PlanetMath: conditional expectation under change of measure

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

We denote with $\E$ the expectation with respect to the measure $\PP$ , and with $\EQ$ the expectation with respect to the measure $\PQ$ .

Theorem 1 If $\PQ$ is restricted to a sub-$\sigma$ -algebra $\sG \subseteq \sF$ , then the restriction has the conditional expectation $\E[Z \mid \sG]$ as its Radon-Nikodym derivative: $d \PQ_{\mid\sG} = \E[ Z \mid \sG ] \, d\PP_{\mid \sG}$ .

In other words, $$ \frac{d\PQ_{\mid \sG}}{d\PP_{\mid \sG}} = \left( \frac{d\PQ}{d\PP} \right)_{\mid \sG}\,. $$

Proof. It is required to prove that, for all $B \in \sG$ , $$ \PQ(B) = \E \bigl[ \E [ Z \mid \sG] \, 1_B \bigr]\,. $$ But this follows at once from the law of iterated conditional expectations:

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

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Theorem 2 Let $\sG \subseteq \sF$ be any sub-$\sigma$ -algebra. For any $\sF$ -measurable random variable $X$ , $$ \E [ Z \mid \sG ] \, \EQ [ X \mid \sG] = \E[ ZX \mid \sG]\,. $$ That is, $$ \left(\frac{d\PQ}{d\PP}\right)_{\mid \sG} \, \EQ [ X \mid \sG ] = \E \left[ \frac{d\PQ}{d\PP} \, X \mid \sG \right]\,. $$

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

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

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

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

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Observe that if $d \PQ / d \PP > 0$ almost surely, then $$ \EQ[X \mid \sG] = \E\left [ \frac{d\PQ}{d\PP} X \mid \sG \right] \Big/ \left( \frac{d\PQ}{d\PP} \right)_{\mid \sG}\,. $$

Theorem 3 If $X_t$ is a martingale with respect to $\PQ$ and some filtration $\{ \sF_t \}$ , then $X_t Z_t$ is a martingale with respect to $\PP$ and $\{ \sF_t\}$ , where $Z_t = \E [ Z \mid \sF_t]$ .

Proof. First observe that $X_t Z_t$ is indeed $\sF_t$ -measurable. Then, we can apply Theorem 2, with $X$ in the statement of that theorem replaced by $X_t$ , $Z$ replaced by $Z_t$ , $\sF$ replaced by $\sF_t$ , and $\sG$ replaced by $\sF_s$ ($s \leq t$ ), to obtain: $$ \E[ X_t Z_t \mid \sF_s ] = Z_s \, \EQ[ X_t \mid \sF_s] = Z_s X_s\,, $$ thus proving that $X_t Z_t$ is a martingale under $\PP$ and $\{ \sF_t \}$ . $\qedsymbol$

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

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