How to derive the formula of a European Libor call option in a Libor Market Model?

Question

I am struggling with the following two mathematical statements. The first is from the book "Term-structure Models: A Graduate Course - Damir Filipović" Suppose we have a deterministic function $\lambda(t,T)$ and we assume the Libor rates $L(t,T)$ has a dynamic

$$dL(t,T)=L(t,T)\lambda(t,T)dW_t$$

where $W$ is a Brownian motion. Hence $L(t,T)$ is given as a stochastic exponential, i.e.

$$L(t,T)=L(s,T)e^{\int_s^t\lambda(u,T)dW_u-\frac{1}{2}\int_s^t\lambda(u,T)^2du}$$ for $s\le t\le T$. If we look at $\log{L(T,T)}=\log{L(s,T)}+\int_s^T\lambda(u,T)dW_u-\frac{1}{2}\int_s^T\lambda(u,T)^2du$. Note: I know that $M_t:=\int_0^t\lambda(u,T)dW_u$ is a martingale, it is normal distributed with mean 0 and variance $\int_0^t\lambda(u,T)^2du$. Now it should be true, that $E[\log{L(T,T)}|\mathcal{F}_t]$ should be gaussian with mean $\log{L(t,T)-\frac{1}{2}\int_t^T\lambda(u,T)^2du}$. Plugging the above into this conditional expectation, we get

$$E[\log{L(T,T)}|\mathcal{F}_t]=E[\log{L(s,T)}+\int_s^T\lambda(u,T)dW_u-\frac{1}{2}\int_s^T\lambda(u,T)^2du|\mathcal{F}_t]=\log{L(s,T)}+E[\int_s^T\lambda(u,T)dW_u|\mathcal{F}_t]-\frac{1}{2}\int_s^T\lambda(u,T)^2du$$ since the first term is $\mathcal{F}_t$ measurable and the last term is deterministic. Using that $M_t$ is a martingale, we end up with $$E[\log{L(T,T)}|\mathcal{F}_t]=\log{L(s,T)}+\int_s^t\lambda(u,T)dW_u-\frac{1}{2}\int_s^T\lambda(u,T)^2du$$ If we put $s=t$ we get $$E[\log{L(T,T)}|\mathcal{F}_t]=\log{L(t,T)}-\frac{1}{2}\int_t^T\lambda(u,T)^2du$$ Of course $\log{L(t,T)}$ would be Gaussian but with the wrong mean. How do I end up with the right distribution?

The second one is a more general question. It deals with calculating certain option type. For simplicity take again $L(t,T)$ as above. Suppose we want to calculate the following: $$E[(L(T,T)-K)^+|\mathcal{F}_t]$$ Without conditioning I have no trouble to calculate this, also for constant $\lambda(t,T)$. Assuming $\lambda(t,T)$ again deterministic but not constant, I would write $$E[(L(T,T)-K)^+|\mathcal{F}_t]=E[L(T,T)\mathbf1_{\{L(T,T)\ge K\}}|\mathcal{F}_t]-KE[\mathbf1_{\{L(T,T)\ge K\}}|\mathcal{F}_t]$$ I have absolutely no idea how to simplify these two expression. It is important for me to understand how you actually calculate this. I know the result, but I am interested in the derivation.

I am very thankful for your help, since this two question are bothering me now quite a while.

Answer 1

If you are able to derive quantities in the LMM with unconditional expectation as functions of $L(0,T)$ and $\lambda(s,T)$ for $0 \leq s \leq T$ then expression of the time $t$ conditional expectation is exactly the same function in $L(t,T)$ and $\lambda(s,T)$ for $t \leq s \leq T$ - just replace $0$ by $t$. This is due to the fact that the model is Markovian (the state in time $t$ if fully described by the initial value $L(t,T)$.

Put differently: You unconditional expectation is also a conditional, conditioned to $\mathcal{F}_0$.

This applies also to the state dependent option price in the second part of the question: they are just function of $L(t,T)$ and e.g., the integrated instanenous vol from $t to $T$.