Pricing a LIBOR Cash Flow with the Payment-Date Forward Measure

# Pricing of LIBOR based CF settled after the LIBOR fixing by switching from risk-neutral to forward-neutral measures


# Pricing of LIBOR based CF settled after the LIBOR fixing by switching from risk-neutral to forward-neutral measures












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 – forward neutral measure, it is equal to:

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

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

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

On the other hand, performing the change of measure using the time-T Radon-Nikodym derivative 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:

- Should the change of measure be done using or ?

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

- If it should be done using , how can we simplify the expression ?

Thanks in advance for your help.

## Answer by byouness (score 1)

https://quant.stackexchange.com/a/68530

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 $t \leq T$!

## Answer by finfree (score 0)

https://quant.stackexchange.com/a/68601

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

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

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

- 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.