conditional expectation under change of measure

Let $\mathbb{P}$ be a given probability measure on some $\sigma$-algebra $\mathcal{F}$. Suppose a new probability measure $\mathbb{Q}$ is defined by $d\mathbb{Q}=Z\,d\mathbb{P}$, using some $\mathcal{F}$-measurable random variable $Z$ as the Radon-Nikodym derivative. (Necessarily we must have $Z\geq 0$ almost surely, and $\mathbb{E}Z=1$.)

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

Theorem 1.

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

In other words,

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

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

 $\mathbb{Q}(B)=\mathbb{E}\bigl{[}\mathbb{E}[Z\mid\mathcal{G}]\,1_{B}\bigr{]}\,.$

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

 $\mathbb{E}\bigl{[}\mathbb{E}[Z\mid\mathcal{G}]\,1_{B}\bigr{]}=\mathbb{E}\bigl{% [}\mathbb{E}[Z1_{B}\mid\mathcal{G}]\bigr{]}=\mathbb{E}[Z1_{B}]=\mathbb{Q}(B)\,.\qed$
Theorem 2.

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

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

That is,

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

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

 $\displaystyle\mathbb{E}^{\mathbb{Q}}\bigl{[}1_{B}\,\mathbb{E}[ZX\mid\mathcal{G% }]\bigr{]}$ $\displaystyle=\mathbb{E}\bigl{[}Y1_{B}\,\mathbb{E}[ZX\mid\mathcal{G}]\bigr{]}$ (since $d\mathbb{Q}_{\mid\mathcal{G}}=Y\,d\mathbb{P}_{\mid\mathcal{G}}$) $\displaystyle=\mathbb{E}\bigl{[}\mathbb{E}[Y1_{B}\,ZX\mid\mathcal{G}]\bigr{]}$ $\displaystyle=\mathbb{E}[Y1_{B}\,ZX]$ $\displaystyle=\mathbb{E}^{\mathbb{Q}}[Y1_{B}\,X]$ (since $d\mathbb{Q}=Z\,d\mathbb{P}$) $\displaystyle=\mathbb{E}^{\mathbb{Q}}\bigl{[}1_{B}\,\mathbb{E}^{\mathbb{Q}}[YX% \mid\mathcal{G}]\bigr{]}\,.$

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

 $\mathbb{E}[ZX\mid\mathcal{G}]=\mathbb{E}^{\mathbb{Q}}[YX\mid\mathcal{G}]=Y% \mathbb{E}^{\mathbb{Q}}[X\mid\mathcal{G}]\,.\qed$

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

 $\mathbb{E}^{\mathbb{Q}}[X\mid\mathcal{G}]=\mathbb{E}\left[\frac{d\mathbb{Q}}{d% \mathbb{P}}X\mid\mathcal{G}\right]\Big{/}\left(\frac{d\mathbb{Q}}{d\mathbb{P}}% \right)_{\mid\mathcal{G}}\,.$
Theorem 3.

If $X_{t}$ is a martingale with respect to $\mathbb{Q}$ and some filtration $\{\mathcal{F}_{t}\}$, then $X_{t}Z_{t}$ is a martingale with respect to $\mathbb{P}$ and $\{\mathcal{F}_{t}\}$, where $Z_{t}=\mathbb{E}[Z\mid\mathcal{F}_{t}]$.

Proof.

First observe that $X_{t}Z_{t}$ is indeed $\mathcal{F}_{t}$-measurable. Then, we can apply Theorem 2, with $X$ in the statement of that theorem replaced by $X_{t}$, $Z$ replaced by $Z_{t}$, $\mathcal{F}$ replaced by $\mathcal{F}_{t}$, and $\mathcal{G}$ replaced by $\mathcal{F}_{s}$ ($s\leq t$), to obtain:

 $\mathbb{E}[X_{t}Z_{t}\mid\mathcal{F}_{s}]=Z_{s}\,\mathbb{E}^{\mathbb{Q}}[X_{t}% \mid\mathcal{F}_{s}]=Z_{s}X_{s}\,,$

thus proving that $X_{t}Z_{t}$ is a martingale under $\mathbb{P}$ and $\{\mathcal{F}_{t}\}$. ∎

Sometimes the random variables $Z_{t}$ in Theorem 3 are 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 $Z_{t}=\mathbb{E}[Z\mid\mathcal{F}_{t}]$ is always a martingale under $\mathbb{P}$ and $\{\mathcal{F}_{t}\}$.)

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

Theorem 4.

Let $Z_{t}=(d\mathbb{Q}/d\mathbb{P})_{t}>0$ almost surely. Then $X_{t}$ is a martingale with respect to $\mathbb{P}$, if and only if $X_{t}/Z_{t}$ is a martingale with respect to $\mathbb{Q}$.

Title conditional expectation under change of measure ConditionalExpectationUnderChangeOfMeasure 2013-03-22 16:54:21 2013-03-22 16:54:21 stevecheng (10074) stevecheng (10074) 9 stevecheng (10074) Derivation msc 60A10 msc 60-00 Martingale ConditionalExpectation