conditional expectation under change of measure

Let $\mathbb{P}$ be a given probability measureon some $\sigma$-algebra $ℱ$.Suppose a new probability measure $\mathbb{Q}$ is definedby $d\mathbb{Q}=Zd\mathbb{P}$, using some $ℱ$-measurablerandom variable $Z$as the Radon-Nikodym derivative.(Necessarily we must have $Z\ge 0$ almost surely, and $𝔼Z=1$.)

We denote with $𝔼$ the expectation with respect to the measure $\mathbb{P}$,and with ${𝔼}^{\mathbb{Q}}$ the expectation with respect to the measure $\mathbb{Q}$.

Theorem 1.

If $\mathrm{Q}$ is restricted to a sub-$\sigma$-algebra$\mathrm{G}\mathrm{\subseteq }\mathrm{F}$,then the restriction has the conditional expectation$\mathrm{E}\mathrm{[}Z\mathrm{\mid }\mathrm{G}\mathrm{]}$ as its Radon-Nikodym derivative:$d{\mathrm{Q}}_{\mathrm{\mid }\mathrm{G}}\mathrm{=}\mathrm{E}\mathrm{[}Z\mathrm{\mid }\mathrm{G}\mathrm{]}d{\mathrm{P}}_{\mathrm{\mid }\mathrm{G}}$.

In other words,

$$\frac{d{\mathbb{Q}}_{\mid 𝒢}}{d{\mathbb{P}}_{\mid 𝒢}}={\left(\frac{d\mathbb{Q}}{d\mathbb{P}}\right)}_{\mid 𝒢}.$$

Proof.

It is required to prove that, for all $B\in 𝒢$,

$$\mathbb{Q}(B)=𝔼\left[𝔼[Z\mid 𝒢]{\mathrm{\hspace{0.17em}1}}_B\right].$$

But this follows at once from the law of iterated conditional expectations:

$$𝔼\left[𝔼[Z\mid 𝒢]{\mathrm{\hspace{0.17em}1}}_B\right]=𝔼\left[𝔼[Z{1}_B\mid 𝒢]\right]=𝔼[Z{1}_B]=\mathbb{Q}(B).\mathit{∎}$$

Theorem 2.

Let $\mathrm{G}\mathrm{\subseteq }\mathrm{F}$ be any sub-$\sigma$-algebra.For any $\mathrm{F}$-measurable random variable $X$,

$$𝔼[Z\mid 𝒢]{𝔼}^{\mathbb{Q}}[X\mid 𝒢]=𝔼[ZX\mid 𝒢].$$

That is,

$${\left(\frac{d\mathbb{Q}}{d\mathbb{P}}\right)}_{\mid 𝒢}{𝔼}^{\mathbb{Q}}[X\mid 𝒢]=𝔼[\frac{d\mathbb{Q}}{d\mathbb{P}}X\mid 𝒢].$$

Proof.

Let $Y=𝔼[Z\mid 𝒢]$, and$B\in 𝒢$. We find:

${𝔼}^{\mathbb{Q}}\left[1_B𝔼[ZX\mid 𝒢]\right]$$=𝔼\left[Y{1}_B𝔼[ZX\mid 𝒢]\right]$(since $d{\mathbb{Q}}_{\mid 𝒢}=Yd{\mathbb{P}}_{\mid 𝒢}$)

$=𝔼\left[𝔼[Y{1}_{B}ZX\mid 𝒢]\right]$

$=𝔼[Y{1}_{B}ZX]$

$={𝔼}^{\mathbb{Q}}[Y{1}_{B}X]$(since $d\mathbb{Q}=Zd\mathbb{P}$)

$={𝔼}^{\mathbb{Q}}\left[1_B{𝔼}^{\mathbb{Q}}[YX\mid 𝒢]\right].$

Since $B\in 𝒢$ is arbitrary, we can equate the $𝒢$-measurable integrands:

$$𝔼[ZX\mid 𝒢]={𝔼}^{\mathbb{Q}}[YX\mid 𝒢]=Y{𝔼}^{\mathbb{Q}}[X\mid 𝒢].\mathit{∎}$$

Observe that if $d\mathbb{Q}/d\mathbb{P}>0$ almost surely,then

$${𝔼}^{\mathbb{Q}}[X\mid 𝒢]=𝔼[\frac{d\mathbb{Q}}{d\mathbb{P}}X\mid 𝒢]/{\left(\frac{d\mathbb{Q}}{d\mathbb{P}}\right)}_{\mid 𝒢}.$$

Theorem 3.

If $X_t$ is a martingale with respect to $\mathrm{Q}$ and somefiltration $\mathrm{\{}{\mathrm{F}}_t\mathrm{\}}$,then $X_t\mathit{}{Z}_t$ is a martingale with respect to $\mathrm{P}$ and $\mathrm{\{}{\mathrm{F}}_t\mathrm{\}}$,where $Z_t\mathrm{=}\mathrm{E}\mathrm{[}Z\mathrm{\mid }{\mathrm{F}}_t\mathrm{]}$.

Proof.

First observe that $X_t{Z}_t$ is indeed ${ℱ}_t$-measurable.Then, we can apply Theorem 2,with $X$ in the statement of that theorem replaced by$X_t$,$Z$ replaced by $Z_t$,$ℱ$ replaced by ${ℱ}_t$, and $𝒢$ replaced by ${ℱ}_s$ ($s\le t$),to obtain:

$$𝔼[X_t{Z}_t\mid {ℱ}_s]=Z_s{𝔼}^{\mathbb{Q}}[X_t\mid {ℱ}_s]=Z_s{X}_s,$$

thus proving that $X_t{Z}_t$ is a martingale under $\mathbb{P}$and $\{{ℱ}_t\}$.∎

Sometimes the random variables $Z_t$ in Theorem 3are written as ${\left(\frac{d\mathbb{Q}}{d\mathbb{P}}\right)}_t$.(This is a Radon-Nikodym derivative process;note that $Z_t$ defined as $Z_t=𝔼[Z\mid {ℱ}_t]$is always a martingaleunder $\mathbb{P}$ and $\{{ℱ}_t\}$.)

Under the hypothesis $Z_t>0$,there is an alternate restatement of Theorem 3that may be more easily remembered:

Theorem 4.

Let $Z_t\mathrm{=}{\mathrm{(}d\mathit{}\mathrm{Q}\mathrm{/}d\mathit{}\mathrm{P}\mathrm{)}}_t\mathrm{>}\mathrm{0}$ almost surely.Then $X_t$ is a martingale with respect to $\mathrm{P}$,if and only if $X_t\mathrm{/}{Z}_t$ is a martingale with respect to $\mathrm{Q}$.