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

I am struggling with the following two mathematical statements. The first is from the book "Term-structure Models: A Graduate Course - Damir FilipoviÄ‡" 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.

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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$. -  I am very thankful, that you are answering my questions frequently! However I am not very familiar with the Markov property (MP). I know the following definition of the MP: A process$X$is said to have the MP if$E[f(X_t)|\mathcal{F}_t]=E[f(X_t)|\sigma(X_s)]$for bounded and measurable$f$. My first question would be why do you see that easily, that$X_t:=L(t,T)$satisfies this? Furthermore for the calculation I do not see how this simplifies things:$E[\log{L(T,T)}|\mathcal{F}_t]=E[\log{L(T,T)}|\sigma(L(t,T))]$. Also for the second question. Moreover I really do not see... – hulik Feb 8 at 9:36 ...where my mistake is in the derivation of$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)^2dt$No matter what i put in for$s$, I do not get the right distribution (with the right mean and variance) for$\log{L(T,T)}$condition on$\mathcal{F}_t$. As I said, this question is bothering me now quite a while ago and I am really interested in understanding it properly. Again thanks for your help and patience. – hulik Feb 8 at 9:41 W.r.t. the Markov property you just have to check: Does the future$T > t$depend on the past$s < t$or is the knowledge of your state variables in$t$enough. In the SDE of the LMM the initial value is L(t), the drift depends on$L(t)$or$L(\tilde{t})$for$\tilde{t} > t$etc., but nothing depends on the past. (There is a little excercise you could try: if you write the LMM as a short rate model, you see that the drift depends not only on$r$, it also depends on that past (to reconstruct the other forward rates). So with respect to$r$LMM is not Markovian. – Christian Fries Feb 8 at 21:35 Ah I see...thanks for explaining the MP in this context. The thing I really think is wrong in the book is the following: They say$Z:=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)dW(u)$. The conditional expectation$E[\log{L(T,T)}|\mathcal{F}_t]$is a r.v.$Z$and this r.v. should be Gaussian with the right mean$\log{L(t,T)} -\frac{1}{2}\int_t^T\lambda(u,T)dW(u)$. However the mean should be a real number! But here appears a stochastic quantity:$\log{L(t,T)}$. I do not understand what mean should be in this context. – hulik Feb 9 at 8:48 The conditional expectation, conditional to$\mathcal{F}_t$, is a random variable. It is$\mathcal{F}_t$-measurable. It is a real number once you look at this random variable at a specific state, namely the states you know at time$t$. The$\log L(t,T)$is just that state... (maybe you like to look at the picture on page 18 and the interpretation on page 17 and 18 here: books.google.de/books?id=q7c8Pi6QGFQC ) - - The expectation conditional to$\mathcal{F}_{0} = \{ \emptyset, \Omega \}$is also a random variable, but it is constant (same for every state =$\mathcal{F}_{0}\$-measurable) – Christian Fries Feb 9 at 16:47