Why is $dQ/dP$ said to be "only notation", when it is exact in finite cases?

134 Views Asked by At

My teacher told me that the notation for the radon nikodym derivative, $dQ/dP$ is just handy notation and that I should ignore it if I don't find it intuitive.

Well, I don't find it intuitive. What is "a small change in Q over a small change in P" even supposed to mean. But then I was playing around with the following finite example:

Let's say we have a probability space with $P$ and $Q$, where $Q$ and $P$ are equivalent. $\Omega$ is discrete and finite, and $P(\omega_i)$ and $Q(\omega_i)$ are both known for all possible $\omega_i$.

The Radon Nikodym derivative is some variable $X$, which also depends on $\omega_i$. Well, fix some $\omega$. Then, look at $Q(\omega)$, then we can write this as $\int_\omega X \, dP = \int X \textbf{1}_\omega \ dP = E^P X \textbf{1}_\omega = P(\omega)X(\omega)$, hence $X(\omega) = \frac{Q(\omega)}{P(\omega)}$.

So, it's not just flimsy notation, it's how we can calculate the derivative in finite cases. So why did my teachers (and certain other posters on this very site) insist on calling it "just notation", when it pretty must describes an exact relationship in the finite case?