Forward LIBOR Dynamics¶

This section explores the principles and methods underlying forward libor dynamics, which form a critical component of modern financial mathematics.

Key Concepts¶

The fundamental concepts in this area include:

Theoretical foundations and mathematical framework
Key definitions and notation
Important theorems and results
Connections to other areas of financial mathematics

Learning Objectives

After completing this section, you should understand:

The core mathematical principles and their financial interpretations
How these concepts connect to practical applications
The relationship between theory and numerical implementation

QuantPie Derivation: LIBOR Market Model Forward Measure¶

Definition of the LIBOR Rate¶

\[\begin{array}{ccccccccccccccc} \displaystyle l_i(t)&=&l&\left(\right.&t&;&T_{i-1}&,&T_{i}&\left.\right)\\ &&&&\uparrow&&\uparrow&&\uparrow&\\ &&&&\text{Now}&&\text{Reset Date}&&\text{Maturity}&\\ &&&&&&\text{Fixing Date}&&&\\ &&&&&&\text{Expiry Date}&&&\\ \end{array}\]

ZCB and LIBOR Relationship¶

\[\begin{array}{ccccccc} \displaystyle P\left(t,T_{i}\right) &=& \displaystyle P\left(t,T_{i-1}\right) &\displaystyle \frac{1}{1+\tau_i l\left(t;T_{i-1},T_{i}\right)}\\ \uparrow&&\uparrow&\uparrow\\ \text{Discount from $T_i$ to $t$}&&\text{Discount from $T_{i-1}$ to $t$}&\text{Discount from $T_i$ to $T_{i-1}$}\\ \text{Observed today $t$}&&\text{Observed today $t$}&\text{Observed today $t$}\\ \end{array}\]

Forward LIBOR Rate¶

\[\begin{array}{lll} \displaystyle l_i\left(t\right) &=&\displaystyle l\left(t;T_{i-1},T_{i}\right)\\ &=&\displaystyle \frac{1}{\tau_i} \left(\frac{P\left(t,T_{i-1}\right)-P\left(t,T_{i}\right)}{P\left(t,T_{i}\right)}\right)\\ &=&\displaystyle \frac{1}{\tau_i} \left(\frac{P\left(t,T_{i-1}\right)}{P\left(t,T_{i}\right)}-1\right)\\ \end{array}\]

Dynamics of the LIBOR Rate¶

\[\begin{array}{ccccccc} \displaystyle dl_i(t) &=& \displaystyle \bar{\mu}_i^\mathbb{P}(t)dt+\bar{\sigma}_i(t)dW_i^\mathbb{P}(t) \end{array}\]

\[\begin{array}{ccccccc} \displaystyle dW_i^\mathbb{P}(t)dW_j^\mathbb{P}(t) =\rho_{ij}dt \end{array}\]

LIBOR is an T_i-Martingale¶

Since $P\left(t,T_{i-1}\right)-P\left(t,T_{i}\right)$ is a price of tradable assets, $l_i\left(t\right)$ is an $T_i$-martingale.

\[\begin{array}{ccccccc} \displaystyle \mathbb{E}^{T_i}\left[\frac{P\left(T_{i-1},T_{i-1}\right)}{P\left(T_{i-1},T_{i}\right)}\Big{|}{\cal F}(t)\right] &=& \displaystyle \frac{P\left(t,T_{i-1}\right)}{P\left(t,T_{i}\right)} \end{array}\]

\[\begin{array}{ccccccc} \displaystyle \mathbb{E}^{T_i}\left[1+\tau_il(T_{i-1};T_{i-1},T_i)\Big{|}{\cal F}(t)\right] &=& \displaystyle 1+\tau_il(t;T_{i-1},T_i) \end{array}\]

\[\begin{array}{ccccccc} \displaystyle \mathbb{E}^{T_i}\left[l(T_{i-1};T_{i-1},T_i)\Big{|}{\cal F}(t)\right] &=& \displaystyle l(t;T_{i-1},T_i) \end{array}\]

LIBOR Dynamics in Forward and Other Measures¶

Under the $T_i$-forward measure:

\[\begin{array}{ccccccc} \displaystyle dl_i(t) = \bar{\sigma}_i(t)dW_i^i(t) \end{array}\]

Under a different forward measure $T_j$:

\[\begin{array}{ccccccc} \displaystyle dl_i(t) = \bar{\mu}_i^j(t)dt+\bar{\sigma}_i(t)dW_i^j(t) \end{array}\]

Lognormal LIBOR Market Model¶

Assume constant proportional volatility:

\[\begin{array}{ccccccc} \displaystyle \bar{\sigma}_i(t) = \sigma_i(t)l_i(t) \quad\Rightarrow\quad \frac{dl_i(t)}{l_i(t)} = \sigma_i(t)dW_i^i(t) \end{array}\]

Change of Measure: Radon-Nikodym Derivative¶

\[\begin{array}{ccccccc} \displaystyle \lambda_i^{i-1}(t) &=&\displaystyle \left.\frac{d\mathbb{Q}^{i-1}}{d\mathbb{Q}^{i}}\Big{|}{\cal F}(t)\right)\\ &=&\displaystyle \frac{P(t,T_{i-1})/P(t_0,T_{i-1})}{P(t,T_{i})/P(t_0,T_{i})}\\ &=&\displaystyle \frac{P(t_0,T_{i})}{P(t_0,T_{i-1})}(\tau_i l_i(t)+1)\\ \end{array}\]

\[\begin{array}{ccccccc} \displaystyle d\lambda_i^{i-1}(t) &=&\displaystyle \frac{P(t_0,T_{i})}{P(t_0,T_{i-1})}\tau_i dl_i(t)\\ &=&\displaystyle \frac{P(t_0,T_{i})}{P(t_0,T_{i-1})}\tau_i \bar{\sigma}_i(t)dW_i^i(t)\\ &=&\displaystyle \frac{\lambda_i^{i-1}(t)}{\tau_il_i(t)+1}\tau_i\bar{\sigma}_i(t)dW_i^i(t)\\ &=&\displaystyle \lambda_i^{i-1}(t)\frac{\tau_i\bar{\sigma}_i(t)}{\tau_il_i(t)+1}dW_i^i(t)\\ \end{array}\]

\[\begin{array}{ccccccc} \displaystyle \frac{d\lambda_i^{i-1}(t)}{\lambda_i^{i-1}(t)} =\frac{\tau_i\bar{\sigma}_i(t)}{\tau_il_i(t)+1}dW_i^i(t)\\ \end{array}\]

Girsanov Transformation¶

\[\begin{array}{ccccccc} \displaystyle dW_i^{i-1}(t) = -\frac{\tau_i\bar{\sigma}_i(t)}{\tau_il_i(t)+1}dt +dW_i^i(t) \end{array}\]

LIBOR Under Different Measures¶

\[\begin{array}{llllll} \displaystyle dl_i(t) &=&\displaystyle \bar{\sigma}_i(t)dW_i^i(t)\\ &=&\displaystyle \bar{\sigma}_i(t)\left(\frac{\tau_i\bar{\sigma}_i(t)}{\tau_il_i(t)+1}dt +dW_i^{i-1}(t)\right)\\ &=&\displaystyle \bar{\sigma}_i(t)\frac{\tau_i\bar{\sigma}_i(t)}{\tau_il_i(t)+1}dt +\bar{\sigma}_i(t)dW_i^{i-1}(t)\\ \end{array}\]

Terminal Measure¶

For the terminal measure $\mathbb{Q}^m$ with numeraire $P(t, T_m)$:

\[\begin{array}{lllll} \displaystyle dW_i^{i}(t) &=&\displaystyle -\frac{\tau_{i+1}\bar{\sigma}_{i+1}(t)}{\tau_{i+1}l_{i+1}(t)+1}dt +dW_i^{i+1}(t)\\ &=&\displaystyle -\frac{\tau_{i+1}\bar{\sigma}_{i+1}(t)}{\tau_{i+1}l_{i+1}(t)+1}dt +\left(-\frac{\tau_{i+2}\bar{\sigma}_{i+2}(t)}{\tau_{i+2}l_{i+2}(t)+1}dt +dW_i^{i+2}(t)\right)\\ &=&\displaystyle -\sum_{k=i+1}^{i+2}\frac{\tau_{k}\bar{\sigma}_{k}(t)}{\tau_{k}l_{k}(t)+1}dt +dW_i^{i+2}(t)\\ &=&\displaystyle -\sum_{k=i+1}^{m}\frac{\tau_{k}\bar{\sigma}_{k}(t)}{\tau_{k}l_{k}(t)+1}dt +dW_i^{m}(t)\\ \end{array}\]

Exercises¶

Exercise 1. Consider two forward LIBOR rates $l_1(t)$ and $l_2(t)$ on an annual grid ($T_0 = 0, T_1 = 1, T_2 = 2$). Under the $T_2$-forward measure, $l_1(t)$ is a martingale. Write down its dynamics $dl_1(t) = \bar{\sigma}_1(t)\,l_1(t)\,dW_1^2(t)$. Now express the dynamics of $l_1(t)$ under the $T_1$-forward measure using the Girsanov drift adjustment. What is the sign of the drift?

Solution to Exercise 1

Under the $T_2$-forward measure, $l_1(t)$ is a martingale (since $T_2$ is the maturity of the bond associated with the second forward rate). In the lognormal LMM, the dynamics are:

\[ dl_1(t) = \sigma_1(t)\,l_1(t)\,dW_1^2(t) \]

where $W_1^2$ is a Brownian motion under $\mathbb{Q}^{T_2}$, and $\bar{\sigma}_1(t) = \sigma_1(t)\,l_1(t)$.

To express the dynamics under the $T_1$-forward measure, we use the Girsanov transformation. From the change-of-measure formula:

\[ dW_1^{1}(t) = -\frac{\tau_2\,\bar{\sigma}_2(t)}{\tau_2\,l_2(t) + 1}\,dt + dW_1^{2}(t) \]

Wait --- this relates $W^1$ to $W^2$ for the second Brownian component. For the first rate's Brownian motion, we need the change from $\mathbb{Q}^{T_2}$ to $\mathbb{Q}^{T_1}$. Since $T_1 < T_2$, moving from $\mathbb{Q}^{T_2}$ to $\mathbb{Q}^{T_1}$, the Girsanov kernel is:

\[ dW_1^{1}(t) = dW_1^{2}(t) + \frac{\tau_2\,\rho_{12}\,\bar{\sigma}_2(t)}{\tau_2\,l_2(t) + 1}\,dt \]

Substituting $dW_1^2(t) = dW_1^1(t) - \frac{\tau_2\,\rho_{12}\,\bar{\sigma}_2(t)}{\tau_2\,l_2(t)+1}\,dt$ into the dynamics of $l_1$:

\[ dl_1(t) = \bar{\sigma}_1(t)\,dW_1^2(t) = \bar{\sigma}_1(t)\left(dW_1^1(t) - \frac{\tau_2\,\rho_{12}\,\bar{\sigma}_2(t)}{\tau_2\,l_2(t)+1}\,dt\right) \]

\[ dl_1(t) = -\frac{\bar{\sigma}_1(t)\,\tau_2\,\rho_{12}\,\bar{\sigma}_2(t)}{\tau_2\,l_2(t)+1}\,dt + \bar{\sigma}_1(t)\,dW_1^1(t) \]

In the lognormal specification with $\bar{\sigma}_k(t) = \sigma_k(t)\,l_k(t)$:

\[ dl_1(t) = -\sigma_1(t)\,l_1(t)\,\frac{\tau_2\,\rho_{12}\,\sigma_2(t)\,l_2(t)}{\tau_2\,l_2(t)+1}\,dt + \sigma_1(t)\,l_1(t)\,dW_1^1(t) \]

The drift is negative. This makes financial sense: moving from the $T_2$-forward measure to the $T_1$-forward measure (an earlier numeraire date), the drift becomes negative. Under $\mathbb{Q}^{T_1}$, the numéraire $P(t,T_1)$ grows faster relative to $P(t,T_2)$, which pushes $l_1$ downward in expectation relative to this measure.

Exercise 2. Under the terminal measure $\mathbb{Q}^m$ (with numéraire $P(t, T_m)$), the forward rate $l_i(t)$ has dynamics

\[ dl_i(t) = -l_i(t)\bar{\sigma}_i(t)\sum_{k=i+1}^{m}\frac{\tau_k\,\bar{\sigma}_k(t)\,l_k(t)}{\tau_k l_k(t) + 1}\,dt + l_i(t)\bar{\sigma}_i(t)\,dW_i^m(t) \]

For $m = 4$, $i = 1$, and constant volatilities $\bar{\sigma}_k = 0.20$, rates $l_k(0) = 5\%$, and $\tau_k = 1$, compute the instantaneous drift of $l_1$ at time 0. Is the drift positive or negative, and why?

Solution to Exercise 2

Under the terminal measure $\mathbb{Q}^4$ with $m = 4$, $i = 1$, constant $\bar{\sigma}_k = 0.20$, $l_k(0) = 0.05$, and $\tau_k = 1$, the drift is:

\[ \mu_1^4(0) = -l_1(0)\,\bar{\sigma}_1\sum_{k=2}^{4}\frac{\tau_k\,\bar{\sigma}_k\,l_k(0)}{\tau_k\,l_k(0)+1} \]

In the lognormal specification, with $\bar{\sigma}_k = 0.20$ as proportional volatilities, we have $\bar{\sigma}_k(t) = 0.20 \cdot l_k(t)$ for the absolute volatility. However, examining the given formula:

\[ dl_i(t) = -l_i(t)\bar{\sigma}_i(t)\sum_{k=i+1}^{m}\frac{\tau_k\,\bar{\sigma}_k(t)\,l_k(t)}{\tau_k l_k(t) + 1}\,dt + l_i(t)\bar{\sigma}_i(t)\,dW_i^m(t) \]

Here $\bar{\sigma}_k$ appears to be the proportional volatility. Let us interpret $\bar{\sigma}_k = 0.20$ as the proportional volatility. Then:

Each term in the sum at $t = 0$:

\[ \frac{\tau_k\,\bar{\sigma}_k\,l_k(0)}{\tau_k\,l_k(0)+1} = \frac{1 \times 0.20 \times 0.05}{1 \times 0.05 + 1} = \frac{0.01}{1.05} = 0.009524 \]

The sum runs from $k = 2$ to $k = 4$, so there are 3 terms, each equal to $0.009524$:

\[ \sum_{k=2}^{4}\frac{\tau_k\,\bar{\sigma}_k\,l_k(0)}{\tau_k\,l_k(0)+1} = 3 \times 0.009524 = 0.028571 \]

The drift of $l_1$ at time 0 is:

\[ \mu_1^4(0) = -l_1(0) \times \bar{\sigma}_1 \times 0.028571 = -0.05 \times 0.20 \times 0.028571 = -0.000286 \]

That is, $\mu_1^4(0) \approx -2.86$ basis points per year (in absolute terms).

The drift is negative. This is because under the terminal measure $\mathbb{Q}^4$, forward rates with indices well below the terminal index acquire negative drifts. The larger the gap between $i$ and $m$, the more terms in the sum and the larger the (negative) drift correction. Intuitively, moving to the terminal measure changes the numéraire to $P(t, T_4)$, which is a product of all forward discount factors; this systematic adjustment pushes earlier rates downward to maintain the no-arbitrage condition.

Exercise 3. The relationship between the bond price ratio and the LIBOR rate is $P(t, T_{i-1})/P(t, T_i) = 1 + \tau_i l_i(t)$. Verify this by expressing $l_i(t)$ in terms of the bond prices. Then show that the forward LIBOR rate is a martingale under $\mathbb{Q}^{T_i}$ by identifying it as a ratio of tradable assets divided by the numéraire $P(t, T_i)$.

Solution to Exercise 3

Expressing $l_i(t)$ in terms of bond prices:

By definition:

\[ l_i(t) = \frac{1}{\tau_i}\left(\frac{P(t, T_{i-1})}{P(t, T_i)} - 1\right) = \frac{P(t, T_{i-1}) - P(t, T_i)}{\tau_i\,P(t, T_i)} \]

Rearranging:

\[ \frac{P(t, T_{i-1})}{P(t, T_i)} = 1 + \tau_i\,l_i(t) \]

which confirms the bond-price-to-LIBOR relationship.

Showing $l_i(t)$ is a martingale under $\mathbb{Q}^{T_i}$:

Under $\mathbb{Q}^{T_i}$ (the forward measure with numéraire $P(t, T_i)$), any price process divided by $P(t, T_i)$ is a martingale (provided the price process represents a tradable asset or a portfolio thereof).

Consider the portfolio: long one $T_{i-1}$-bond, short one $T_i$-bond. Its value at time $t$ is:

\[ V(t) = P(t, T_{i-1}) - P(t, T_i) \]

This is the value of a self-financing trading strategy involving tradable zero-coupon bonds. Dividing by the numéraire $P(t, T_i)$:

\[ \frac{V(t)}{P(t, T_i)} = \frac{P(t, T_{i-1}) - P(t, T_i)}{P(t, T_i)} = \frac{P(t, T_{i-1})}{P(t, T_i)} - 1 = \tau_i\,l_i(t) \]

Since $V(t)/P(t, T_i)$ is a martingale under $\mathbb{Q}^{T_i}$, and $\tau_i$ is a constant:

\[ \mathbb{E}^{T_i}\bigl[l_i(T_{i-1}) \mid \mathcal{F}(t)\bigr] = l_i(t) \]

Therefore $l_i(t)$ is a $\mathbb{Q}^{T_i}$-martingale.

Exercise 4. Explain the relationship between the forward measure change formula $dW_i^{i}(t) = \frac{\tau_{i+1}\bar{\sigma}_{i+1}(t)\,l_{i+1}(t)}{\tau_{i+1}l_{i+1}(t)+1}\,dt + dW_i^{i+1}(t)$ and Girsanov's theorem. Identify the Girsanov kernel and explain why it depends on the forward rate $l_{i+1}(t)$.

Solution to Exercise 4

Girsanov's theorem states that when changing from probability measure $\mathbb{Q}$ to $\tilde{\mathbb{Q}}$ via Radon--Nikodym derivative $\Lambda_t = d\tilde{\mathbb{Q}}/d\mathbb{Q}|_{\mathcal{F}_t}$, if $d\Lambda_t/\Lambda_t = \gamma_t\,dW_t$ (where $\gamma_t$ is the Girsanov kernel), then $\tilde{W}_t = W_t - \int_0^t \gamma_s\,ds$ is a Brownian motion under $\tilde{\mathbb{Q}}$.

The Radon--Nikodym derivative from $\mathbb{Q}^{i+1}$ to $\mathbb{Q}^i$ is:

\[ \lambda_{i+1}^{i}(t) = \frac{P(t, T_i)/P(0, T_i)}{P(t, T_{i+1})/P(0, T_{i+1})} = \frac{P(0, T_{i+1})}{P(0, T_i)}\bigl(1 + \tau_{i+1}\,l_{i+1}(t)\bigr) \]

Computing the dynamics:

\[ \frac{d\lambda_{i+1}^{i}(t)}{\lambda_{i+1}^{i}(t)} = \frac{\tau_{i+1}\,\bar{\sigma}_{i+1}(t)}{\tau_{i+1}\,l_{i+1}(t) + 1}\,dW_{i+1}^{i+1}(t) \]

The Girsanov kernel is:

\[ \gamma_{i+1}(t) = \frac{\tau_{i+1}\,\bar{\sigma}_{i+1}(t)}{\tau_{i+1}\,l_{i+1}(t) + 1} \]

By Girsanov's theorem, the Brownian motion transforms as:

\[ dW_i^{i}(t) = -\rho_{i,i+1}\,\gamma_{i+1}(t)\,dt + dW_i^{i+1}(t) = -\frac{\tau_{i+1}\,\rho_{i,i+1}\,\bar{\sigma}_{i+1}(t)}{\tau_{i+1}\,l_{i+1}(t) + 1}\,dt + dW_i^{i+1}(t) \]

(The correlation $\rho_{i,i+1}$ enters because $W_i$ and $W_{i+1}$ are correlated.)

Why the kernel depends on $l_{i+1}(t)$: The numéraire ratio $P(t,T_i)/P(t,T_{i+1}) = 1 + \tau_{i+1}l_{i+1}(t)$ depends directly on the forward rate $l_{i+1}$. Since the Girsanov kernel is the volatility of the numéraire ratio (relative to itself), it inherits the dependence on $l_{i+1}(t)$. The factor $\tau_{i+1}l_{i+1}(t)/(1 + \tau_{i+1}l_{i+1}(t))$ represents the fraction of the bond price ratio that is attributable to the forward rate --- it is the "leverage" of the forward rate in the bond price ratio.

Exercise 5. In a three-rate LMM ($l_1, l_2, l_3$), write down the drift of each rate under the spot (rolling) measure, where the numéraire is the discretely-compounded money market account. Compare the drift structure with that under the terminal measure. Which measure leads to simpler simulation and why?

Solution to Exercise 5

Three-rate LMM under the spot (rolling) measure:

The spot measure $\mathbb{Q}^M$ uses the discretely-compounded money market account as numéraire. The drift of each rate is:

\[ dl_i(t) = \bar{\sigma}_i(t)\sum_{k=\bar{m}(t)+1}^{i}\frac{\tau_k\,\bar{\sigma}_k(t)}{\tau_k\,l_k(t)+1}\,dt + \bar{\sigma}_i(t)\,dW_i^M(t) \]

where $\bar{m}(t) = \min\{j : t \leq T_j\} - 1$.

For $t < T_1$ (before the first reset), $\bar{m}(t) = 0$, so the drifts are:

Rate $l_1$ ($i = 1$): Sum from $k = 1$ to $1$:

\[ dl_1(t) = \bar{\sigma}_1(t)\frac{\tau_1\,\bar{\sigma}_1(t)}{\tau_1\,l_1(t)+1}\,dt + \bar{\sigma}_1(t)\,dW_1^M(t) \]

Rate $l_2$ ($i = 2$): Sum from $k = 1$ to $2$:

\[ dl_2(t) = \bar{\sigma}_2(t)\left(\frac{\tau_1\,\rho_{21}\,\bar{\sigma}_1(t)}{\tau_1\,l_1(t)+1} + \frac{\tau_2\,\bar{\sigma}_2(t)}{\tau_2\,l_2(t)+1}\right)dt + \bar{\sigma}_2(t)\,dW_2^M(t) \]

Rate $l_3$ ($i = 3$): Sum from $k = 1$ to $3$:

\[ dl_3(t) = \bar{\sigma}_3(t)\left(\frac{\tau_1\,\rho_{31}\,\bar{\sigma}_1(t)}{\tau_1\,l_1(t)+1} + \frac{\tau_2\,\rho_{32}\,\bar{\sigma}_2(t)}{\tau_2\,l_2(t)+1} + \frac{\tau_3\,\bar{\sigma}_3(t)}{\tau_3\,l_3(t)+1}\right)dt + \bar{\sigma}_3(t)\,dW_3^M(t) \]

All drifts are positive under the spot measure.

Under the terminal measure $\mathbb{Q}^3$:

Rate $l_3$ ($i = 3$): Martingale (drift = 0).

Rate $l_2$ ($i = 2$): Sum from $k = 3$ to $3$:

\[ dl_2(t) = -\bar{\sigma}_2(t)\frac{\tau_3\,\rho_{23}\,\bar{\sigma}_3(t)}{\tau_3\,l_3(t)+1}\,dt + \bar{\sigma}_2(t)\,dW_2^3(t) \]

Rate $l_1$ ($i = 1$): Sum from $k = 2$ to $3$:

\[ dl_1(t) = -\bar{\sigma}_1(t)\left(\frac{\tau_2\,\rho_{12}\,\bar{\sigma}_2(t)}{\tau_2\,l_2(t)+1} + \frac{\tau_3\,\rho_{13}\,\bar{\sigma}_3(t)}{\tau_3\,l_3(t)+1}\right)dt + \bar{\sigma}_1(t)\,dW_1^3(t) \]

All non-martingale drifts are negative under the terminal measure.

Comparison: Under the spot measure, the drift sum for $l_i$ runs from $k = 1$ (or $\bar{m}(t)+1$) up to $i$. Under the terminal measure, the sum runs from $k = i+1$ up to $m$. The spot measure is simpler for simulation because:

Rates can be simulated in order $l_1, l_2, l_3$: to compute $l_i$'s drift, one only needs $l_1, \ldots, l_i$, which have already been updated at the current step
Under the terminal measure, $l_1$'s drift depends on $l_2$ and $l_3$, creating a coupling that complicates the stepping order

Exercise 6. Show that as the tenor $\tau_i \to 0$ (continuous limit), the LMM forward rate dynamics converge to the HJM instantaneous forward rate dynamics. Specifically, identify $l_i(t)\tau_i \approx \int_{T_{i-1}}^{T_i} f(t,u)\,du$ and show that the drift condition becomes the standard HJM drift condition $\alpha(t,T) = \sigma(t,T)\int_t^T \sigma(t,u)\,du$.

Solution to Exercise 6

Continuous limit of the LMM:

Identify the discrete forward rate with the integral of the instantaneous forward rate:

\[ \tau_i\,l_i(t) \approx \int_{T_{i-1}}^{T_i} f(t, u)\,du \]

As $\tau_i \to 0$, $l_i(t) \to f(t, T_{i-1})$ (the instantaneous forward rate at maturity $T_{i-1}$).

The bond price ratio becomes:

\[ \frac{P(t, T_{i-1})}{P(t, T_i)} = 1 + \tau_i\,l_i(t) \approx 1 + \tau_i\,f(t, T_i) \approx \exp\left(\int_{T_{i-1}}^{T_i} f(t, u)\,du\right) \]

The key term in the LMM drift under the terminal measure is:

\[ \frac{\tau_k\,\bar{\sigma}_k(t)\,l_k(t)}{1 + \tau_k\,l_k(t)} = \frac{\tau_k\,\sigma_k(t)\,l_k(t)\,l_k(t)}{1 + \tau_k\,l_k(t)} \]

For the proportional volatility specification $\bar{\sigma}_k = \sigma_k\,l_k$, as $\tau_k \to 0$:

\[ \frac{\tau_k\,\sigma_k\,l_k^2}{1 + \tau_k\,l_k} \approx \tau_k\,\sigma_k\,l_k^2 \to 0 \]

Instead, work with the absolute volatility. In the HJM framework, $f(t, T)$ satisfies:

\[ df(t, T) = \alpha(t, T)\,dt + \sigma(t, T)\,dW(t) \]

The LMM drift under the terminal measure for a single Brownian motion factor is:

\[ dl_i(t) = -\bar{\sigma}_i(t)\sum_{k=i+1}^{m}\frac{\tau_k\,\bar{\sigma}_k(t)}{\tau_k\,l_k(t)+1}\,dt + \bar{\sigma}_i(t)\,dW_i^m(t) \]

In the continuous limit, the sum becomes an integral. Let $\sigma^{\text{HJM}}(t, T) = \bar{\sigma}_i(t)/\tau_i$ for $T \in [T_{i-1}, T_i)$. Then the drift term:

\[ \sum_{k=i+1}^{m}\frac{\tau_k\,\bar{\sigma}_k(t)}{\tau_k\,l_k(t)+1} \to \int_{T_i}^{T_m} \frac{\sigma^{\text{HJM}}(t, u)}{1 + 0}\,du = \int_{T_i}^{T_m}\sigma^{\text{HJM}}(t, u)\,du \]

(since $\tau_k\,l_k \to 0$ as $\tau_k \to 0$). Under the terminal measure $\mathbb{Q}^{T_m}$, the drift of the instantaneous forward rate becomes:

\[ \alpha^{T_m}(t, T) = -\sigma(t, T)\int_T^{T_m}\sigma(t, u)\,du \]

Converting to the risk-neutral measure (where $\mathbb{Q}^{T_m}$ becomes $\mathbb{Q}$ in the continuous limit), the standard HJM no-arbitrage drift condition is recovered:

\[ \alpha(t, T) = \sigma(t, T)\int_t^T \sigma(t, u)\,du \]

This confirms that the LMM is a discretization of the HJM framework, and its drift structure converges to the HJM drift condition in the continuous tenor limit.