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.

  • $\begingroup$ I have edited your post: since you have $+\frac{1}{2}\sigma_B^2\left(s, T\right)ds + \sigma_B\left(s, T\right)dW_T$ in the bond price dynamics, the corresponding term must be negative in the forward rate dynamics. Not only does it express the negative correlation between bond price and forward rate, but it also has its importance for your problem (more in my comment below). $\endgroup$ – siou0107 Dec 13 '19 at 13:34
  • $\begingroup$ Changed $$\frac{dB(t,T)}{B(t,T)}=(r_t + \sigma_B^2(t,T) ) dt+σ_B (t,T)dW_Q$$ to $$\frac{dB(t,T)}{B(t,T)}=(r_t + \sigma_B^2(t,T) ) dt+σ_B (t,T)dW_T.$$ $\endgroup$ – Gordon Dec 13 '19 at 14:56

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*}

  • $\begingroup$ From your Brownian conversion, you can see that there is a $T$ sensitivity in $W_T$. Hence, you have a "$-d\partial_TW_T$" (any more correct notation is welcome!) term which is $+\partial_T\sigma_B(t, T)$. This cancels the drift of your forward rate under the $T$-forward measure. $\endgroup$ – siou0107 Dec 13 '19 at 12:21
  • $\begingroup$ Thanks @siou0107: Changed the signs. $\endgroup$ – Gordon Dec 13 '19 at 14:55

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