An example of how measure theory has obfuscated something once simple to me.
Consider tossing a coin and observing the face-up side, either $H$ or $T$. The sample space is clearly $\Omega = \{H,T\}$ and the sigma algebra is $\mathcal{F} = 2^\Omega$. We'll also have a probability measure such that $P(H) = p$.
Now consider tossing the coin twice and again observing the face-up side. Now the sample space for this experiment is $\Omega^\prime = \{HH, HT, TH, TT\}$ and the sigma algebra is $\mathcal{F}^\prime = 2^{\Omega^\prime}$. Again, we have some probability measure $P^\prime$ on this measurable space $(\Omega^\prime, \mathcal{F}^\prime)$.
The two tosses are clearly independent, and so it should be that, e.g., $P^\prime(HT) = p(1-p)$. But, it seems this doesn't have to be the case for two reasons:
- How would we actually obtain the equality $P^\prime(HT) = p(1-p)$? Usually, we might say $P(H \cap T) = P(H) \cdot P(T) = p(1-p)$ for independent events $H$ and $T$. But, the set $H \cap T = \emptyset$ in $\mathcal{F}$, so this we should get $P(H \cap T) = 0$. Furthermore, how would we get to measuring with $P$ from $P^\prime$?
- The set $\{H\} \notin \mathcal{F}^\prime$, so how does $P^\prime(H)$ even make sense?
I've really confused myself, and would appreciate some extra eyes on this.
Best Answer
In $\Omega'$, the event $\{H\}$ (or entity $H$ alone) is ambiguous, because we don't know if you're referring to the first toss or the second toss. To be clear you need to use notation like $H_1$ to denote the event that the first toss is heads. Then for sure $H_1 = \{HH, HT\}$ is in your sigma algebra. Similarly $T_2 := \{HT, TT\}$ is in your sigma algebra, and $H_1$ and $T_2$ are independent, since $$P'(H_1 \cap T_2)=P'(\{HT\})=p(1-p)=P'(H_1)P'(T_2).$$ IMO the measure theory hasn't obfuscated this; rather, it's making us aware that sometimes our notation can be sloppy and needs more precision.