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I am reading Shreve's "Stochastic Calculus for Finance II". In it, he states (Theorem 1.6.1) that if $Z$ is an almost-surely strictly positive random variable on a probability space $(\Omega, \mathcal F, P)$ with $E(Z) = 1$, then the probability measure defined as $\widetilde P(A) := \int_A Z d P$ satisfies $$ E(Y) = \widetilde E(Y/Z) $$ for any random variable Y (here E and $\widetilde E$ are the expected values in the respective measures). I don't understand the meaning of this expression; $Y/Z$ is not a random variable since there may be $\omega \in \Omega$ upon which $Z$ vanishes.

Is $Y/Z$ a valid expression, or is it shorthand for something more rigorous?

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up vote 5 down vote accepted

EDIT: Replacing $\tilde \cdot$ (tilde) by $\widetilde \cdot$ (wide tilde); for some reason still does not appear like a (wide) tilde.

To complete Nate's answer, it is important to consider here $Y/Z$ and $\tilde E(Y/Z):=\int_\Omega {(Y/Z)d\tilde P} $ with respect to the probability measure $\tilde P$. As Nate noted, there is no problem with $Y/Z$ with respect to the probability measure $P$ (since $P(Z=0)=0$). Let's see that there is also no problem with respect to $\tilde P$, which amounts to showing that $\tilde P(Z=0)=0$. Indeed, let $A$ denote the event $\{Z=0\}:=\{ \omega \in \Omega : Z(\omega) = 0\}$. Then, $$ \tilde P(Z=0) = \tilde P(A) := \int_A {ZdP} = \int_A {Z(\omega )dP(\omega )}. $$ So, there is no problem here for two (independent) reasons. First, by definition, $Z(\omega) = 0$ for $\omega \in A$, hence $$ \int_A {Z(\omega )dP(\omega )} = \int_A {0dP(\omega )} = 0. $$ Second, by assumption, $P(Z=0)=0$, that is $P(A)=0$; hence, $\int_A {ZdP} = 0$ (recall from measure theory that $ \int_E {fd\mu } = 0$ if $\mu(E)=0$).

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Thanks for adding this. I wrote mine in haste and forgot to address this very important point. – Nate Eldredge May 17 '11 at 18:26
@Nate: Thanks... – Shai Covo May 17 '11 at 18:32
So Y/Z is not a random variable (a function), but it is integrable with respect to this measure? Is there a name for this kind of "function"? – James Davidoff May 22 '11 at 2:18
James, if $Y/Z = \pm \infty$ on some parts of $\Omega$, it may be called an "extended random variable". More precisely, see Definition 2.30 in…. – Shai Covo May 22 '11 at 5:17
However, when $Y/Z = \pm \infty$ on a set $A$ of probability zero, you can consider a modification of $Y/Z$ which is equal to $0$ (or any finite number) on $A$ (and to $Y/Z$ on $\Omega-A$), which is a random variable having the same probabilistic properties of the original $Y/Z$. – Shai Covo May 22 '11 at 5:26

You can think of it as the following shorthand: $$(Y/Z)(\omega) := \begin{cases} Y(\omega)/Z(\omega), & Z(\omega) \ne 0 \\ 43.7, & Z(\omega) = 0 \end{cases} $$

(Exercise: check that $Y/Z$ thusly defined is measurable.)

The point being: you can consider $Y/Z$ to be whatever you want on the event $\{Z=0\}$. Since the event has probability zero, it doesn't matter what you pick.

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