maybe a bit of a basic question but I don't see how to formally solve it right now.

In order to analyze a stochastic process, I want to change my probability space to a more convenient one (we can assume all sample spaces are discrete and finite). So let's assume we are given the following equation. For all $A,F$, it holds $$\Pr[A = \omega \mid f(\omega)=F \colon \omega \leftarrow \Omega_1] = \Pr[A = \omega \mid \pi'(\omega) \colon \omega \leftarrow \Omega_2(F)]$$ In other words, whether I draw my sample $\omega$ from $\Omega_1$ or $\Omega_2(F)$ is irrelevant, as long as I condition on some function $f(\omega) = F$, and some predicate $\pi'(\omega)$, respectively.

Now I want to apply this theorem to my analysis. I am given a statement of the form $\Pr[\pi(\omega) \colon \omega \leftarrow \Omega_1]$, but for some reason I prefer computing in the probability space $\Omega_2$. Therefore, I'd like to apply my equation from above, for instance as follows: \begin{align}\Pr[\pi(\omega) \colon \omega \leftarrow \Omega_1] &= \sum_{F = -\infty}^{\infty} \Pr[\pi(\omega) \mid f(\omega)=F\colon \omega \leftarrow \Omega_1]\Pr[f(\omega) = F \colon \omega \leftarrow \Omega_1] \\ &= \sum_{F = -\infty}^{\infty} \Pr[\pi(\omega) \mid \pi'(\omega) \colon \omega \leftarrow \Omega_2(F)]\Pr[f(\omega) = F \colon \omega \leftarrow \Omega_1] \end{align}

I am now concerned how to prove this transformation. Intuitively it is clear, since the likelihood of drawing $\omega$ from one or the other distribution is the same, as long as I condition on the appropriate events. However, I cannot think of a formal proof for that intuition. Is anyone able to help?



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