When deriving the LIBOR-based swap rate formula in any interest rate model, expressions of the following types appear naturally:


Literature tells us that, switching to the formula – forward neutral measure, it is equal to:


where the expressions formula and formula represent the time-t forward and time-T spot Libor rates respectively.

On one hand, performing the change of measure using the time-formula Radon-Nikodym derivative formula leads directly to the desired result.

However, it seems to me that we are not entitled to do so because formula is defined only up to time T and it wouldn’t make sense to apply formula to a security that is no longer defined at time formula, would it?

On the other hand, performing the change of measure using the time-T Radon-Nikodym derivative formula makes more sense to me, but leads to the presence of terms that I don't know how to simplify in the thus obtained forward-neutral expectation.

Hence the following questions:

  1. Should the change of measure be done using formula or formula ?

  2. If it should be done using formula, how is it compatible with the fact that formula is no longer defined fot t > T?

  3. If it should be done using formula, how can we simplify the expression formula ?

Thanks in advance for your help.

  • $\begingroup$ Hello finfree and welcome to SE. There are some typos in your formulas, the $\delta$ is sometimes there and sometimes not. In my answer below, I considered a flow of $L_T(T, T+\delta)$ paid at $T + \delta$. $\endgroup$
    – byouness
    Oct 26, 2021 at 20:55

2 Answers 2


The numéraire changing formula tells us that for a tradeable asset $X$, and two numéraires $N$ and $M$ we can write (by $\mathbb{E}^N$ I denote the expectation under the martingale measure associated to numéraire $N$): $$ N_t\times\mathbb{E}^N_t \left[\frac{X_{T_1}}{N_{T_1}} \right] = M_t\times\mathbb{E}^M_t \left[\frac{X_{T_1}}{M_{T_1}} \right] $$

This formula is not necessarily valid on $\sigma$-algebra $\mathcal{F}_{T_1}$! In general, it will be valid on a $\sigma$-algebra $\mathcal{F}_{T_2}$ with $T_2 \leq T_1$.

On the one hand, your LIBOR flow is $\mathcal{F}_T$-measurable, so we will work with $t \leq T$.

On the other hand, the natural numéraire to use is the zero-coupon bond with same maturity as your LIBOR flow: $T+\delta$. This is because the LIBOR flow can be seen as a basket of zero-coupon bonds, expressed in terms of a this numéraire: $$ L_T(T,T+\delta) = \frac{1}{\delta} \left(\frac{P_T^T - P_T^{T+\delta}}{P_T^{T+\delta}}\right) $$ In mathematical terms, this means that the LIBOR flow is a martingale under the measure associated to this numéraire.

So, applying the above formula for your LIBOR flow, one gets: $$ B_t \times \mathbb{E}_t^\mathbb{Q} \left[\frac{L_T(T, T+\delta)}{B_{T+\delta}} \right] = P_t^{T+\delta} \times \mathbb{E}_t^{\mathbb{Q}_{T+\delta}}\left[\frac{L_T(T, T+\delta)}{P_{T+\delta}^{T+\delta}} \right] $$ which simplifies to: $$ \begin{aligned} \mathbb{E}_t^\mathbb{Q} \left[ e^{-\int_t^{T+\delta} r(u)du }L_T(T, T+\delta) \right] &= P(t, T+\delta) \times \mathbb{E}_t^{\mathbb{Q}_{T+\delta}} \left[L_T(T, T+\delta)\right] \\ &= P(t, T+\delta) \times L_t(T,T+\delta) \end{aligned} $$

However, as noted above, this is valid only in $\mathcal{F}_T$, so only for $t \leq T$!

  • $\begingroup$ Hello byouness, as required, I deleted my initial comment posted erroneously as an answer and, with it, all the associated comments which followed... So, in your answer, I understand you prove the result by applying the changing-numeraire formula considering $N_t=B_t$, $M_t=P_t^{T+\delta}$, $X_t=L_t(T,T+\delta)$ and $T_1={T+\delta}$. Is it correct?! If so, could you explain why it is valid to apply a $T+\delta$-Radon-Nikodym derivative ($T_1={T+\delta}$) to $L_t(T,T+\delta)$ whereas this latter is defined only up to time T? Thanks. $\endgroup$
    – finfree
    Oct 31, 2021 at 10:08
  • $\begingroup$ You can see it this way, instead of considering $L_t(T, T + \delta)$, consider $X_t$ defined by: $X_t = L_t(T, T+\delta), \quad t \leq T$ and $X_t = L_T(T, T+\delta), \quad T \leq t \leq T + \delta$, and apply your reasoning as usual for $0 \leq t \leq T + \delta$, then you will get the result. I hope this helps ! $\endgroup$
    – byouness
    Oct 31, 2021 at 12:38
  • $\begingroup$ Thanks for your answer byouness! I had thought of this way of seeing things, but I must admit (and maybe I'm totally wrong about that) that I'm not fully convinced... (in fact that's the reason why I posted my question in the first place...) For this reason, and for the sake of completeness, I'll post another answer inspired by the exercise 10.12 from Shreve's book stochastic calculus for finance II and which allows to use the $\frac{dQ^{T+\delta}}{dQ}|_T$ Radon-Nikodym derivative instead of the $\frac{dQ^{T+\delta}}{dQ}|_{T+\delta}$ RN derivative to perform the needed change of measure. $\endgroup$
    – finfree
    Nov 1, 2021 at 12:19

here is a solution inspired by the exercise 10.12 from Shreve's book "stochastic calculus for finance II":

  1. change of measure, the trick consisting in applying the law of iterated expectation:

$$E_t^Q\left[e^{-\int_{u=t}^{T+\delta}r_udu}\times L_T(T,T+\delta)\right]=E_t^Q\left[E_T^Q\left[e^{-\int_{u=t}^{T+\delta}r_udu}\times L_T(T,T+\delta)\right]\right]=E_t^Q\left[e^{-\int_{u=t}^{T}r_udu}\times L_T(T,T+\delta)\times E_T^Q\left[e^{-\int_{u=T}^{T+\delta}r_udu}\right]\right]=E_t^Q\left[\frac{B_t}{B_T}\times L_T(T,T+\delta)\times P_T^{T+\delta}\right]=E_t^Q\left[\left(\frac{B_t}{B_T}\times \frac{P_T^{T+\delta}}{P_t^{T+\delta}}\right)\times L_T(T,T+\delta)\right]\times P_t^{T+\delta}=E_t^Q\left[\frac{dQ^{T+\delta}}{dQ}|_T\times L_T(T,T+\delta)\right]\times P_t^{T+\delta}=E_t^{Q_{T+\delta}}\left[L_T(T,T+\delta)\right]\times P_t^{T+\delta}$$

  1. proof that $L_t(T,T+\delta)$ is a martingale under the $T+\delta$-forward measure:

$$1+\delta L_t(T,T+\delta)=\frac{P_t^T}{P_t^{T+\delta}}=\frac{E_t^Q\left[\frac{B_t}{B_T}\right]}{P_t^{T+\delta}}=E_t^Q\left[\left(\frac{B_t}{B_T}\times \frac{P_T^{T+\delta}}{P_t^{T+\delta}}\right)\times \frac{P_T^T}{P_T^{T+\delta}}\right]=E_t^Q\left[\frac{dQ^{T+\delta}}{dQ}|_T\times \left(1+\delta L_T(T,T+\delta)\right)\right]=E_t^{Q_{T+\delta}}\left[1+\delta L_T(T,T+\delta)\right]$$

we thus have:  $L_t(T,T+\delta)=E_t^{Q_{T+\delta}}\left[L_T(T,T+\delta)\right]$

Combining the both equalities leads to the result.


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