Numeraires — Stochastic Calculus Notes

Chapter 4 Numeraires

In these notes we introduce numeraires and theorems related to change of numeraires along with their applications.

4.1 Change of Measure

Definition 4.1 (Radon-Nikodym Derivative).

Consider two equivalent probability measures $\mathbb{P}$ and $\hat{\mathbb{P}}$ on a measurable space $(\Omega,\Sigma)$ . The Radon-Nikodym derivate $d\hat{\mathbb{P}}/d\mathbb{P}:\Omega\to\mathbb{R}$ is defined such that for any subset $A,\Omega\supseteqq A\in\Sigma$ ,

\int_{A}d\hat{\mathbb{P}}=\int_{A}(d\hat{\mathbb{P}}/d\mathbb{P})d\mathbb{P}.

(4.1)

Suppose $(\Omega,\mathcal{F},\mathcal{F}_{t})$ is a filtered probability space, then note from the above definition we have

(d\hat{\mathbb{P}}/d\mathbb{P})|_{\mathcal{F}_{t}}=\mathbb{E}^{\mathbb{P}}[(d% \hat{\mathbb{P}}/d\mathbb{P})|\mathcal{F}_{t}]

(4.2)

$(d\hat{\mathbb{P}}/d\mathbb{P})|_{\mathcal{F}_{t}}$ is also written as $(d\hat{\mathbb{P}}/d\mathbb{P})_{t}$ and is thus a martingale stochastic process (by iterated conditioning) in $\mathbb{P}$ .

Example 4.1 Let’s consider a simple example to better understand Radom Nikodym derivative. Consider a die roll. We can assign mulitple probability distributions to the outcomes.

$\omega$	$\mathbb{P}$	$\hat{\mathbb{P}}$	$d\hat{\mathbb{P}}/d\mathbb{P}$
1	1/6	1/2	3
2	1/6	1/4	3/2
3	1/6	1/8	3/4
4	1/6	1/16	3/8
5	1/6	1/32	3/16
6	1/6	1/32	3/16

The probability in $\hat{\mathbb{P}}$ of getting an odd number is $\int_{\omega\in\{1,3,5\}}d\hat{\mathbb{P}}=1/2+1/8+1/32=21/32=3*1/6+3/4*1/6+3/% 16*1/6=\int_{\omega\in\{1,3,5\}}(d\hat{\mathbb{P}}/d\mathbb{P})d\mathbb{P}$

Theorem 4.2 (Abstract Bayes’ Theorem).

Let $\mathbb{P}$ and $\hat{\mathbb{P}}$ be two measures on measurable space $(\Omega,\mathcal{F})$ . Let $\mathcal{G}\subset\mathcal{F}$ be another sigma algebra on $\Omega$ . Then for any $A\in G$ and random variable X

\mathbb{E}^{\hat{\mathbb{P}}}[X|G]=\frac{\mathbb{E}^{\mathbb{P}}[(d\hat{% \mathbb{P}}/d\mathbb{P})X|G]}{\mathbb{E}^{\mathbb{P}}[(d\hat{\mathbb{P}}/d% \mathbb{P})|G]}.

(4.3)

Proof.

We show that for any $A\in G$

\mathbb{E}^{\hat{\mathbb{P}}}[X|G]\mathbb{E}^{\mathbb{P}}[(d\hat{\mathbb{P}}/d% \mathbb{P})|G]=\mathbb{E}^{\mathbb{P}}[(d\hat{\mathbb{P}}/d\mathbb{P})X|G]

Since the random variables involved are constant over A, we can check equality on integrals over A.

$\displaystyle\int_{A}\mathbb{E}^{\hat{\mathbb{P}}}[X\|G]\mathbb{E}^{\mathbb{P}}% [(d\hat{\mathbb{P}}/d\mathbb{P})\|G]d\mathbb{P}$	$\displaystyle=\int_{A}\mathbb{E}^{\mathbb{P}}[(d\hat{\mathbb{P}}/d\mathbb{P})% \mathbb{E}^{\hat{\mathbb{P}}}[X\|G]\|G]d\mathbb{P}$	$\displaystyle(\text{$\mathbb{E}^{\hat{\mathbb{P}}}[X\|G]$ is G measurable})$
	$\displaystyle=\int_{A}(d\hat{\mathbb{P}}/d\mathbb{P})\mathbb{E}^{\hat{\mathbb{% P}}}[X\|G]d\mathbb{P}$	$\displaystyle(\text{definition of conditional expectation})$
	$\displaystyle=\int_{A}\mathbb{E}^{\hat{\mathbb{P}}}[X\|G]d\hat{\mathbb{P}}$	$\displaystyle(\text{definition of Radon Nikodym derivative})$
	$\displaystyle=\int_{A}Xd\hat{\mathbb{P}}$	$\displaystyle(\text{definition of conditional expectation})$
	$\displaystyle=\int_{A}(d\hat{\mathbb{P}}/d\mathbb{P})Xd\mathbb{P}$	$\displaystyle(\text{definition of Radon Nikodym derivative})$
	$\displaystyle=\int_{A}\mathbb{E}^{\mathbb{P}}[(d\hat{\mathbb{P}}/d\mathbb{P})X% \|G]d\mathbb{P}$	$\displaystyle(\text{definition of conditional expectation}).$

∎

Remark.

Taking $X=V_{T}$ where $V_{s}$ is $\mathcal{F}_{s}$ adapted and $G=\mathcal{F}_{t}$ for $t<T$ in the above theorem, we get the very useful formula for measure change for conditional expectations on filtered spaces,

\mathbb{E}^{\hat{\mathbb{P}}}[V_{T}|\mathcal{F}_{t}]=\mathbb{E}^{\mathbb{P}}[% \frac{(d\hat{\mathbb{P}}/d\mathbb{P})_{T}}{(d\hat{\mathbb{P}}/d\mathbb{P})_{t}% }V_{T}|\mathcal{F}_{t}]

(4.4)

Proof.

$\displaystyle\mathbb{E}^{\hat{\mathbb{P}}}[V_{T}\|\mathcal{F}_{t}]$	$\displaystyle=\frac{\mathbb{E}^{\mathbb{P}}[(d\hat{\mathbb{P}}/d\mathbb{P})V_{% T}\|\mathcal{F}_{t}]}{\mathbb{E}^{\mathbb{P}}[(d\hat{\mathbb{P}}/d\mathbb{P})\|% \mathcal{F}_{t}]}$	$\displaystyle(\text{abstract Bayes' theorem})$
	$\displaystyle=\frac{\mathbb{E}^{\mathbb{P}}[\mathbb{E}^{\mathbb{P}}[(d\hat{% \mathbb{P}}/d\mathbb{P})V_{T}\|\mathcal{F}_{T}]\|\mathcal{F}_{t}]}{\mathbb{E}^{% \mathbb{P}}[(d\hat{\mathbb{P}}/d\mathbb{P})\|\mathcal{F}_{t}]}$	$\displaystyle(\text{iterated conditioning})$
	$\displaystyle=\frac{\mathbb{E}^{\mathbb{P}}[\mathbb{E}^{\mathbb{P}}[(d\hat{% \mathbb{P}}/d\mathbb{P})\|\mathcal{F}_{T}]V_{T}\|\mathcal{F}_{t}]}{\mathbb{E}^{% \mathbb{P}}[(d\hat{\mathbb{P}}/d\mathbb{P})\|\mathcal{F}_{t}]}$	$\displaystyle(\text{$V_{T}$ is $\mathcal{F}_{T}$ measureable})$
	$\displaystyle=\frac{\mathbb{E}^{\mathbb{P}}[(d\hat{\mathbb{P}}/d\mathbb{P})_{T% }V_{T}\|\mathcal{F}_{t}]}{(d\hat{\mathbb{P}}/d\mathbb{P})_{t}}$	$\displaystyle(\text{martigale property of Radon Nikodym derivative})$
	$\displaystyle=\mathbb{E}^{\mathbb{P}}[\frac{(d\hat{\mathbb{P}}/d\mathbb{P})_{T% }}{(d\hat{\mathbb{P}}/d\mathbb{P})_{t}}V_{T}\|\mathcal{F}_{t}]$	$\displaystyle(\text{$(d\hat{\mathbb{P}}/d\mathbb{P})_{t}$ is $\mathcal{F}_{t}$% measurable}).$

∎

We consider processes until terminal time S i.e. $\mathcal{F}=\mathcal{F}_{S}$ and $(d\hat{\mathbb{P}}/\mathbb{P})=(d\hat{\mathbb{P}}/\mathbb{P})_{S}$ . We take a strictly positive martingale process to be the Random Nikodym derivative:

df_{t}=f_{t}\sigma(t)dW_{t};\,\,\,f_{t}=e^{-\int_{0}^{t}\frac{1}{2}\sigma^{2}(% s)ds+\int_{0}^{t}\sigma(s)dW_{s}}.

(4.5)

Theorem 4.3 (Girsanov Theorem).

If $W_{1,t}$ is a Brownian motion in $\mathbb{P}$ and the Radon-Nikodym derivative $(d\hat{\mathbb{P}}/d\mathbb{P})_{t}=f_{t}$ is given by $f_{t}=e^{-\int_{0}^{t}\frac{1}{2}\sigma^{2}(s)ds+\int_{0}^{t}\sigma(s)dW_{2,s}% ^{\mathbb{P}}}$ then $W_{1,t}-\int_{0}^{t}\sigma(s)dW_{1,s}dW_{2,s}$ is a Brownian motion in $\hat{\mathbb{P}}$ .

Proof.

We show that $X_{t}=W_{1,t}-\int_{0}^{t}\sigma(s)dW_{1,s}dW_{2,s}$ follows normal distribution with mean 0 and variance t in $\hat{\mathbb{P}}$ . Other required properties can be verified easily. We show that the moment generating function of $X_{t}$ is same as that of normal distribution with mean 0 and variance t.

	$\displaystyle\mathbb{E}^{\hat{\mathbb{P}}}[e^{-yX_{t}}]$	$\displaystyle=\mathbb{E}^{\mathbb{P}}[(d\hat{\mathbb{P}}/d\mathbb{P})_{t}e^{-% yX_{t}}]$
		$\displaystyle=\mathbb{E}^{\mathbb{P}}[e^{-\int_{0}^{t}\frac{1}{2}\sigma^{2}(s)% ds+\int_{0}^{t}\sigma(s)dW_{2,s}}e^{-yX_{t}}]$
		$\displaystyle=e^{-\int_{0}^{t}\frac{1}{2}\sigma^{2}(s)ds}\mathbb{E}^{\mathbb{P% }}[e^{\int_{0}^{t}\sigma(s)dW_{2,s}-yX_{t}}]$
		$\displaystyle=e^{-\int_{0}^{t}\frac{1}{2}\sigma^{2}(s)ds}\mathbb{E}^{\mathbb{P% }}[e^{\int_{0}^{t}\sigma(s)dW_{2,s}-y(W_{1,t}-\int_{0}^{t}\sigma(s)dW_{1,s}dW_% {2,s})}]$
		$\displaystyle=e^{-\int_{0}^{t}\frac{1}{2}\sigma^{2}(s)ds}\mathbb{E}^{\mathbb{P% }}[e^{\int_{0}^{t}(\sigma(s)dW_{2,s}-ydW_{1,s}+y\sigma(s)dW_{1,s}dW_{2,s})}].$

Define the martingale $Z_{y,t}=\int_{0}^{t}(\sigma(s)dW_{2,s}-ydW_{1,s})$ . Then $e^{Z_{y,t}-\frac{1}{2}\int_{0}^{t}dZ_{y,s}dZ_{y,s}}$ is a martingale as well, with

dZ_{y,s}dZ_{y,s}=(\sigma^{2}(s)+y^{2})ds-2y\sigma dW_{1,s}dW_{2,s}.

We then have

	$\displaystyle\mathbb{E}^{\hat{\mathbb{P}}}[e^{-yX_{t}}]$	$\displaystyle=e^{-\int_{0}^{t}\frac{1}{2}\sigma^{2}(s)ds}\mathbb{E}^{\mathbb{P% }}[e^{Z_{y,t}-\frac{1}{2}\int_{0}^{t}dZ_{y,s}dZ_{y,s}+\frac{1}{2}\int_{0}^{t}(% \sigma^{2}(s)+y^{2})ds}]$
		$\displaystyle=e^{\frac{1}{2}y^{2}t}\mathbb{E}^{\mathbb{P}}[e^{Z_{y,t}-\frac{1}% {2}\int_{0}^{t}dZ_{y,s}dZ_{y,s}}]$
		$\displaystyle=e^{\frac{1}{2}y^{2}t}(e^{Z_{y,t}-\frac{1}{2}\int_{0}^{t}dZ_{y,s}% dZ_{y,s}})\|_{t=0}$
		$\displaystyle=e^{\frac{1}{2}y^{2}t}.$

∎

4.2 Numeraires

Definition 4.4 (Numeraire).

A Numeraire is a strictly positive price process of a tradable relative to which prices of all other tradables are expressed.

A savings account which earns interest at the instantaneous interest rate can be taken as a numeraire. The value of the savings account at any point is given by

A_{t}=e^{\int_{0}^{t}r(t)dt}=1/D(t)

(4.6)

where $D(t)$ is the discount factor.

The fact that discounted trade prices are martingales in risk-neutral measure can then also be stated as: tradable prices in savings account (or money market) numeraire are martingales.

Consider another numeraire $N_{t}$ . Since the numeraire is itself a price process, $D_{t}N_{t}$ is a martingale and we can take it as a Radon-Nikodym derivative, with a suitable normalization so that $\mathbb{E}^{\mathbb{P}}[(d\mathbb{N}/d\mathbb{P})]=1$ , giving

\mathbb{E}^{\mathbb{N}}[\frac{X_{T}}{N_{T}}|\mathcal{F}_{t}]=\mathbb{E}^{% \mathbb{P}}[\frac{(d\mathbb{N}/d\mathbb{P})_{T}}{(d\mathbb{N}/d\mathbb{P})_{t}% }\frac{X_{T}}{N_{T}}|\mathcal{F}_{t}]=\mathbb{E}^{\mathbb{P}}[\frac{D_{T}N_{T}% }{D_{t}N_{t}}\frac{X_{T}}{N_{T}}|\mathcal{F}_{t}]=\frac{1}{D_{t}N_{t}}\mathbb{% E}^{\mathbb{P}}[D_{T}X_{T}|\mathcal{F}_{t}]=\frac{X_{t}}{N_{t}}

(4.7)

where $\mathbb{P}$ is the risk neutral measure and $\mathbb{N}$ is the measure corresponding to the numeraire $N_{t}$ .

Remark.

The above equation shows that tradable prices with respect to a numeraire are martingales in the measure corresponding to the numeraire.

Risk neutral measure corresponds to the savings account (or money market) numeraire.

Definition 4.5 (T-forward measure).

If we take $N_{t}$ to be the price of a riskless bond maturing at time T, the corresponding measure is the T-forward measure.

X_{t}/B(t,T)=\mathbb{E}^{T}[X_{T}/B(T,T)]=\mathbb{E}^{T}[X_{T}]

(4.8)

We see that that expectation of $X_{T}$ in the T-forward measure gives the forward price $X_{t}/B(t,T)$ of the trade with expiration date T, hence the name T-forward measure. From the above equation we also notice that expiration T forward price process on an asset is martingale in T-forward measure.

Example 4.2 Option Pricing To see how measure changes can be used in pricing, let’s take the example of Black Scholes option pricing. For an option on stock with expiry at T and strike K, we have the valuation forumula

V_{t}=\frac{1}{D_{t}}\mathbb{E}^{\mathbb{P}}[D_{T}\max(S_{T}-K,0)|\mathcal{F}_% {t}]

where $\mathbb{P}$ is the risk neutral probability measure, $D_{t}$ is the discount factor and $S_{t}$ is the stock price which follows the Black Scholes diffusion

\frac{dS_{t}}{S_{t}}=rdt+\sigma dW_{t}

where $r$ is the riskfree interest rate and $W_{t}$ is a Brownian motion in $\mathbb{P}$ .

Denote by $\mathbb{I}_{S_{T}>K}$ the indicator variable which takes value 1 is $S_{T}>K$ and 0 otherwise. We then have

	$\displaystyle V_{t}$	$\displaystyle=\frac{1}{D_{t}}\mathbb{E}^{\mathbb{P}}[D_{T}\mathbb{I}_{S_{T}>K}% (S_{T}-K)\|\mathcal{F}_{t}]$
		$\displaystyle=\frac{1}{D_{t}}\mathbb{E}^{\mathbb{P}}[D_{T}\mathbb{I}_{S_{T}>K}% S_{T}\|\mathcal{F}_{t}]-\frac{1}{D_{t}}\mathbb{E}^{\mathbb{P}}[D_{T}\mathbb{I}_% {S_{T}>K}K\|\mathcal{F}_{t}]$

Taking $\mathbb{S}$ to be the measure corresponding to stock price as numeraire, and taking $\mathbb{T}$ to be the measure corresponding to T-expiry bond as numeraire, we have

	$\displaystyle V_{t}$	$\displaystyle=\frac{1}{D_{t}}\mathbb{E}^{\mathbb{S}}[D_{T}\frac{(d\mathbb{P}/d% \mathbb{S})_{T}}{(d\mathbb{P}/d\mathbb{S})_{t}}\mathbb{I}_{S_{T}>K}S_{T}\|% \mathcal{F}_{t}]-\frac{1}{D_{t}}\mathbb{E}^{\mathbb{T}}[D_{T}\frac{(d\mathbb{P% }/d\mathbb{T})_{T}}{(d\mathbb{P}/d\mathbb{T})_{t}}\mathbb{I}_{S_{T}>K}K\|% \mathcal{F}_{t}]$
		$\displaystyle=\frac{1}{D_{t}}\mathbb{E}^{\mathbb{S}}[D_{T}\frac{(d\mathbb{S}/d% \mathbb{P})_{t}}{(d\mathbb{S}/d\mathbb{P})_{T}}\mathbb{I}_{S_{T}>K}S_{T}\|% \mathcal{F}_{t}]-\frac{1}{D_{t}}\mathbb{E}^{\mathbb{T}}[D_{T}\frac{(d\mathbb{T% }/d\mathbb{P})_{t}}{(d\mathbb{T}/d\mathbb{P})_{T}}\mathbb{I}_{S_{T}>K}K\|% \mathcal{F}_{t}]$
		$\displaystyle=\frac{1}{D_{t}}\mathbb{E}^{\mathbb{S}}[D_{T}\frac{D_{t}S_{t}}{D_% {T}S_{T}}\mathbb{I}_{S_{T}>K}S_{T}\|\mathcal{F}_{t}]-\frac{1}{D_{t}}\mathbb{E}^% {\mathbb{T}}[D_{T}\frac{D_{t}(B(t,T))}{D_{T}B(T,T)}\mathbb{I}_{S_{T}>K}K\|% \mathcal{F}_{t}]$
		$\displaystyle=S_{t}\mathbb{E}^{\mathbb{S}}[\mathbb{I}_{S_{T}>K}\|\mathcal{F}_{t% }]-KB(t,T)\mathbb{E}^{\mathbb{T}}[\mathbb{I}_{S_{T}>K}\|\mathcal{F}_{t}]$

where $B(t,T)$ is the bond price with the diffusion

dB(t,T)/B(t,T)=rdt+\sigma_{2}dW_{2,t}

where $W_{2,t}$ is another Brownian motion in $\mathbb{P}$ such that $dW_{t}dW_{2,t}=\rho dt$ .

Using Girsaonov’s theorem, we have

	$\displaystyle dW_{t}^{\mathbb{S}}$	$\displaystyle=dW_{t}^{\mathbb{P}}-\sigma dW_{t}^{\mathbb{P}}dW_{t}^{\mathbb{P}}$	$\displaystyle=dW_{t}^{\mathbb{P}}-\sigma dt$
	$\displaystyle dW_{t}^{\mathbb{T}}$	$\displaystyle=dW_{t}^{\mathbb{P}}-\sigma_{2}dW_{t}^{\mathbb{P}}dW_{2,t}^{% \mathbb{P}}$	$\displaystyle=dW_{t}^{\mathbb{P}}-\sigma_{2}\rho dt$

as Brownian motions in the measures corresponding to S, and B(t,T) numeraires respectively.

\frac{dS_{t}}{S_{t}}=rdt+\sigma(dW_{t}^{\mathbb{S}}+\sigma dt)=rdt+\sigma(dW_{% t}^{\mathbb{T}}+\sigma_{2}\rho dt)

(4.9)

where $dW_{t}^{\mathbb{S}}=dW_{t}-\sigma dt$ is a Brownian motion in $\mathbb{S}$ and $dW_{t}^{\mathbb{T}}=dW_{t}^{\mathbb{S}}-\sigma_{2}\rho dt$ is a Brownian motion in $\mathbb{T}$ . Rearranging,

\frac{dS_{t}}{S_{t}}=(r+\sigma^{2})dt+\sigma dW_{t}^{\mathbb{S}}=(r+\sigma% \sigma_{2}\rho)dt+\sigma dW_{t}^{\mathbb{T}}

(4.10)

{S_{T}}=S_{t}e^{(r+\frac{1}{2}\sigma^{2})\tau+\sigma W_{\tau}^{\mathbb{S}}}=S_% {t}e^{(r+\sigma(\sigma_{2}\rho-\frac{1}{2}\sigma)\tau+\sigma W_{\tau}^{\mathbb% {T}}}

(4.11)

where $\tau=T-t$ .

	$\displaystyle\mathbb{E}^{\mathbb{S}}[\mathbb{I}_{S_{T}>K}\|\mathcal{F}_{t}]$	$\displaystyle=\mathbb{E}^{\mathbb{S}}[\mathbb{I}_{S_{t}e^{(r+\frac{1}{2}\sigma% ^{2})\tau+\sigma W_{\tau}^{\mathbb{S}}}>K}\|\mathcal{F}_{t}]$
		$\displaystyle=\mathbb{E}^{\mathbb{S}}[\mathbb{I}_{(r+\frac{1}{2}\sigma^{2})% \tau+\sigma W_{\tau}^{\mathbb{S}}>\log(K/S_{t})}\|\mathcal{F}_{t}]$
		$\displaystyle=\mathbb{E}^{\mathbb{S}}[\mathbb{I}_{\sigma W_{\tau}^{\mathbb{S}}% >\log(K/S_{t})-(r+\frac{1}{2}\sigma^{2})\tau}\|\mathcal{F}_{t}]$
		$\displaystyle=\mathbb{E}^{\mathbb{S}}[\mathbb{I}_{\frac{1}{\sqrt{\tau}}W_{\tau% }^{\mathbb{S}}>\frac{1}{\sigma\sqrt{\tau}}(\log(K/S_{t})-(r+\frac{1}{2}\sigma^% {2})\tau)})\|\mathcal{F}_{t}]$
		$\displaystyle=N(-\frac{1}{\sigma\sqrt{\tau}}(\log(K/S_{t})-(r+\frac{1}{2}% \sigma^{2})\tau))$
		$\displaystyle=N(d_{1})$

where $N$ is the cumulative normal distribution, and $d_{1}=\frac{1}{\sigma\sqrt{\tau}}(\log(S_{t}/K)+(r+\frac{1}{2}\sigma^{2})\tau)$ .

Similarly we have,

	$\displaystyle\mathbb{E}^{\mathbb{T}}[\mathbb{I}_{S_{T}>K}\|\mathcal{F}_{t}]$	$\displaystyle=\mathbb{E}^{\mathbb{T}}[\mathbb{I}_{S_{t}e^{(r+\sigma(\sigma_{2}% \rho-\frac{1}{2}\sigma)\tau+\sigma W_{\tau}^{\mathbb{S}}}>K}\|\mathcal{F}_{t}]$
		$\displaystyle=\mathbb{E}^{\mathbb{T}}[\mathbb{I}_{(r+\sigma(\sigma_{2}\rho-% \frac{1}{2}\sigma)\tau+\sigma W_{\tau}^{\mathbb{S}}>\log(K/S_{t})}\|\mathcal{F}% _{t}]$
		$\displaystyle=\mathbb{E}^{\mathbb{T}}[\mathbb{I}_{\sigma W_{\tau}^{\mathbb{S}}% >\log(K/S_{t})-(r+\sigma(\sigma_{2}\rho-\frac{1}{2}\sigma)\tau}\|\mathcal{F}_{t}]$
		$\displaystyle=\mathbb{E}^{\mathbb{T}}[\mathbb{I}_{\frac{1}{\sqrt{\tau}}W_{\tau% }^{\mathbb{S}}>\frac{1}{\sigma\sqrt{\tau}}(\log(K/S_{t})-(r+\sigma(\sigma_{2}% \rho-\frac{1}{2}\sigma)\tau)}\|\mathcal{F}_{t}]$
		$\displaystyle=N(-\frac{1}{\sigma\sqrt{\tau}}(\log(K/S_{t})-(r+\sigma(\sigma_{2% }\rho-\frac{1}{2}\sigma)\tau))$
		$\displaystyle=N(d_{2})$

where $N$ is the cumulative normal distribution, and $d_{2}=\frac{1}{\sigma\sqrt{\tau}}(\log(S_{t}/K)+(r+\sigma\sigma_{2}\rho-\frac{% 1}{2}\sigma^{2})\tau)$ .

And finally we have,

	$\displaystyle V_{t}$	$\displaystyle=S_{t}N(d_{1})-KB(t,T)N(d_{2})$
		$\displaystyle=S_{t}N(d+\sigma\sqrt{\tau})-KB(t,T)N(d+\sigma_{2}\rho\sqrt{\tau})$

where $d=\frac{1}{\sigma\sqrt{\tau}}(\log(S_{t}/K)+(r-\frac{1}{2}\sigma^{2})\tau)$ .

Note that $\sigma d\tau$ and $\sigma_{2}\rho d\tau$ are the drift correction terms we get to stock price Brownian motion using Girsanov’s theorem to change measure to stock and bond numeraires respectively.

$\displaystyle\mathbb{E}^{\hat{\mathbb{P}}}[V_{T}\|\mathcal{F}_{t}]$	$\displaystyle=\frac{\mathbb{E}^{\mathbb{P}}[(d\hat{\mathbb{P}}/d\mathbb{P})V_{% T}\|\mathcal{F}_{t}]}{\mathbb{E}^{\mathbb{P}}[(d\hat{\mathbb{P}}/d\mathbb{P})\|% \mathcal{F}_{t}]}$	$\displaystyle(\text{abstract Bayes' theorem})$
	$\displaystyle=\frac{\mathbb{E}^{\mathbb{P}}[\mathbb{E}^{\mathbb{P}}[(d\hat{% \mathbb{P}}/d\mathbb{P})V_{T}\|\mathcal{F}_{T}]\|\mathcal{F}_{t}]}{\mathbb{E}^{% \mathbb{P}}[(d\hat{\mathbb{P}}/d\mathbb{P})\|\mathcal{F}_{t}]}$	$\displaystyle(\text{iterated conditioning})$
	$\displaystyle=\frac{\mathbb{E}^{\mathbb{P}}[\mathbb{E}^{\mathbb{P}}[(d\hat{% \mathbb{P}}/d\mathbb{P})\|\mathcal{F}_{T}]V_{T}\|\mathcal{F}_{t}]}{\mathbb{E}^{% \mathbb{P}}[(d\hat{\mathbb{P}}/d\mathbb{P})\|\mathcal{F}_{t}]}$	$\displaystyle(\text{$V_{T}$ is $\mathcal{F}_{T}$ measureable})$
	$\displaystyle=\frac{\mathbb{E}^{\mathbb{P}}[(d\hat{\mathbb{P}}/d\mathbb{P})_{T% }V_{T}\|\mathcal{F}_{t}]}{(d\hat{\mathbb{P}}/d\mathbb{P})_{t}}$	$\displaystyle(\text{martigale property of Radon Nikodym derivative})$
	$\displaystyle=\mathbb{E}^{\mathbb{P}}[\frac{(d\hat{\mathbb{P}}/d\mathbb{P})_{T% }}{(d\hat{\mathbb{P}}/d\mathbb{P})_{t}}V_{T}\|\mathcal{F}_{t}]$	$\displaystyle(\text{$(d\hat{\mathbb{P}}/d\mathbb{P})_{t}$ is $\mathcal{F}_{t}$% measurable}).$

	$\displaystyle V_{t}$	$\displaystyle=\frac{1}{D_{t}}\mathbb{E}^{\mathbb{S}}[D_{T}\frac{(d\mathbb{P}/d% \mathbb{S})_{T}}{(d\mathbb{P}/d\mathbb{S})_{t}}\mathbb{I}_{S_{T}>K}S_{T}\|% \mathcal{F}_{t}]-\frac{1}{D_{t}}\mathbb{E}^{\mathbb{T}}[D_{T}\frac{(d\mathbb{P% }/d\mathbb{T})_{T}}{(d\mathbb{P}/d\mathbb{T})_{t}}\mathbb{I}_{S_{T}>K}K\|% \mathcal{F}_{t}]$
		$\displaystyle=\frac{1}{D_{t}}\mathbb{E}^{\mathbb{S}}[D_{T}\frac{(d\mathbb{S}/d% \mathbb{P})_{t}}{(d\mathbb{S}/d\mathbb{P})_{T}}\mathbb{I}_{S_{T}>K}S_{T}\|% \mathcal{F}_{t}]-\frac{1}{D_{t}}\mathbb{E}^{\mathbb{T}}[D_{T}\frac{(d\mathbb{T% }/d\mathbb{P})_{t}}{(d\mathbb{T}/d\mathbb{P})_{T}}\mathbb{I}_{S_{T}>K}K\|% \mathcal{F}_{t}]$
		$\displaystyle=\frac{1}{D_{t}}\mathbb{E}^{\mathbb{S}}[D_{T}\frac{D_{t}S_{t}}{D_% {T}S_{T}}\mathbb{I}_{S_{T}>K}S_{T}\|\mathcal{F}_{t}]-\frac{1}{D_{t}}\mathbb{E}^% {\mathbb{T}}[D_{T}\frac{D_{t}(B(t,T))}{D_{T}B(T,T)}\mathbb{I}_{S_{T}>K}K\|% \mathcal{F}_{t}]$
		$\displaystyle=S_{t}\mathbb{E}^{\mathbb{S}}[\mathbb{I}_{S_{T}>K}\|\mathcal{F}_{t% }]-KB(t,T)\mathbb{E}^{\mathbb{T}}[\mathbb{I}_{S_{T}>K}\|\mathcal{F}_{t}]$

	$\displaystyle\mathbb{E}^{\mathbb{S}}[\mathbb{I}_{S_{T}>K}\|\mathcal{F}_{t}]$	$\displaystyle=\mathbb{E}^{\mathbb{S}}[\mathbb{I}_{S_{t}e^{(r+\frac{1}{2}\sigma% ^{2})\tau+\sigma W_{\tau}^{\mathbb{S}}}>K}\|\mathcal{F}_{t}]$
		$\displaystyle=\mathbb{E}^{\mathbb{S}}[\mathbb{I}_{(r+\frac{1}{2}\sigma^{2})% \tau+\sigma W_{\tau}^{\mathbb{S}}>\log(K/S_{t})}\|\mathcal{F}_{t}]$
		$\displaystyle=\mathbb{E}^{\mathbb{S}}[\mathbb{I}_{\sigma W_{\tau}^{\mathbb{S}}% >\log(K/S_{t})-(r+\frac{1}{2}\sigma^{2})\tau}\|\mathcal{F}_{t}]$
		$\displaystyle=\mathbb{E}^{\mathbb{S}}[\mathbb{I}_{\frac{1}{\sqrt{\tau}}W_{\tau% }^{\mathbb{S}}>\frac{1}{\sigma\sqrt{\tau}}(\log(K/S_{t})-(r+\frac{1}{2}\sigma^% {2})\tau)})\|\mathcal{F}_{t}]$
		$\displaystyle=N(-\frac{1}{\sigma\sqrt{\tau}}(\log(K/S_{t})-(r+\frac{1}{2}% \sigma^{2})\tau))$
		$\displaystyle=N(d_{1})$

	$\displaystyle\mathbb{E}^{\mathbb{T}}[\mathbb{I}_{S_{T}>K}\|\mathcal{F}_{t}]$	$\displaystyle=\mathbb{E}^{\mathbb{T}}[\mathbb{I}_{S_{t}e^{(r+\sigma(\sigma_{2}% \rho-\frac{1}{2}\sigma)\tau+\sigma W_{\tau}^{\mathbb{S}}}>K}\|\mathcal{F}_{t}]$
		$\displaystyle=\mathbb{E}^{\mathbb{T}}[\mathbb{I}_{(r+\sigma(\sigma_{2}\rho-% \frac{1}{2}\sigma)\tau+\sigma W_{\tau}^{\mathbb{S}}>\log(K/S_{t})}\|\mathcal{F}% _{t}]$
		$\displaystyle=\mathbb{E}^{\mathbb{T}}[\mathbb{I}_{\sigma W_{\tau}^{\mathbb{S}}% >\log(K/S_{t})-(r+\sigma(\sigma_{2}\rho-\frac{1}{2}\sigma)\tau}\|\mathcal{F}_{t}]$
		$\displaystyle=\mathbb{E}^{\mathbb{T}}[\mathbb{I}_{\frac{1}{\sqrt{\tau}}W_{\tau% }^{\mathbb{S}}>\frac{1}{\sigma\sqrt{\tau}}(\log(K/S_{t})-(r+\sigma(\sigma_{2}% \rho-\frac{1}{2}\sigma)\tau)}\|\mathcal{F}_{t}]$
		$\displaystyle=N(-\frac{1}{\sigma\sqrt{\tau}}(\log(K/S_{t})-(r+\sigma(\sigma_{2% }\rho-\frac{1}{2}\sigma)\tau))$
		$\displaystyle=N(d_{2})$