Correcting the HJM Measure Change Between Risk-Neutral and Forward Measures

# HJM framework: Girsanov transformation between forward measure and risk neutral measure


# HJM framework: Girsanov transformation between forward measure and risk neutral measure












Consider a HJM framework $$d f(t, T) = \sigma (t, T) d W_t^T$$ which is a SDE of instantaneous forward rates on $T$-forward measure, and let $$P (t, T) = \exp (-\int_t^T f (t, u) d u)$$ $$B (t) = \exp (\int_0^t f(u, u) d u)$$ be a $T$-discount bond price and a continuously compounded money market account.

By definition, \begin{align} P(t, T) &= \exp (-\int_t^T f(0, u) d u - \int_t^T \int_0^t \sigma (s, u) d W_s^T d u) \\ &= \exp (-\int_t^T f(0, u) d u - \int_0^t \int_t^T \sigma (s, u) d u d W_s^T) \end{align} and \begin{align} B(t) &= \exp (\int_0^t f(0, u) d u + \int_0^t \int_0^u \sigma (s, u) d W_s^T d u) \\ &= \exp (\int_0^t f(0, u) d u + \int_0^t \int_s^t \sigma (s, u) d u d W_s^T) \end{align} therefore \begin{align} Z(t, T) &:= B(t)^{-1} P(t, T) \\ &= \exp (-\int_0^T f(0, u) d u - \int_0^t \int_s^T \sigma (s, u) d u d W_s^T) \end{align}

By Ito's formula for the exponential function, $\frac{d \exp (X_t)}{X_t} = d X_t + \frac{1}{2} d X_t^2$ holds. By applying this to the previous result, $$\frac{d Z(t, T)}{Z (t, T)} = -b (t, T) d W_t^T + \frac{1}{2} b^2 (t, T) d t$$ where $$b(s, T) := \int_s^T \sigma (s, u) d u$$

Since $Z (t, T)$ is a price of a tradable $P(t, T)$ discounted by the money market account, it is a martingale under the risk neutral measure $\mathbb{Q}$. Thus, by changing the numeraire, \begin{align} \frac{d Z(t, T)}{Z (t, T)} &= -b (t, T) d W_t^T + \frac{1}{2} b^2 (t, T) d t \\ &= -b (t, T) d W_t^\mathbb{Q} \end{align} and it implies that $$d W_t^T = d W_t^\mathbb{Q} + \frac{1}{2} b (t, T) d t$$

However, many books say $$d W_t^T = d W_t^\mathbb{Q} + b (t, T) d t$$ which means I'm wrong somewhere. Can anyone help with this?

## Answer by ir7 (score 1)

https://quant.stackexchange.com/a/85626

Your Ito application to $\exp(X_t)$ is wrong (also you don't explicitly say how $X_t$ and $Z(t,T)$ relate):

$$ d\exp(X_t) = \exp(X_t) dX_t + 0.5 \exp(X_t) (dX_t)^2. $$

A clearer path is to first agree with the following (no T-forward measure):

\begin{align*} df(t,T) = \sigma(t,T) b(t,T) dt +\sigma(t,T) \,dW_t \end{align*} \begin{align*} f(t,T) = f(0,T) + \int_0^t \sigma(v,T)b(v,T)\,dv + \int_0^t \sigma(v,T)\,dW_v \end{align*} \begin{align*} P(t,T)=& \frac{P(0,T)}{P(0,t)}\exp\left(-0.5 \int_0^t (b(v,T)^2 - b(v,t)^2)\,dv\right.\\ &\left. - \int_0^t (b(v,T) - b(v,t))\,dW_v\right) \end{align*}

\begin{align*} B_t^{-1} = P(0,T) \exp \left( - 0.5 \int_0^t b(v,t)^2\,dv -\int_0^t b(v,t)\,dW_v \right) \end{align*}

Then, this gives us the exponential martingale needed for the density process in order to apply Girsanov's theorem: \begin{align*} \rho^{\frac{dQ^T}{dQ}}_t =& \frac{P(t,T)/P(0,T)}{B_t/B_0} \\ =& \exp \left( - 0.5 \int_0^t b(v,T)^2\,dv -\int_0^t b(v,T)\,dW_v \right) \end{align*} Consequently: $$dW^T_t = dW_t + b(t,T)dt, $$ is a Brownian motion in the $Q^T$ measure. Finally, we get $P$ and $B$ in terms of $W^T$: $$ P(t,T) = \frac{P(0,T)}{P(0,t)}\exp\left(-0.5\int_0^t b(v,t)^2 \, dv + 0.5\int_0^t b(v,T)^2 \, dv - \int_0^t (b(v,T) - b(v,t))dW_v^T\right) $$ $$ B_t^{-1} = P(0,t)\exp\left(-0.5\int_0^t b(v,t)^2 \, dv + \int_0^t b(v,t)b(v,T)\, dv - \int_0^t b(v,t)dW_v^T\right) $$

## Answer by nessy (score 0)

https://quant.stackexchange.com/a/85621

These equations are not true: \begin{align} P(t, T) = \exp (-\int_t^T f(0, u) d u - \int_t^T \int_0^t \sigma (s, u) d W_s^T d u) \end{align} \begin{align} B(t) = \exp (\int_0^t f(0, u) d u + \int_0^t \int_0^u \sigma (s, u) d W_s^T d u) \end{align}

The right equations are: \begin{align} P(t, T) = \exp (-\int_t^T f(0, u) d u - \int_t^T \int_0^t \sigma (s, u) d W_s^u d u) \end{align} \begin{align} B(t) = \exp (\int_0^t f(0, u) d u + \int_0^t \int_0^u \sigma (s, u) d W_s^u d u) \end{align}

This is where the things get wrong.