Why Forward Rates Are Martingales Under the T-Forward Measure

# Forward rates are martingale under the T-forward measure


# Forward rates are martingale under the T-forward measure












Forward rates are martingale under the $T$-forward measure but this derivation is suggesting otherwise. Could anyone please point out the mistake ?

Let $dW_Q$ be a Brownian Motion in the risk neutral measure. Let $B(t,T)$ be a bond starting at time $t$ and paying at $T$. Assume: $$\frac{dB(t,T)}{B(t,T)}=r_t dt+σ_B (t,T)dW_Q$$ Converting the Brownian into corresponding one in $T$-forward measure: $$dW_T = dW_Q - σ_B (t,T)dt$$ Therefore in $T$-forward measure: $$\frac{dB(t,T)}{B(t,T)}=(r_t + \sigma_B^2(t,T) ) dt+σ_B (t,T)dW_T$$ By Ito's lemma: $$\ln(B(t,T)) = \ln(B(0,T)) + \int_0^t[r(s)+\sigma_B^2(s,T)-\frac{1} {2}\sigma_B^2(s,T)]ds + \int_0^t\sigma_B(s,T)dW_T$$ Now, $f(t,T)=-\frac{\partial \ln(B(t,T))}{\partial T}$ hence: $$f(t,T)=f(0,T) - \int_0^t[\sigma_B(s,T).\partial_T\sigma_B(s,T)]ds -\int_0^t\partial_T\sigma_B(s,T)dW_T$$ Thus: $$df(t,T) = -\sigma_B(s,T).\partial_T\sigma_B(s,T)dt -\partial_T\sigma_B(s,T)dW_T$$

which is not driftless and therefore is not coming out to be a martingale.

## Answer by Gordon (score 4)

https://quant.stackexchange.com/a/50263

Note that, the Brownian motion $W_T$ also depends on $T$. When you are taking the derivative with respect to $T$, you also need to consider the change of the Brownian motion family.

In order to avoid such difficulty, what you can do is to first derive the dynamics under the risk-neutral measure, and then change to the $T$-forward measure. Specifically, along with the line of your derivation, under the risk-neutral measure, \begin{align*} df(t, T) = \sigma_B(s,T).\partial_T\sigma_B(s,T)dt -\partial_T\sigma_B(s,T)dW_Q. \end{align*} Then, from the numeraire change, $W_T$ is a Brownian motion under the $T$-forward measure $P_T$, where $dW_T= dW_Q-\sigma_B(t, T)dt$. Moreover, under $P_T$, \begin{align*} df(t, T) = -\partial_T\sigma_B(s,T)dW_T. \end{align*}