In an example of Shreve, he begins with a random variable $X\sim N(0,1)$ under a measure $\mathbb{P}$ and defines a variable $Y = X+\theta$ and sets $Z = e^{-\theta X – \frac{1}{2}\theta^{2}}$ to define a new measure
$$\widetilde{\mathbb{P}}(A) = \int_{A}Z(\omega)d\mathbb{P}(\omega)$$
for all $A\in \mathcal{F}$ and shows that $Y$ is standard normal. In an exercise at the end of the chapter, he explains that $X$ is related to $Y$ and if you take $\hat{Z} = e^{\theta Y-\frac{\theta^{2}}{2}} = 1/Z$ and define
$$\hat{\mathbb{P}}(A) = \int_{A}\hat{Z}(\omega)d\widetilde{\mathbb{P}}(\omega)$$
for all $A\in\mathcal{F}$, he uses this to ask you to show that $\hat{\mathbb{P}} =\mathbb{P}$.
However, to show $\mathbb{P} = \hat{\mathbb{P}}$ I think it depends on the following equation (1) to make sense :
$$\hat{P}(A) = \int_{A}\hat{Z}d\widetilde{\mathbb{P}}(\omega) = \int_{A}\hat{Z}Zd\mathbb{P}(\omega) \quad (1)$$
but I cannot figure out whether or not I can do this. Earlier in the change of measure section, he references another book he authored that I do not own and uses it as the basis for writing
$$Z(\omega)\mathbb{P}(\omega) =\widetilde{\mathbb{P}}(\omega)\quad (2)$$
which I think is supposed to be used as a non-rigorous mnemonic to help manipulate equations like (1).
My question is then, is what I wrote for equation (1) correct? If it is correct, how might I justify it? Can (2) be informally viewed as a change of variables formula of some kind in a manner that is reminiscent of how people manipulate $dy/dx$ in differential equations or things like $dW_{t}dt = 0$ in the quadratic variation of Brownian Motion?
Best Answer
Generally, if $(X,\mathcal{E},\mu)$ is a measure space and $\nu$ is a another measure on $(X,\mathcal{E})$ with density $g$ with respect to $\mu$, i.e. $$ \nu(A)=\int_A g\,\mathrm d\mu,\quad A\in\mathcal{E}, $$ for some non-negative measurable function $g$, then $$ \int f\,\mathrm d\nu=\int fg\,\mathrm d\mu $$ for integrable $f$.
To show this, just note that this obviously holds when $f=\mathbf{1}_B$ for any Borel set $B$, and then the result follows using standard arguments in measure theory.
The notation $Z(\omega)\mathbb{P}(\omega)=\tilde{\mathbb{P}}(\omega)$ is shorthand notation for saying that $\tilde{\mathbb{P}}$ has density $Z$ with respect to $\mathbb{P}$. Another common notation for this is $$ \frac{\mathrm d \tilde{\mathbb{P}}}{\mathrm d\mathbb{P}}(\omega)=Z(\omega). $$