Does arbitrary reweighting of the mass function of a random variable produces a well-defined probability space?

Question

I am given a probability space with probability measure P. The space contains a few random variables, some of these are discrete, some are continuous. Say that X is discrete over {1, ... n}.

If given this setup, I decide to arbitrarily reweight the probability mass function of X, such that the new weights are different from the original ones, ie. I set

$$P(X = k) := p_k$$

where $\sum p_k = 1$ and $p_k \geq 0$.

Will this reassigning of the probabilities break the probability space? What happens to variables that are not independent of X? Are all Kolmogorov axioms preserved, and is the resulting new space still a well-defined probability space?

Answer 1

If measurable space $(\Omega,\mathcal A)$ has been chosen to serve as a model for some situation that asks for finding probabilities, then it is determined completely which functions $X:\Omega\to\mathbb R$ are measurable and which are not.

If next to that we make $(\Omega,\mathcal A)$ a probability space by adding a probability measure $P$ then this fixes all probabilities like $P(X=2)$ or $P(Y\leq 5)$ for these random variable.

Now if you "meet" e.g. something like $P(X=2)=0.3$ and you want to change $0.3$ into $0.4$ then only one thing can be done: replace $P$ by another probability measure $Q$ such that: $$Q(\{\omega\mid X(\omega)=2\})=0.4\neq0.3=P(\{\omega\in\Omega\mid X(\omega)=2\})$$

But this has "enormous" consequences. Probability space $(\Omega,\mathcal A,Q)$ is really a different one than $(\Omega,\mathcal A,P)$.

In order to diminish the consequences you can at most try to restrict the adaptions to events that are in the $\sigma$-algebra that is generated by $X$.

If e.g. $\Omega=\mathbb R^2$ and $P(A\times B)=P_1(A)P_2(B)$ for every pair of measurable subsets $A,B\subseteq\mathbb R$ where $P_1$ and $P_2$ are probability measures on $(\mathbb R,\mathcal B)$ then you can adapt by only changing $P_1$. This gives new probabilities for random variable $X:\mathbb R^2\to\mathbb R$ prescribed by $\langle x,y\rangle\mapsto x$. The probabilities for $Y$ prescribed by $\langle x,y\rangle\mapsto y$ will stay untouched.

My main point is that you cannot escape from choosing another probability measure.

Answer 2

I suspect the problem is that you have multiple choices of a new probability space meeting these conditions. We can try to pin one down

Clearly the probability measure in the probability space will change, let's say from $P_0$ to $P_1$, so giving you a new probability space even if you keep the same sample space $\Omega$ and $\sigma$-algebra $\mathcal{F}$

Let's show the original probability mass function for $X$ with $P_0(X=k)=p_{0,k}$ and the new one with $P_1(X=k)=p_{1,k}$

With event $A$ having original probability $P_0(A)= \sum_k P_0(A \cap X=k)= \sum_k p_{0,k}\,P_0(A \mid X=k)$ with the conditional probability $P_0(A \cap X=k) = \frac{P_0(A \cap X=k)}{p_{0,k}}$ when $p_{0,k} \gt 0$, you could then define the new probability measure as $P_1(A)= \sum_k p_{1,k}\,P_0(A \mid X=k)$ and this will give you the results you expect, not changing the probability of events independent of $X$ in the original probability space but typically changing the probabilities of other events

There is a potential difficulty if there is a $k$ with $p_{0,k}=0$ and $p_{1,k}\gt 0$ which would require more work