} - e^{\beta t}) \tag{6}$$ equating HJM drift with drift from eq (3) and substituting for $\tau=T-t$ yields the following equity $$\frac{\rho(z_t)^2}{\beta} e^{\beta \tau} (e^{\beta \tau} - 1) = -h'(\tau)+e^{\beta \tau}(b(z_t)-\beta z) \tag{7}$$ assuming $\tau=0$ results in $$b(z)=h'(0)+\beta z \tag{8}$$ comparing with the given equation $$b(z)=b+ \beta z \tag{9}$$ we can infer that $b=h'(0)$ finally plugging $b(z)$ (8) into (7) gives the $alpha$ value $$\rho^2= \frac{\beta(e^{\beta \tau} h'(0)- h'(t))}{e^{\beta \tau}(e^{\beta \tau}-1)} \tag{10}$$ $$\alpha=\rho^2$$

Further questions to this problem
b and $\alpha$ have been derived from the drift condition. I don't understand in what sense does it prove the initial question?
Does it implicitly mean that the RHS of (10) is a constant?
what is the relation between the HJM drift and the coefficient of $dW_t^ *$ in (3)?

• Should $h(T-1)$ be $h(T-t)$ in (1)? – Gordon May 9 '16 at 17:57
• @Gordon , you are right, it should be h(T-t) there I corrected the typo, sorry – Michal May 9 '16 at 20:59
• I made some further typo corrections. – Gordon May 9 '16 at 22:44

You almost get there. However, you ca not conclude that $\rho^2$ is a constant based on $(10)$. Note that, from your $(7)$ and $(8)$, \begin{align*} \frac{\rho(z_t)^2}{\beta} e^{\beta \tau} (e^{\beta \tau} - 1) = -h'(\tau)+e^{\beta \tau}h'(0). \end{align*} Taking derivative with respect to $\tau$ on both sides, we obtain that \begin{align*} \frac{\rho(z_t)^2}{\beta} \left(2\beta e^{2\beta \tau} -\beta e^{\beta \tau}\right) = -h''(\tau)+\beta e^{\beta \tau}h'(0). \end{align*} Then, setting $\tau=0$, \begin{align*} \rho(z_t)^2 = -h''(0)+\beta h'(0). \end{align*} That is, $\rho(z_t)^2=\alpha$ is a constant, where $\alpha = -h''(0)+\beta h'(0)$.