Numeraire

Study guide: Exam INV-201

A numeraire is the unit in which prices are measured: any traded asset whose price $N_t$ is always positive can serve, such as the money-market account, a zero-coupon bond or a stock. For each numeraire there is an equivalent martingale measure $\mathbb{Q}^N$ under which every traded price divided by $N_t$ is a martingale, so a derivative can be priced as an expectation in whichever unit makes that expectation easiest. Choosing the numeraire well is the standard device for interest-rate derivatives, where discounting and the payoff are both random.

$\frac{V_t}{N_t} = E^{\mathbb{Q}^N}\!\left[\frac{V_T}{N_T}\,\Big|\,\mathcal{F}_t\right]$

$V_0 = e^{-rT}E^{\mathbb{Q}}\left[V_T\right]$

$V_0 = P(0,T)\,E^{\mathbb{Q}^T}\left[V_T\right]$

$\frac{d\mathbb{Q}^N}{d\mathbb{Q}^M}\bigg|_{\mathcal{F}_T} = \frac{N_T/N_0}{M_T/M_0}$

Symbols: $V_t$ is the price of any traded claim, $N_t$ and $M_t$ are two numeraires, and $P(t,T)$ is the price at $t$ of a zero-coupon bond paying 1 at $T$ . With constant rates the money-market account is $B_t = e^{rt}$ , and in general it is $B_t = \exp\int_0^t r_s\,ds$ .
Money-market numeraire: Using $B_t$ gives the risk-neutral measure $\mathbb{Q}$ and the second block. With random rates, the discount factor $B_T^{-1}$ sits inside the expectation, as $E^{\mathbb{Q}}\left[e^{-\int_0^T r_s\,ds}\,V_T\right]$ , and is correlated with any rate-dependent payoff.
Forward measure: Using the zero-coupon bond maturing at $T$ gives the $T$ -forward measure $\mathbb{Q}^T$ . Because $P(T,T) = 1$ , the discount factor comes outside the expectation (third block), and the forward price of any asset for delivery at $T$ , $F_t = V_t/P(t,T)$ , is a martingale under $\mathbb{Q}^T$ . Hull calls this a world that is forward risk neutral with respect to $P(t,T)$ .
Forward rates: The forward rate for the period from $T$ to $T + \tau$ is a martingale under the $(T + \tau)$ -forward measure. That is why a caplet, which pays at $T + \tau$ , is priced with Black's model on the forward rate, and why an interest rate derivative that pays a rate at the end of its period is valued at today's forward rate.
Other numeraires: An annuity (the value of a stream of zero-coupon bonds) is the numeraire that makes a forward swap rate a martingale, which gives Black's model for a swaption. A stock used as numeraire prices an option to exchange one asset for another.
Change of numeraire: The fourth block converts expectations between two measures. In Hull's form, moving from numeraire $g$ to numeraire $h$ raises the expected growth rate of any variable $v$ by $\rho\,\sigma_v\,\sigma_w$ , where $w = h/g$ and $\rho$ is the correlation between $v$ and $w$ . Volatilities do not change, as Girsanov's theorem requires.

Valuing a Claim That Pays a Bond Under the Forward Measure Continuously compounded zero rates are 3.0% for 2 years and 3.5% for 5 years. A claim pays at $T = 2$ the price at that time of a zero-coupon bond with face $1,000 that matures at year 5. Value the claim under the 2-year forward measure, without modelling interest rates.
Answer The bond's forward price for delivery at 2 is $P(t,5)/P(t,2)$ , which is a martingale under $\mathbb{Q}^2$ , and at $t = 2$ it equals $P(2,5)$ .
 $\begin{align*} P(0,2) &= e^{-0.06} = 0.941765 \\ P(0,5) &= e^{-0.175} = 0.839457 \\ E^{\mathbb{Q}^2}\left[P(2,5)\right] &= \frac{0.839457}{0.941765} \\ &= 0.891366 \\ V_0 &= 1{,}000\,P(0,2)\,E^{\mathbb{Q}^2}\left[P(2,5)\right] \\ &= 1{,}000(0.941765)(0.891366) \\ &= 839.46 \end{align*}$ 
The claim is worth $839.46, which is exactly today's price of the 5-year bond, as it must be, since holding the bond delivers the same thing at time 2. Under $\mathbb{Q}$ this would require the joint distribution of $e^{-\int_0^2 r\,ds}$ and $P(2,5)$ , but under $\mathbb{Q}^2$ only the forward price is needed.

A Call on a Zero-Coupon Bond by Black's Model Using the same curve, a 2-year European call on the 5-year bond above has strike 0.88 per unit of face, and the forward bond price is lognormal under $\mathbb{Q}^2$ with volatility $\sigma = 0.03$ . Find the call's value per $1,000 of face.
Answer Under $\mathbb{Q}^2$ the payoff is $\max\left(P(2,5) - K, 0\right)$ , the bond price at 2 equals its forward price $F_2$ , and $F_t$ is a lognormal martingale starting at $F_0 = 0.891366$ . Discounting at $P(0,2)$ gives Black's formula.
 $\begin{align*} d_1 &= \frac{\ln(0.891366/0.88) + \tfrac12(0.03)^2(2)}{0.03\sqrt{2}} \\ &= 0.32370 \\ d_2 &= 0.32370 - 0.042426 \\ &= 0.28127 \\ c &= P(0,2)\left[F_0N(d_1) - KN(d_2)\right] \\ &= 0.941765\left[0.891366(0.62692) - 0.88(0.61075)\right] \\ &= 0.021353 \times 0.941765 \\ &= 0.020110 \end{align*}$ 
The call is worth about $20.11 per $1,000 of face. The forward measure removes the stochastic discount factor, so the drift of the bond price and the dynamics of $r$ never enter.

How it appears on the exam

INV-201: The syllabus lists change of numeraire among its stochastic calculus techniques, and its next objective asks candidates to apply the risk-neutral measure, the forward measure and normalization. The readings are Hull's chapter on martingales and measures, including its sections on alternative numeraires and on changing the numeraire.

Choosing a numeraire: A question asks which normalized prices are martingales under which measure. Examples are a forward price under the forward measure, a forward rate under the measure of the bond that pays at the end of its period, and a swap rate under the annuity measure.

Pricing: Candidates price a bond option, caplet or claim paying a bond price with Black's model under the forward measure, or show why $E^{\mathbb{Q}^T}[S_T]$ equals the forward price.

Drift adjustments: A question may ask for the drift of a variable after a change of numeraire, using $\rho\,\sigma_v\,\sigma_w$ , or explain why a convexity or timing adjustment arises when a rate is paid at the wrong time for its natural measure.

Exam traps

The wrong discount factor: Under $\mathbb{Q}^T$ the discount factor is today's bond price $P(0,T)$ , a known number. Putting the random factor $e^{-\int_0^T r_s\,ds}$ inside a $\mathbb{Q}^T$ expectation, or using $\mathbb{Q}^T$ probabilities with the money-market account, mixes two measures and counts the discounting twice or not at all.

Measure and payment date must match: A forward rate for the period from $T$ to $T + \tau$ is a martingale under the $(T + \tau)$ -forward measure, not the $T$ -forward measure. Pricing a rate paid at $T$ at today's forward rate ignores a timing adjustment.

Non-traded numeraires: The numeraire must be a traded asset with a positive price. A rate, an index level or a payoff that can be zero cannot serve.

Which volatility: A change of numeraire changes drifts, never volatilities, but Black's model needs the volatility of the forward price. In a one-factor model the forward bond price $P(t,5)/P(t,2)$ has the difference of the two bonds' volatilities, not the 5-year bond's own, so use the volatility given for the forward price.