HJM drift condition problem: Show that the HJM drift condition implies $b(t) \equiv b, \rho^{2}(t) \equiv a$

Question

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:

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: $h(x)=-\frac{a}{2} x^{2}+b x+c$.

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.