Unit 9, lesson 65

Numeraires and forward measures

How a strictly positive traded numeraire defines its own pricing measure through a density process, why numeraire invariance keeps currency prices fixed, and how the bond numeraire gives the forward measure.

  • Video 65
  • Numeraires and short-rate models

A claim pays 100 at a future date TT only in the states where interest rates have been high along the way. With a random money-market account, its price is the expectation of the payoff divided by the account, and the payoff and the discount factor move together, so neither can be taken out of the expectation on its own. Can the price be written as today’s bond price times a plain average of the payoff, and under which probability? This lesson builds the pricing measure attached to any strictly positive traded asset, proves with the conditional Bayes formula that changing the unit of account leaves every currency price unchanged, and specializes to the zero-coupon bond, whose measure is the TT-forward measure.

Key terms

Numeraire

Definition Work on a filtered probability space (Ω,F,(Ft)0≤t≤T,Q)(\Omega, \mathcal{F}, (\mathcal{F}_t)_{0 \le t \le T}, \mathbb{Q}) in which F0\mathcal{F}_0 is trivial up to Q\mathbb{Q}-null sets, so that every time-00 value is a known number. Let Bt=exp⁡(∫0trs ds)B_t = \exp\bigl(\int_0^t r_s \, ds\bigr) be the money-market account, a strictly positive process, and let Q\mathbb{Q} be a pricing measure for the numeraire BB. A claim that pays HH at TT, where HH is FT\mathcal{F}_T-measurable with EQ[∣H∣/BT]<∞\mathbb{E}^{\mathbb{Q}}[\lvert H \rvert / B_T] < \infty, then has the time-tt value

Vt=Bt EQ[HBT | Ft],V_t = B_t \, \mathbb{E}^{\mathbb{Q}}\left[\frac{H}{B_T} \,\middle\vert\, \mathcal{F}_t\right],

and the discounted price of every traded asset without payouts is a Q\mathbb{Q}-martingale. A numeraire on [0,T][0, T] is an adapted process NN such that

  1. it is the value of a traded asset or of a self-financing portfolio that makes no payouts on [0,T][0, T], any income being reinvested in the position,
  2. it is strictly positive, Nt>0N_t > 0 for every t∈[0,T]t \in [0, T] almost surely, including at TT,
  3. the ratio N/BN/B is a true Q\mathbb{Q}-martingale on [0,T][0, T], not only a local martingale.

The account BB is a numeraire, with N/B=1N/B = 1. A price measured in units of NN is the ratio Vt/NtV_t/N_t, the number of units of NN that VtV_t buys. A positive quantity that is not the value of a position anyone can hold, such as the short rate or a statistical index, is not a numeraire, however convenient it would be as a unit.

In plain words A numeraire is a positive asset used as the unit of account. Prices are quoted as so many units of that asset instead of so many currency units, and the asset must be something one can actually hold, with all its income kept inside the position.

Example Take Black-Scholes dynamics under Q\mathbb{Q} with constant rr, a constant dividend yield q>0q > 0 and a constant volatility σ>0\sigma > 0. The stock price itself is not a numeraire, because e−rtSte^{-rt} S_t drifts down at the rate qq and is not a Q\mathbb{Q}-martingale. The position that starts with one share and reinvests every dividend holds eqte^{qt} shares at time tt, so its value is Nt=eqtSt>0N_t = e^{qt} S_t > 0 and

NtBt=S0exp⁡(σWtQ−12σ2t),\frac{N_t}{B_t} = S_0 \exp\Bigl(\sigma W^{\mathbb{Q}}_t - \tfrac12 \sigma^2 t\Bigr),

an exponential martingale with constant volatility, so this position is a numeraire. A call option fails as a numeraire on [0,T][0, T] even though its price is positive before expiry, because its value at TT is (ST−K)+(S_T - K)^{+}, which is zero with positive Q\mathbb{Q}-probability.

Density process

Definition For a numeraire NN on [0,T][0, T], the density process with respect to Q\mathbb{Q} is

ZtN=Nt/BtN0/B0Z_t^N = \frac{N_t / B_t}{N_0 / B_0}

for 0≤t≤T0 \le t \le T. It is a strictly positive true Q\mathbb{Q}-martingale with Z0N=1Z_0^N = 1, so EQ[ZTN]=1\mathbb{E}^{\mathbb{Q}}[Z_T^N] = 1. The measure

QN(A)=EQ[ZTN1A],\mathbb{Q}^N(A) = \mathbb{E}^{\mathbb{Q}}\bigl[Z_T^N \mathbf{1}_A\bigr],

defined for A∈FTA \in \mathcal{F}_T, is a probability on FT\mathcal{F}_T, equivalent to Q\mathbb{Q} because ZTN>0Z_T^N > 0. For A∈FtA \in \mathcal{F}_t the tower property gives QN(A)=EQ[ZtN1A]\mathbb{Q}^N(A) = \mathbb{E}^{\mathbb{Q}}[Z_t^N \mathbf{1}_A], which is the statement

dQNdQ∣Ft=ZtN.\left.\frac{d\mathbb{Q}^N}{d\mathbb{Q}}\right|_{\mathcal{F}_t} = Z_t^N .

The conditional Bayes formula follows. For every FT\mathcal{F}_T-measurable XX with EQN[∣X∣]<∞\mathbb{E}^{\mathbb{Q}^N}[\lvert X \rvert] < \infty,

EQN[X∣Ft]=EQ[ZTNX∣Ft]ZtN.\begin{aligned} &\mathbb{E}^{\mathbb{Q}^N}[X \mid \mathcal{F}_t] \\[0.4em] &\quad = \frac{\mathbb{E}^{\mathbb{Q}}[Z_T^N X \mid \mathcal{F}_t]}{Z_t^N} . \end{aligned}

Each hypothesis on NN has a role here. A positive local martingale is a supermartingale, so if N/BN/B were only a local martingale, the normalized terminal value could have EQ[ZTN]<1\mathbb{E}^{\mathbb{Q}}[Z_T^N] < 1 and QN\mathbb{Q}^N would have total mass below one. If NN could reach zero, then ZTNZ_T^N would vanish on an event of positive Q\mathbb{Q}-probability, QN\mathbb{Q}^N would only be absolutely continuous with respect to Q\mathbb{Q}, not equivalent to it, and a payoff divided by NTN_T would be undefined there.

In plain words The density reweights each outcome by how much the numeraire has grown relative to the money-market account, scaled so that the weights average to one. Outcomes in which the numeraire did well become more likely under its own measure.

Example For the reinvested share position of the previous card, the density is

ZtN=exp⁡(σWtQ−12σ2t),Z_t^N = \exp\Bigl(\sigma W^{\mathbb{Q}}_t - \tfrac12 \sigma^2 t\Bigr),

with mean 11 and, for σ=0.2\sigma = 0.2 and t=1t = 1, variance e0.04−1≈0.040811e^{0.04} - 1 \approx 0.040811. It is the stochastic exponential of ∫σ dWQ\int \sigma \, dW^{\mathbb{Q}}, and Girsanov’s theorem for a density E(∫θ dWQ)\mathcal{E}(\int \theta \, dW^{\mathbb{Q}}), which makes WtQ−∫0tθs dsW^{\mathbb{Q}}_t - \int_0^t \theta_s \, ds a Brownian motion, shows with θ=σ\theta = \sigma that WtS=WtQ−σtW^{S}_t = W^{\mathbb{Q}}_t - \sigma t is a Brownian motion under the share measure QS=QN\mathbb{Q}^{S} = \mathbb{Q}^N. Substituting WTQ=WTS+σTW^{\mathbb{Q}}_T = W^{S}_T + \sigma T gives

ln⁡ST=ln⁡S0+σWTS+(r−q+12σ2)T,\begin{aligned} \ln S_T &= \ln S_0 + \sigma W^{S}_T \\[0.4em] &\quad + \bigl(r - q + \tfrac12 \sigma^2\bigr) T , \end{aligned}

so under the share measure the log price has the drift r−q+σ2/2r - q + \sigma^2/2 instead of r−q−σ2/2r - q - \sigma^2/2. With r=0.05r = 0.05, q=0.02q = 0.02 and σ=0.2\sigma = 0.2, that drift is 0.050.05 per year.

Numeraire invariance

Definition Change of numeraire theorem. Let Q\mathbb{Q} be a pricing measure for BB, let NN be a numeraire on [0,T][0, T] with density process ZNZ^N and measure QN\mathbb{Q}^N, and let HH be FT\mathcal{F}_T-measurable with EQ[∣H∣/BT]<∞\mathbb{E}^{\mathbb{Q}}[\lvert H \rvert / B_T] < \infty. Then EQN[∣H∣/NT]=(B0/N0) EQ[∣H∣/BT]<∞\mathbb{E}^{\mathbb{Q}^N}[\lvert H \rvert / N_T] = (B_0/N_0)\,\mathbb{E}^{\mathbb{Q}}[\lvert H \rvert / B_T] < \infty, and for every t∈[0,T]t \in [0, T]

Vt=Bt EQ[HBT | Ft]=Nt EQN[HNT | Ft].\begin{aligned} V_t &= B_t \, \mathbb{E}^{\mathbb{Q}}\left[\frac{H}{B_T} \,\middle\vert\, \mathcal{F}_t\right] \\[0.4em] &= N_t \, \mathbb{E}^{\mathbb{Q}^N}\left[\frac{H}{N_T} \,\middle\vert\, \mathcal{F}_t\right]. \end{aligned}

The proof is the conditional Bayes formula with X=H/NTX = H/N_T. The terminal density contains NTN_T, which cancels the NTN_T in the payoff, so ZTNH/NT=(B0/N0) H/BTZ_T^N H / N_T = (B_0/N_0)\, H/B_T. The current density is ZtN=(B0/N0) Nt/BtZ_t^N = (B_0/N_0)\, N_t/B_t. Hence

Nt EQ[ZTNH/NT∣Ft]ZtN=Nt BtNt EQ[HBT | Ft],\begin{aligned} &N_t \, \frac{\mathbb{E}^{\mathbb{Q}}[Z_T^N H / N_T \mid \mathcal{F}_t]}{Z_t^N} \\[0.4em] &\quad = N_t \, \frac{B_t}{N_t} \, \mathbb{E}^{\mathbb{Q}}\left[\frac{H}{B_T} \,\middle\vert\, \mathcal{F}_t\right], \end{aligned}

which is VtV_t. Every cancellation pairs factors with the same date, a terminal NTN_T with a terminal NTN_T and a current NtN_t with a current NtN_t. The theorem also shows that V/NV/N is a QN\mathbb{Q}^N-martingale, and the same argument makes S/NS/N a QN\mathbb{Q}^N-martingale for every traded price SS without payouts. The unit and the measure change together, so mixing them, for example averaging in units of NN under Q\mathbb{Q}, gives wrong prices in general.

In plain words Measure the payoff in units of the numeraire, average it with the numeraire’s own probabilities, and convert back to currency with today’s value of the numeraire. Any positive traded asset that satisfies the hypotheses gives the same currency price, because the change of units and the change of probabilities cancel exactly.

Example An asset-or-nothing call pays STS_T at TT if ST>KS_T > K and nothing otherwise. In the share-numeraire setting above, N0=S0N_0 = S_0 and H/NT=e−qT1{ST>K}H/N_T = e^{-qT} \mathbf{1}_{\{S_T > K\}}, so the theorem gives

V0=S0e−qT QS(ST>K)=S0e−qT Φ(d1),\begin{aligned} V_0 &= S_0 e^{-qT} \, \mathbb{Q}^{S}(S_T > K) \\[0.4em] &= S_0 e^{-qT} \, \Phi(d_1), \end{aligned}

where d1=(ln⁡(S0/K)+(r−q+σ2/2)T)/(σT)d_1 = \bigl(\ln(S_0/K) + (r - q + \sigma^2/2)T\bigr)/(\sigma\sqrt{T}) comes from the share-measure drift of the previous card. For the constructed values S0=K=100S_0 = K = 100, r=0.05r = 0.05, q=0.02q = 0.02, σ=0.2\sigma = 0.2 and T=1T = 1, the parameter is d1=0.25d_1 = 0.25, Φ(0.25)≈0.598706\Phi(0.25) \approx 0.598706 and V0≈98.019867×0.598706≈58.6851V_0 \approx 98.019867 \times 0.598706 \approx 58.6851. A direct numerical integration of e−rTST1{ST>K}e^{-rT} S_T \mathbf{1}_{\{S_T > K\}} under Q\mathbb{Q} gives the same value.

Bond numeraire

Definition Assume EQ[1/BT]<∞\mathbb{E}^{\mathbb{Q}}[1/B_T] < \infty, and take B0=1B_0 = 1. The default-free zero-coupon bond that pays 11 at TT has the time-tt price

P(t,T)=Bt EQ[1BT | Ft]P(t,T) = B_t \, \mathbb{E}^{\mathbb{Q}}\left[\frac{1}{B_T} \,\middle\vert\, \mathcal{F}_t\right]

for 0≤t≤T0 \le t \le T. It is strictly positive because 1/BT>01/B_T > 0, it satisfies P(T,T)=1P(T,T) = 1, and P(t,T)/BtP(t,T)/B_t is the conditional expectation of the integrable variable 1/BT1/B_T, so it is a true Q\mathbb{Q}-martingale on [0,T][0, T]. The bond is therefore a numeraire on [0,T][0, T] without any further check, and its density process is

ZtT=P(t,T)Bt P(0,T),ZTT=1BT P(0,T).\begin{gathered} Z_t^{T} = \frac{P(t,T)}{B_t \, P(0,T)}, \\[0.4em] Z_T^{T} = \frac{1}{B_T \, P(0,T)} . \end{gathered}

The bond matures at TT, so it serves as a numeraire on [0,T][0, T] only. When the rates are deterministic, BTB_T is a number, P(0,T)=1/BTP(0,T) = 1/B_T and ZT≡1Z^{T} \equiv 1, and the bond price process is the discount factor D(t,T)=Bt/BTD(t,T) = B_t/B_T. With stochastic rates P(t,T)P(t,T) is a random process, and a deterministic discount function cannot stand in for it.

In plain words A zero-coupon bond is positive up to and including its maturity and is worth exactly one at maturity. That makes it a natural unit of account up to that date, and in bond units a payment at TT needs no discounting at all.

Example In the course’s constructed two-state model, B0=1B_0 = 1 and BTB_T equals 1.021.02 or 1.081.08 with Q\mathbb{Q}-probability 12\tfrac12 each. The bond costs

P(0,T)=0.51.02+0.51.08=875918≈0.953159.\begin{aligned} P(0,T) &= \frac{0.5}{1.02} + \frac{0.5}{1.08} \\[0.4em] &= \frac{875}{918} \approx 0.953159 . \end{aligned}

The density 1/(BTP(0,T))1/(B_T P(0,T)) equals 36/35≈1.02857136/35 \approx 1.028571 in the first state and 34/35≈0.97142934/35 \approx 0.971429 in the second, with mean 12(36/35+34/35)=1\tfrac12(36/35 + 34/35) = 1. If instead BT=1.05B_T = 1.05 in both states, then P(0,T)=1/1.05P(0,T) = 1/1.05 and the density is 11 in both states.

Forward measure

Definition The TT-forward measure QT\mathbb{Q}^{T} is the measure of the bond numeraire, defined on FT\mathcal{F}_T by

dQTdQ∣FT=1BT P(0,T),\left.\frac{d\mathbb{Q}^{T}}{d\mathbb{Q}}\right|_{\mathcal{F}_T} = \frac{1}{B_T \, P(0,T)} ,

with B0=1B_0 = 1. Since P(T,T)=1P(T,T) = 1, numeraire invariance gives, for every FT\mathcal{F}_T-measurable HH with EQ[∣H∣/BT]<∞\mathbb{E}^{\mathbb{Q}}[\lvert H \rvert / B_T] < \infty and every t∈[0,T]t \in [0, T],

Vt=P(t,T) EQT[H∣Ft].V_t = P(t,T) \, \mathbb{E}^{\mathbb{Q}^{T}}[H \mid \mathcal{F}_t] .

The random discount is absorbed into the measure, not dropped. If moreover EQ[∣H∣]<∞\mathbb{E}^{\mathbb{Q}}[\lvert H \rvert] < \infty, the two expectations at time 00 are linked by

EQT[H]−EQ[H]=Cov⁡Q(H,1/BT)P(0,T),\begin{aligned} &\mathbb{E}^{\mathbb{Q}^{T}}[H] - \mathbb{E}^{\mathbb{Q}}[H] \\[0.4em] &\quad = \frac{\operatorname{Cov}^{\mathbb{Q}}(H, 1/B_T)}{P(0,T)} , \end{aligned}

because EQ[H/BT]=P(0,T) EQ[H]+Cov⁡Q(H,1/BT)\mathbb{E}^{\mathbb{Q}}[H/B_T] = P(0,T)\,\mathbb{E}^{\mathbb{Q}}[H] + \operatorname{Cov}^{\mathbb{Q}}(H, 1/B_T). So the formula P(0,T) EQ[H]P(0,T)\,\mathbb{E}^{\mathbb{Q}}[H] is right only when HH and 1/BT1/B_T are uncorrelated under Q\mathbb{Q}, for example when rates are deterministic. For a traded asset SS without payouts on [0,T][0, T], the ratio St/P(t,T)S_t/P(t,T) is a QT\mathbb{Q}^{T}-martingale that ends at STS_T, so the forward price for delivery at TT is St/P(t,T)=EQT[ST∣Ft]S_t/P(t,T) = \mathbb{E}^{\mathbb{Q}^{T}}[S_T \mid \mathcal{F}_t]. Each maturity has its own forward measure, and two forward measures differ in general.

In plain words Under the forward measure, the price of a payment at TT is today’s bond price times a plain average of the payment. States in which the account grows slowly, so that discounting is mild, get more weight, and that reweighting carries the correlation between rates and the payoff.

Example In the two-state model the forward weights are 12⋅3635=1835≈0.514286\tfrac12 \cdot \tfrac{36}{35} = \tfrac{18}{35} \approx 0.514286 and 1735≈0.485714\tfrac{17}{35} \approx 0.485714. They add up to 11 but are not the equal Q\mathbb{Q}-weights, and the state with the larger account gets less weight. Let the claim pay 00 in the first state and 100100 in the second. Cash-account pricing gives

V0=12⋅1001.08=125027≈46.296296,\begin{aligned} V_0 &= \frac12 \cdot \frac{100}{1.08} \\[0.4em] &= \frac{1250}{27} \approx 46.296296 , \end{aligned}

and the forward route gives EQT[H]=100⋅1735≈48.571429\mathbb{E}^{\mathbb{Q}^{T}}[H] = 100 \cdot \tfrac{17}{35} \approx 48.571429 and P(0,T)⋅48.571429≈46.296296P(0,T) \cdot 48.571429 \approx 46.296296, the same price. The tempting shortcut P(0,T) EQ[H]=0.953159⋅50≈47.6580P(0,T)\,\mathbb{E}^{\mathbb{Q}}[H] = 0.953159 \cdot 50 \approx 47.6580 is too high by about 1.36171.3617, which is minus the covariance Cov⁡Q(H,1/BT)≈−1.3617\operatorname{Cov}^{\mathbb{Q}}(H, 1/B_T) \approx -1.3617.

Drift under a forward measure

Definition Let WQW^{\mathbb{Q}} be a dd-dimensional Q\mathbb{Q}-Brownian motion for (Ft)(\mathcal{F}_t), and let the bond have the Q\mathbb{Q}-dynamics

dP(t,T)P(t,T)=rt dt+σP(t,T)⊤dWtQ\begin{aligned} \frac{dP(t,T)}{P(t,T)} &= r_t \, dt \\[0.4em] &\quad + \sigma_P(t,T)^{\top} dW^{\mathbb{Q}}_t \end{aligned}

on [0,T][0, T], where the bond loading σP(⋅,T)\sigma_P(\cdot, T) is predictable with values in Rd\mathbb{R}^d and ∫0T∣σP(s,T)∣2 ds<∞\int_0^T \lvert \sigma_P(s,T) \rvert^2 \, ds < \infty almost surely. The drift is rtr_t because P/BP/B is a Q\mathbb{Q}-martingale. Then dZtT=ZtT σP(t,T)⊤dWtQdZ^{T}_t = Z^{T}_t \, \sigma_P(t,T)^{\top} dW^{\mathbb{Q}}_t, so the density is the stochastic exponential of ∫σP⊤dWQ\int \sigma_P^{\top} dW^{\mathbb{Q}}, and it is a true martingale because it equals P(t,T)/(BtP(0,T))P(t,T)/(B_t P(0,T)). Girsanov’s theorem, in the form for a density E(∫θ⊤dWQ)\mathcal{E}(\int \theta^{\top} dW^{\mathbb{Q}}) with θ=σP(⋅,T)\theta = \sigma_P(\cdot, T), makes

WtT=WtQ−∫0tσP(s,T) ds\begin{aligned} W^{T}_t &= W^{\mathbb{Q}}_t \\[0.4em] &\quad - \int_0^t \sigma_P(s,T) \, ds \end{aligned}

a QT\mathbb{Q}^{T}-Brownian motion on [0,T][0, T]. So dWtQ=dWtT+σP(t,T) dtdW^{\mathbb{Q}}_t = dW^{T}_t + \sigma_P(t,T)\,dt, and an Itô process with dXt=μt dt+σX,t⊤dWtQdX_t = \mu_t \, dt + \sigma_{X,t}^{\top} dW^{\mathbb{Q}}_t has under QT\mathbb{Q}^{T} the dynamics

dXt=(μt+σX,t⊤σP(t,T))dt+σX,t⊤ dWtT.\begin{aligned} dX_t &= \bigl(\mu_t + \sigma_{X,t}^{\top} \sigma_P(t,T)\bigr) dt \\[0.4em] &\quad + \sigma_{X,t}^{\top} \, dW^{T}_t . \end{aligned}

The volatility is unchanged and the drift moves by the instantaneous covariation of XX with the log bond price. In the Heath-Jarrow-Morton framework the bond loading is minus the integrated forward-rate volatility, so the adjustment there carries a minus sign. Working out the loading and the density convention first is safer than remembering a drift sign.

In plain words Moving to the forward measure tilts each Brownian motion toward the direction in which the bond gains value. A quantity whose moves are negatively correlated with the bond price, such as the short rate in the Vasicek model below, picks up a negative drift adjustment.

Example Let the short rate follow the Vasicek dynamics drt=κ(θ−rt) dt+σ dWtQdr_t = \kappa(\theta - r_t)\,dt + \sigma \, dW^{\mathbb{Q}}_t in one dimension. The affine bond formula of Vasicek and CIR short rate models gives the bond loading σP(t,T)=−σB(T−t)\sigma_P(t,T) = -\sigma B(T - t) with B(u)=(1−e−κu)/κB(u) = (1 - e^{-\kappa u})/\kappa. Under QT\mathbb{Q}^{T},

drt=κ(θ−rt) dt−σ2B(T−t) dt+σ dWtT.\begin{aligned} dr_t &= \kappa(\theta - r_t) \, dt \\[0.4em] &\quad - \sigma^2 B(T - t) \, dt \\[0.4em] &\quad + \sigma \, dW^{T}_t . \end{aligned}

For the constructed values κ=0.5\kappa = 0.5 per year and σ=0.01\sigma = 0.01 per square-root year with five years to maturity, B(5)≈1.835830B(5) \approx 1.835830 and the drift adjustment is about −0.0001⋅1.835830≈−0.000184-0.0001 \cdot 1.835830 \approx -0.000184 per year. It shrinks to zero as tt approaches TT, since B(0)=0B(0) = 0.

Check yourself

  1. In a constructed one-period model, B0=1B_0 = 1 and BTB_T equals 1.011.01 or 1.051.05 with Q\mathbb{Q}-probabilities 0.40.4 and 0.60.6. A claim pays 00 in the first state and 100100 in the second. Find P(0,T)P(0,T), the forward-measure probabilities and the price of the claim by both routes.
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The bond costs P(0,T)=0.4/1.01+0.6/1.05=684/707≈0.967468P(0,T) = 0.4/1.01 + 0.6/1.05 = 684/707 \approx 0.967468. The density 1/(BTP(0,T))1/(B_T P(0,T)) is about 1.0233921.023392 in the first state and 0.9844050.984405 in the second. Multiplying by the Q\mathbb{Q}-probabilities gives the forward probabilities 0.4⋅1.023392≈0.4093570.4 \cdot 1.023392 \approx 0.409357 and 0.6⋅0.984405≈0.5906430.6 \cdot 0.984405 \approx 0.590643, exactly 70/17170/171 and 101/171101/171. Cash-account pricing gives 0.6⋅100/1.05=400/7≈57.1428570.6 \cdot 100/1.05 = 400/7 \approx 57.142857. The forward route gives EQT[H]≈59.064327\mathbb{E}^{\mathbb{Q}^{T}}[H] \approx 59.064327 and 0.967468⋅59.064327≈57.1428570.967468 \cdot 59.064327 \approx 57.142857, the same price. The forward measure moves weight toward the first state, where the account grew less.

  1. For the claim of the previous question, a colleague computes P(0,T) EQ[H]P(0,T)\,\mathbb{E}^{\mathbb{Q}}[H], the bond price times the average payoff under the cash-account measure. How far off is this, why, and when would the shortcut be correct?
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The shortcut gives 0.967468⋅60≈58.0480910.967468 \cdot 60 \approx 58.048091, which is about 0.9052330.905233 above the correct 57.14285757.142857. The bond price is the right discount only together with the forward measure. Under Q\mathbb{Q} the exact price is

V0=P(0,T) EQ[H]+Cov⁡Q(H,1/BT),\begin{aligned} V_0 &= P(0,T)\,\mathbb{E}^{\mathbb{Q}}[H] \\[0.4em] &\quad + \operatorname{Cov}^{\mathbb{Q}}(H, 1/B_T), \end{aligned}

and here the covariance is about −0.905233-0.905233, negative because the claim pays only in the state where the account grew more and the discount factor 1/BT1/B_T is smaller. The shortcut is correct exactly when HH and 1/BT1/B_T are uncorrelated under Q\mathbb{Q}, for example when rates are deterministic, in which case QT\mathbb{Q}^{T} and Q\mathbb{Q} agree on FT\mathcal{F}_T.

  1. Under Q\mathbb{Q} the short rate follows drt=κ(θ−rt) dt+σ dWtQdr_t = \kappa(\theta - r_t)\,dt + \sigma\,dW^{\mathbb{Q}}_t, and the bond maturing at TT satisfies dP(t,T)/P(t,T)=rt dt−σB(T−t) dWtQdP(t,T)/P(t,T) = r_t\,dt - \sigma B(T - t)\,dW^{\mathbb{Q}}_t with B(u)=(1−e−κu)/κB(u) = (1 - e^{-\kappa u})/\kappa. Find the drift of rr under QT\mathbb{Q}^{T}, then compute EQ[rT]\mathbb{E}^{\mathbb{Q}}[r_T] and EQT[rT]\mathbb{E}^{\mathbb{Q}^{T}}[r_T] for r0=0.03r_0 = 0.03, κ=0.5\kappa = 0.5, θ=0.04\theta = 0.04, σ=0.01\sigma = 0.01 and T=5T = 5. Which is larger, and why?
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The bond loading is σP(t,T)=−σB(T−t)\sigma_P(t,T) = -\sigma B(T - t), and the forward-measure Brownian motion is WtT=WtQ−∫0tσP(s,T) dsW^{T}_t = W^{\mathbb{Q}}_t - \int_0^t \sigma_P(s,T)\,ds, so dWtQ=dWtT−σB(T−t) dtdW^{\mathbb{Q}}_t = dW^{T}_t - \sigma B(T - t)\,dt. Substituting gives the QT\mathbb{Q}^{T}-drift κ(θ−rt)−σ2B(T−t)\kappa(\theta - r_t) - \sigma^2 B(T - t). The mean m(t)m(t) of a linear equation solves m′=κθ−σ2B(T−t)−κmm' = \kappa\theta - \sigma^2 B(T - t) - \kappa m, and with ∫0Te−κuB(u) du=(1−e−κT)2/(2κ2)\int_0^T e^{-\kappa u} B(u)\,du = (1 - e^{-\kappa T})^2/(2\kappa^2) this gives

EQT[rT]=θ+(r0−θ)e−κT−σ22κ2(1−e−κT)2.\begin{aligned} &\mathbb{E}^{\mathbb{Q}^{T}}[r_T] \\[0.4em] &\quad = \theta + (r_0 - \theta)e^{-\kappa T} \\[0.4em] &\qquad - \frac{\sigma^2}{2\kappa^2}\bigl(1 - e^{-\kappa T}\bigr)^2 . \end{aligned}

The first two terms are EQ[rT]=0.04−0.01 e−2.5≈0.039179\mathbb{E}^{\mathbb{Q}}[r_T] = 0.04 - 0.01\,e^{-2.5} \approx 0.039179. The last term is 0.0002⋅0.842568≈0.0001690.0002 \cdot 0.842568 \approx 0.000169, so EQT[rT]≈0.039011\mathbb{E}^{\mathbb{Q}^{T}}[r_T] \approx 0.039011, which is smaller. The rate and the bond price move in opposite directions, so the forward measure, which favors states where the bond did well, gives less weight to high-rate paths. The value 0.0390110.039011 equals the instantaneous forward rate −∂Tln⁡P(0,T)-\partial_T \ln P(0,T) of this model’s bond curve, an instance of the identity f(0,T)=EQT[rT]f(0,T) = \mathbb{E}^{\mathbb{Q}^{T}}[r_T], which holds whenever differentiation under the expectation is justified.

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