Exercises 8.1 Markov LMM As seen in this chapter, the MC simulation of a LMM is time-consuming.

The main problem comes from the drifts which are reminiscent of the non-

Markovian nature of the HJM model. To circumvent this diculty, one can try to nd a low-dimensional Markov representation of the LMM. By this, we assume that each Libor can be written as a functional of the time t and a low-dimensional It^o diusion process Yt Li  Li(t; Yt) ; i = 1;


; n In this problem, we make the additional assumption that Yt is a one dimensional process.

The process follows a general It^o SDE dY = (t; Y )dt + (t; Y )dWt 1: Prove that it is always possible to assume that the process Yt can be written as Yt = f(t;Xt) with Xt a driftless process. By redening the functional Li, we have Li = Li(t;Xt).

Note: The process Xt has no nancial interpretation. However, for a short rate model, Xt can be seen as the instantaneous interest rate rt.
2: For an ane short-rate model such as the HW2 model (see 8.2), obtain the explicit map Li(t; xt; yt).
3: Prove that the Markovian property imposes from (8.6) the following constraint i = (t;X)@XLi (8.27)
????(t;X)2@XLi Xn j=i+1 j@XLj 1 + jLj = @tLi +
1 2(t;X)2@2X Li (8.28)
In the following, we explicitly check that the BGM model does not have a Markov representation. The Libors are driven by a log-normal process i = i(t)Li.

The equations (8.27,8.28) reduce to i(t) = (t;X)@X lnLi (8.29)
????(t;X)2@XLi Xn j=i+1 j@XLj 1 + jLj = @tLi +
1 2(t;X)2@2X Li (8.30)
Without loss of generality, we take Ln = X.
4: Prove that the rst constraint (8.29) is equivalent to Li = i(t)L n(t)
i(t)
n 5: Using this solution, prove that the second constraint (8.30) reduces ???? 2n i(t)
Xn j=i+1 jj(t)L n(t)
j (t)
n 1 + jj(t)L n(t)
j (t)
n = i(t)0 + i(t)

n(t)
i(t)
0 L????1 n +
1 2 n(t)2i(t)

n(t)
i(t)
2 6: Specifying this equation for i = n ???? 1, show that this equation has only a solution in the degenerate case n????1(t) = 0.
So, although every short-rate models admit a Markovian representation by construction, it is not the case for a particular LMM. To be able to do that, we must assume that the volatility of each Libor is a general Dupire local volatility i(t;Li). In the following, we explain how to construct a MF introduced rst in [110].
In order to be able to simulate exactly the process Xt, we take Xt as a Gaussian process dX = (t)dWt (8.31)
with the initial condition X0 = 0.
We introduce the de ated bond i(t) = PtTi PtTn i = 1;


n We recall that a Libor Li(t) is given by iLi(t) = i????1(t)
i(t) ???? 1 (8.32)
7: Prove that i  i(t;X) satises the PDE @ti(t;X) +
1 2(t)2@2X i(t;X) = 0 8: Doing a change of local time, prove that this is equivalent to the heat kernel on R and the fundamental solution is i(t0;X) = Z 1 ????1 i(T0 i ; Y )pG(T0 i ; Y jt0;X)dY (8.33)
where t0 < T0 i .
Therefore the process i(t;X) is completely characterized if we specify its terminal value at t = Ti. Its terminal value will be determined in order to calibrate exactly the implied volatility of a unique swaption starting at Ti.
9: Prove that the fair value of a digital swaption option with an underlying swap s is in the measure Pn D (K) = ????PtTnEP n[
X i=+1 ii(T;X)1(s (T;X) > K)jFt]
(8.34)
10: Assuming that s (T;X) is a monotone function in X, prove that the digital fair value (8.34) becomes D (K) = ????PtTn X i=+1 i Z 1 X
i(T;X)pG(T0 ;Xj0)dX (8.35)
11: In particular for = n ???? 1, = n, prove that (8.35) becomes Dn????1;n(K) = ????PtTnn(1 ???? N(X))
with N(X) the cumulative Gaussian distribution. Given the caplet implied volatility BS n????1;n(K), we can determine X as a function of K by Ln(Tn????1;X) = K
12: Now, let us assume that i(t;X) is known for 8i > . Then, identifying in (8.35) D (K) with the market fair value, we determine K for a given X.
We obtain s (T;X) = K
13: Prove the relation (T;X) = (T;X) + K X i=+1 ii(T;X) (8.36)
By integrating (8.33) with the terminal condition (T;X) given by (8.36), we obtain (t;X) for all t  T and the Libors via (8.32).

that the terminal correlation of two swap rates is approximatively < ln si(Ti); ln si(Tj) >
vuut R Ti 0 (s)2ds R Tj 0 (s)2ds By taking (t) = eat where a is called the mean-reverting coecient (see question 2.), we have in particular < ln si(Ti); ln si(Tj) >
s e2aTi ???? 1 e2aTj ???? 1 Exercises 8.2 Almost Markov LMM For a generic LV LMM, the Libor dynamics in the spot Libor measure Ps is dLi(t) = i(t)i(Li)
Xi j=(t)
k;j(t)jj(t)j(Lj)
1 + jLj(t) dt +i(t)i(Li)dZi R1: To get rid of the local volatility i(Li), we dene the new variables qi(t) = Li L0i dx i(x) . Show that in these new variables, the SDEs above become dqi(t) = i(t)

i(t;L)dt ????
1 2i(t)0 i(Li)dt + dZi 
The diculty in making this SDE Markovian comes from the drift term. The simplest way to deal with this problem is to replace it with its value at the spot Libor dqi(t) = i(t)

i(t;L0)dt ????
1 2i(t)0 i(L0i )dt + dZi 
(8.37)
This is equivalent in spirit to the Hull-White-Rebonato freezing argument.
2: Integrate (8.37) to qi(t) = di(t) +
Z t 0 i(t)dZi with di(t) a deterministic function.
3: Deduce that a Libor Li can be written as a function of a time-changed Brownian motion Li(t) = Li(t;Zi;t0 )

4: Assuming the separability condition of the volatility i(t)
i(t) = i(t)
prove that the Libors fLig can be mapped to a unique time-changed Brownian motion Li(t) = Li(t;Wt0 )