I need your help with understanding and solving the HJM framework. I am hoping I can get some help as I feel so lost with HJM and learning online because of the pandemic is adding more stress. Anyway this is the problem:


An HJM forward curve evolution by parallel shifts is then of the form $$ f(t, T)=h(T-t)+Z(t) $$ \begin{aligned} &\text { for some deterministic initial curve } f(0, T)=h(T) \text { and some Itô process } d Z(t)=\\ &b(t) d t+\rho(t) d W^*(t) \text { with } Z(0)=0 \end{aligned}

Show that the HJM drift condition implies $b(t) \equiv b, \rho^{2}(t) \equiv a$, and $$ h(x)=-\frac{a}{2} x^{2}+b x+c $$ for some constants $a \geq 0$, and $b, c \in \mathbb{R}$.

My attempt:

We take the derivative of $f(t,T)$

$$d f((t, T))=\left(-h^{\prime}(T-t)+b(t)\right) d t+\rho(t) d w^{*}(t)$$

The Q dynamics of the forward rates of the HJM framework is in the form of:

$$f(t, T)=f(0, T)+\int_{0}^{t}\left(\sigma(s, T) \int_{S}^{T} \sigma(s, u) d u\right) d s+\int_{0}^{t} \sigma(s, T) d u_{t}^{*}$$

$$d f(t, T)=\sigma(t, T) \int_{t}^{T} \sigma(t, u) d u+\sigma\left(s, T\right) d \omega_{t}^{*}$$

Hence the HJM drift equals:

$$d f(t, T)=\rho(t) \int_{t}^{T} \rho(t) d u$$

$$d f(t, T)=\rho^{2}(t)(T-t)$$

Setting $x = T- t$ we get:

$\rho^{2}(t) x=-h(x)+b(t)$ <- not sure about this part.

Taking the derivative with respect to $x$ on both sides we get:

$$\rho^{2}(t)=-h^{\prime \prime}(x)$$

Setting $x = 0$ we have:

$$\rho^{2}(t)=-h^{\prime \prime}(0)=a $$

Now to show that $$b(t) \equiv b$, we know $\rho^{2} = a$ which is a constant hence:

$$a \cdot x=-h(x)+b(t)$$

Setting $x = 0$, we get: $b(t)=h(0)=b$.

I am not sure how to show this part: $h(x)=-\frac{a}{2} x^{2}+b x+c$.


1 Answer 1


Fixing again some typos of yours, we know that in HJM under the risk-neutral measure $$ f(t, T)=f(0, T)+\int_0^t\left(\sigma(s, T) \int_s^T \sigma(s, u) \,du\right)\,ds+\int_0^t \sigma(s,T)\,dW_t^* $$ always holds. This implies $$ h(T-t)+\int_0^tb(s)\,ds=f(0,T)+\int_0^t\left(\sigma(s, T) \int_s^T \sigma(s, u) \,du\right)\,ds\,. $$ Taking the derivative w.r.t. $t$ gives $$ -h'(T-t)+b(t)=\sigma(t,T)\int_t^T\sigma(t,u)\,du\,. $$ Using $\sigma(t,T)=\rho(t)$ gives $$ -h'(T-t)+b(t)=\rho^2(t)\,(T-t)\,. $$ Writing $x=T-t$ gives $$ -h'(x)+b(t)=\rho^2(t)\,x\,,\quad\quad x,t\ge 0\,.\quad\quad\quad(1) $$ This implies the two identities: $$ h(x)=-\rho^2(t)\frac{x^2}{2}+b(t)\,x+c\,, $$ and $$ -h''(x)=\rho^2(t)\,. $$ It follows that $\rho$ cannot depend on $t$ (must be constant). From (1) it follows now also that $b$ must be constant.

  • $\begingroup$ Thank you so much for replying. I am a bit confused about this part: $h(T-t)+\int_{0}^{t} b(s) d s=f(0, T)+\int_{0}^{t}\left(\sigma(s, T) \int_{s}^{T} \sigma(s, u) d u\right) d s$. Do you take $h(T-t)$ equals to the left hand side because it is a constant like $f(0,T)$ or because if u take the derivative wrt to t, it will be a $dt$ term? $\endgroup$ Aug 25, 2021 at 2:09
  • 1
    $\begingroup$ You started with $f(t,T)=h(T-t)+Z(t)\,.$ Just equate this to the equation you got from the HJM framework (first eq. in my answer). $\endgroup$
    – Kurt G.
    Aug 25, 2021 at 8:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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