Dartboard paradox and understanding independence

1. • By definition, events $A$ and $B$ are independent if $$P(A \cap B) = P(A)\:P(B).$$ Thus if an event $A$ happens almost never $\left(P(A)=0\right),$ then $A$ is independent of all events, including itself. So, hitting the exact centre of a dartboard (happens almost never) and hitting within its inner ring are independent events.

   • On the other hand, it is standard to characterise independence as follows:

     Two events are independent if the occurrence of one does not affect the probability of occurrence of the other.

     Since hitting the exact centre of a dartboard guarantees hitting within its inner ring (which is otherwise not guaranteed), the two events are dependent.

   Isn't this a contradiction?? If yes, then do we accept that the definition of independence is not meant to fully correspond to its verbal/intuitive characterisation?

2. Is there any semantic difference between the following two versions?

   $(i)$ Two events are independent if the occurrence of one does not affect the probability of occurrence of the other.

   $(ii)$ Two events are independent if the occurrence of one does not affect the occurrence of the other.

ADDENDUM

On further pondering, I have resolved both questions:

2.
Consider this experiment: flip two fair coins, letting $H_1$ be the event that the first coin lands on Heads, and $X$ be the event that the coins land on different sides. Then $$ P\left(H_1 \cap X\right)=\frac14=P(H_1)\:P(X);$$ i.e., $H_1$ and $X$ are independent events.

Knowledge that $H_1$ occurs reduces the known number of ways that $X$ can eventuate—from $2$ (HT, TH) to $1$ (HT)—but does not alter the probability $(\frac12)$ assigned to $X.$

1.
The following revision characterises pairwise independence more clearly and accurately:

Let $P(A)\neq0.$

Events $A$ and $B$ are independent iff. the probability of $B$ is unaffected by the knowledge that $A$ occurs.

Almost-never events are now excluded from being conditioned on. (What does it even mean to say that an almost-never event has occurred: in what sense have I hit the exact centre of a dartboard?) It motivates the definition of pairwise independence, which does allow both events to be impossible.

Solution 1:

Events $A$ and $B$ are indeed defined to be independent if $P[AB] = P[A]P[B]$. This definition holds for all cases, including cases when $P[A]=0$ and/or $P[B]=0$. The definition leads to your desired interpretation "knowing if $B$ occurred does not affect the likelihood of $A$" in the special case $P[B]>0$, since it means $P[A|B] = P[A]$ in that case. It does not allow your interpretation in the case when $P[B]=0$, since we cannot condition on a zero-probability event. So indeed, there are peculiarities when $P[B]=0$ since intuition fails in that case. Nevertheless, your intuition is still useful since it applies when $P[B]>0$.

It may also be helpful to observe that if $P[B]=0$ then $P[A|B^c] = P[A]$, which is intuitively what we would want when $A$ and $B^c$ are independent.

A situation about darts similar to the one you posed can be addressed by the definitions of conditioning with respect to a random variable. For a dartboard, we can imagine $(X,Y)$ uniformly distributed over a 2-d ball of radius 1. Then define $R = \sqrt{X^2+Y^2}$ as the distance from the center. Consider the events: $$ A = \{R \leq 1/10\}, B = \{R = 1/10\}$$ Now $P[B]=0$ so $A$ and $B$ are independent. Nevertheless this is similar to the situation you describe because $B \subseteq A$, so "intuitively" we would want $P[A|B]=1$. This holds if we use: $$ P[A|R=r] = \frac{f_{R|A}(r)P[A]}{f_R(r)} $$ which shall be applied when $f_R$ and $f_{R|A}$ are continuous at point $r$ and $f_R(r)\neq 0$, and we observe: \begin{align} f_R(r) &= \left\{\begin{array}{cc} 2r & \mbox{ if $r \in [0,1]$}\\ 0 & \mbox{else} \end{array}\right.\\ f_{R|A}(r) &= \left\{\begin{array}{cc} \frac{2r}{P[A]} & \mbox{ if $r \in [0,1/10]$} \\ 0 & \mbox{ else} \end{array}\right. \end{align} so $$ P[A|R=r] = \left\{\begin{array}{cc} 0 & \mbox{if $1/10 < r<1$} \\ 1 & \mbox{ if $0<r\leq 1/10$} \end{array}\right.$$

Our expression $P[A|R=r]$ cannot be viewed as an instance of the definition of conditional probability $P[A|B]$ for events $A$ and $B$, since the event $\{R=r\}$ has probability 0. Rather, it is an expression related to the definition of conditioning on a random variable $R$. Formally, you can say the random variable $Z$ defined by $$ Z = \left\{\begin{array}{cc} 0 & \mbox{if $R>1/10$} \\ 1 & \mbox{ if $0\leq R \leq 1/10$} \end{array}\right.$$ is a "version" of the conditional expectation $E[1_A|R]$. As such, it is almost surely the same as any other version. Thus, while this version takes the liberty to define $Z=1$ when $R=0$, and $Z=0$ when $R\geq 1$, this is fine because it only affects a set of measure 0.

Solution 2:

"On the other hand, it is standard to characterise pairwise independence as follows:

Two events are independent if the occurrence of one does not affect the probability of occurrence of the other."

Errr no, I disagree; the usual and informal notion of independence is slightly different:

Two events are said to be independent if knowledge that one of them has occurred does not cause you to assign a different value to the probability of the other.

That is, if $A$ and $B$ the two events in question and we know that $A$ had occurred, then we should use $P(B\mid A)$ are instead of $P(B)$ as the probability of $B$ in our calculations. Then, $A$ and $B$ are said to be independent events if $P(B\mid A)$ happens to be equal to $P(B)$. Thus, informally, $A$ and $B$ are said to be independent if $P(B\mid A) = P(B)$, or if one prefers, $P(A\mid B) = P(A)$. This is an informal definition because the conditional probabilities are not necessarily defined: the usual definition $\displaystyle P(B\mid A) = \frac{P(B \cap A)}{P(A)}$ assumes that $P(A) > 0$. It is much better to use the standard definition of (mutual) independence $P(A\cap B) = P(A)P(B)$ which shows the symmetry of the relationship in contrast to $P(B\mid A) = P(B)$ which is asymmetric and if we stick in $P(A\mid B) = P(A)$ also to show symmetry, then we commit the sin of prolixity. Worse yet, both $A$ and $B$ might be events of probability $0$ making both $P(B\mid A)$ and $P(A\mid B)$ be undefined. In contrast, the definition of independence as $P(A\cap B) = P(A)P(B)$ has no such issues and is vastly to be preferred. $P(B\mid A) = P(B)$ is fine for informal use and for intuition, but not as a definition. See also this answer of mine over on stats.SE.