Leland Morrow

2022-06-24

The following is the definition of a random variable:

Let $(\mathrm{\Omega},\mathbb{F},P)$ be a probability space. A random variable is a real-valued function X on $\mathrm{\Omega}$ such that for all $x\in \mathbb{R},\{\omega :X(\omega )\in B\}\in \mathbb{F},\forall B\in B(\mathbb{R})$

I don't understand how this makes sense if our choice of $F$ can be arbitrary.

Let's say we are rolling a biased dice such that each side has a different probability, and the information we know about the system is $F=\{\mathrm{\varnothing},\{1\},\{2,...,6\},\mathrm{\Omega}\}$ -- then for this $F$, according to the definition, $X$ is not a random variable (because you could construct a (disjoint) Borel set which covers the probabilities of e.g. $\{1,3,5\}$, and $\{1,3,5\}$ is not in $F$). How does this make sense?

Let $(\mathrm{\Omega},\mathbb{F},P)$ be a probability space. A random variable is a real-valued function X on $\mathrm{\Omega}$ such that for all $x\in \mathbb{R},\{\omega :X(\omega )\in B\}\in \mathbb{F},\forall B\in B(\mathbb{R})$

I don't understand how this makes sense if our choice of $F$ can be arbitrary.

Let's say we are rolling a biased dice such that each side has a different probability, and the information we know about the system is $F=\{\mathrm{\varnothing},\{1\},\{2,...,6\},\mathrm{\Omega}\}$ -- then for this $F$, according to the definition, $X$ is not a random variable (because you could construct a (disjoint) Borel set which covers the probabilities of e.g. $\{1,3,5\}$, and $\{1,3,5\}$ is not in $F$). How does this make sense?

jarakapak7

Beginner2022-06-25Added 14 answers

Well sure, then the biased dice roll would not be a random variable in the sense you're used to thinking about it. However, it is quite meaningless to claim that it should be a traditional random variable in the first place, since $\mathbb{F}$, or often: $\mathcal{F}$ or $\mathrm{\Sigma}$, is (intuitively) the sigma-algebra of possible measurable events which we can consider.

With your choice of $\mathbb{F}$, only the events: nothing, 1, {2⋯6} or everything, can be considered in terms of probability. It is meaningless to say that the probability of 3 is different to that of 5, since neither 3 nor 5 are measurable events and as such do not have any sense of probability. You can only ask questions such as: what is the probability $X\ge 2$, or the probability $X=1$. You cannot talk of the probability $X\ge 3$ even, since $\{3,4,5,6\}$ is not a measurable event and as such has no probability defined for it.

So: yes, the sigma-algebra of events is arbitrary, but so is the probability space itself. You keep the theory general, and make specific definitions and examples when it suits.

With your choice of $\mathbb{F}$, only the events: nothing, 1, {2⋯6} or everything, can be considered in terms of probability. It is meaningless to say that the probability of 3 is different to that of 5, since neither 3 nor 5 are measurable events and as such do not have any sense of probability. You can only ask questions such as: what is the probability $X\ge 2$, or the probability $X=1$. You cannot talk of the probability $X\ge 3$ even, since $\{3,4,5,6\}$ is not a measurable event and as such has no probability defined for it.

So: yes, the sigma-algebra of events is arbitrary, but so is the probability space itself. You keep the theory general, and make specific definitions and examples when it suits.

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