Forward Measure and Caplet Pricing

Options Pricing · Hard · Free problem

Let $P(t, T)$ be the price at time $t$ of a zero-coupon bond maturing at $T$, and let $Q$ denote the risk-neutral measure with the money-market account $B(t)$ as numeraire.

Define the $T$-forward measure $Q^T$ using $P(t, T)$ as numeraire.

  1. Write down the Radon-Nikodym derivative $\frac{dQ^T}{dQ}$ on $\mathcal{F}_T$.
  1. Prove that any traded asset price $X_t$, when discounted by $P(t, T)$, is a martingale under $Q^T$. That is, show that $\frac{X_t}{P(t, T)}$ is a $Q^T$-martingale.
  1. A caplet pays $(L(T, T+\delta) - K)^{+} \cdot \delta$ at time $T + \delta$, where $L(T, T+\delta)$ is the LIBOR rate set at $T$ for the period $[T, T+\delta]$. Use the forward measure machinery to express the caplet price at time $0$ as an expectation under $Q^{T+\delta}$. State exactly which random variable you would simulate in a Monte Carlo pricer under $Q^{T+\delta}$.

Hints

  1. Think about what happens when you change the numeraire from $B(t)$ to $P(t, T)$ -- the Radon-Nikodym derivative is the ratio of the new numeraire to the old, normalized appropriately.
  2. To prove the martingale property under $Q^T$, use the abstract Bayes formula: $E^{Q^T}[Y | \mathcal{F}_s] = E^Q[Y \cdot L_t | \mathcal{F}_s] / L_s$, and notice that the $P(t,T)$ terms cancel, leaving you with a $Q$-martingale.
  3. For the caplet, choose $P(t, T+\delta)$ as the numeraire (not $P(t,T)$) since the payoff occurs at $T+\delta$. The forward LIBOR rate $L(t, T, T+\delta) = \frac{1}{\delta}(P(t,T)/P(t,T+\delta) - 1)$ is then a martingale under $Q^{T+\delta}$.

Worked Solution

How to Think About It: The forward measure is the single most useful change-of-numeraire trick in fixed-income derivatives. The core idea: pricing under the risk-neutral measure $Q$ with the money-market account as numeraire forces you to deal with stochastic discount factors inside expectations. That is painful when rates are random. Switching to the $T$-forward measure -- where the zero-coupon bond $P(t, T)$ is the numeraire -- kills the stochastic discounting. The forward rate becomes a martingale under this measure, and pricing reduces to computing an expectation of a payoff function of that forward rate. This is exactly how the Black caplet formula works.

Quick Sanity Checks: The Radon-Nikodym derivative should equal 1 in expectation under $Q$ (it is a valid density). At time $T$, the bond pays 1, so the numeraire collapses and things simplify. The forward LIBOR rate $L(0, T, T+\delta)$ should be a martingale under $Q^{T+\delta}$ -- this is the whole point of choosing that measure.

Derivation:

*Part (i): Radon-Nikodym derivative*

Under $Q$, the money-market account $B(t) = \exp\left(\int_0^t r_s \, ds\right)$ is the numeraire, and asset prices divided by $B(t)$ are $Q$-martingales. We want to switch to the numeraire $P(t, T)$.

The Radon-Nikodym derivative on $\mathcal{F}_T$ is:

$\frac{dQ^T}{dQ}\bigg|_{\mathcal{F}_T} = \frac{P(T, T)}{B(T)} \cdot \frac{B(0)}{P(0, T)} = \frac{1}{B(T) \cdot P(0, T)}$

since $P(T, T) = 1$ and $B(0) = 1$. Equivalently, this is the terminal value of the ratio of the new numeraire to the old numeraire, normalized so the density integrates to 1.

More generally, the density process (for $t \leq T$) is:

$\left.\frac{dQ^T}{dQ}\right|_{\mathcal{F}_t} = \frac{P(t, T)}{B(t) \cdot P(0, T)}$

You can verify $E^Q\left[\frac{dQ^T}{dQ}\right] = \frac{E^Q[P(T,T)/B(T)]}{P(0,T)} = \frac{P(0,T)}{P(0,T)} = 1$ because $P(T,T)/B(T)$ discounted is a $Q$-martingale with initial value $P(0,T)/B(0) = P(0,T)$.

*Part (ii): Martingale property of $X_t / P(t, T)$ under $Q^T$*

We need to show that for $s \leq t \leq T$:

$E^{Q^T}\left[\frac{X_t}{P(t, T)} \bigg| \mathcal{F}_s\right] = \frac{X_s}{P(s, T)}$

Using the abstract Bayes formula for the change of measure, for any $\mathcal{F}_t$-measurable random variable $Y$:

$E^{Q^T}[Y | \mathcal{F}_s] = \frac{E^Q[Y \cdot L_t | \mathcal{F}_s]}{L_s}$

where $L_t = \frac{P(t,T)}{B(t) \cdot P(0,T)}$ is the density process.

Applying this with $Y = X_t / P(t, T)$:

$E^{Q^T}\left[\frac{X_t}{P(t,T)} \bigg| \mathcal{F}_s\right] = \frac{E^Q\left[\frac{X_t}{P(t,T)} \cdot \frac{P(t,T)}{B(t) \cdot P(0,T)} \bigg| \mathcal{F}_s\right]}{\frac{P(s,T)}{B(s) \cdot P(0,T)}}$

The numerator simplifies:

$E^Q\left[\frac{X_t}{B(t) \cdot P(0,T)} \bigg| \mathcal{F}_s\right] = \frac{1}{P(0,T)} E^Q\left[\frac{X_t}{B(t)} \bigg| \mathcal{F}_s\right] = \frac{1}{P(0,T)} \cdot \frac{X_s}{B(s)}$

The last step uses the fact that $X_t / B(t)$ is a $Q$-martingale (fundamental theorem of asset pricing). Dividing by the denominator:

$\frac{X_s / (B(s) \cdot P(0,T))}{P(s,T) / (B(s) \cdot P(0,T))} = \frac{X_s}{P(s,T)}$

This completes the proof. $\square$

*Part (iii): Caplet pricing under the forward measure*

The caplet pays $(L(T, T+\delta) - K)^{+} \cdot \delta$ at time $T + \delta$. To price it, we use $P(t, T+\delta)$ as numeraire and work under $Q^{T+\delta}$.

The time-0 price of the caplet is:

$V_0 = P(0, T+\delta) \cdot E^{Q^{T+\delta}}\left[\delta \cdot (L(T, T+\delta) - K)^{+}\right]$

This follows from part (ii): the caplet payoff at $T+\delta$ divided by $P(T+\delta, T+\delta) = 1$ is a $Q^{T+\delta}$-martingale when normalized by the numeraire.

The key fact is that the forward LIBOR rate $L(t, T, T+\delta)$ defined by:

$L(t, T, T+\delta) = \frac{1}{\delta}\left(\frac{P(t, T)}{P(t, T+\delta)} - 1\right)$

is a $Q^{T+\delta}$-martingale. This is because $P(t, T) / P(t, T+\delta)$ is a ratio of a traded asset to the numeraire, hence a martingale by part (ii), and $L$ is just an affine transformation of this ratio.

Since $L(T, T, T+\delta) = L(T, T+\delta)$ (the forward rate equals the spot rate at fixing), the caplet price becomes:

$V_0 = P(0, T+\delta) \cdot \delta \cdot E^{Q^{T+\delta}}\left[(L(T, T, T+\delta) - K)^{+}\right]$

Monte Carlo implementation: Under $Q^{T+\delta}$, simulate the forward LIBOR rate $L(t, T, T+\delta)$. Since it is a martingale under this measure, a common model is to assume it follows a driftless geometric Brownian motion:

$dL = \sigma L \, dW^{T+\delta}$

where $W^{T+\delta}$ is a Brownian motion under $Q^{T+\delta}$. This gives $L(T) = L(0) \exp\left(-\frac{\sigma^2 T}{2} + \sigma W^{T+\delta}_T\right)$ with $L(0)$ observable from today's yield curve. The random variable to simulate is $L(T, T, T+\delta)$ -- the terminal value of the forward LIBOR rate. For each path, compute $\delta \cdot (L(T) - K)^{+}$, average across paths, and multiply by $P(0, T+\delta)$. Under the lognormal assumption this recovers Black's formula in closed form.

Practical Interpretation: The forward measure eliminates the need to model the entire short-rate process or deal with stochastic discount factors. You only need to model the dynamics of the single forward rate relevant to your payoff. This is exactly the logic behind the LIBOR Market Model (BGM model): each forward rate is modeled as a martingale under its own forward measure, and caplets are priced one at a time using Black's formula. Swaptions and exotic products require correlating multiple forward rates, but for vanilla caplets, the single-rate forward measure approach is both elegant and practical.

Answer: (i) $\frac{dQ^T}{dQ}\big|_{\mathcal{F}_T} = \frac{1}{B(T) \cdot P(0,T)}$. (ii) $X_t / P(t,T)$ is a $Q^T$-martingale by the abstract Bayes formula and the fact that $X_t / B(t)$ is a $Q$-martingale. (iii) The caplet price is $V_0 = P(0, T+\delta) \cdot \delta \cdot E^{Q^{T+\delta}}[(L(T, T, T+\delta) - K)^{+}]$. In Monte Carlo, simulate the forward LIBOR rate $L(t, T, T+\delta)$ as a driftless process under $Q^{T+\delta}$ and evaluate the call payoff on the terminal rate.

Intuition

The forward measure is the workhorse of fixed-income derivatives pricing. The deep insight is that every pricing problem becomes simpler when you pick the right numeraire. Under the risk-neutral measure with the money-market numeraire, you have to take expectations of payoffs multiplied by stochastic discount factors -- and when rates are random, that discount factor is correlated with the payoff, making the expectation hard. Switching to the forward measure kills this correlation by absorbing the discounting into the measure itself. The payoff and the discount factor decouple.

In practice, this is why Black's formula works for caplets even though interest rates are stochastic. Under the $(T+\delta)$-forward measure, the forward LIBOR rate is a martingale, so if you model it as lognormal, you get a Black-Scholes-style formula with zero drift. The entire complexity of the term structure collapses into a single volatility parameter. This same principle scales up to the LIBOR Market Model, where each forward rate lives under its own forward measure, and you only need to specify correlations between rates to price exotic products. The common mistake in interviews is choosing the wrong maturity for the forward measure -- always match the numeraire bond maturity to the payment date of the derivative.

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