Are “mutually exclusiveness” and “independency” orthogonal

Question

Background

It will be easier if we distinguish "mutually exclusiveness" from "independency" by considering the sample space in mind.

• Two events that are compared for mutually exclusiveness must be from a single sample space. For example,

  • Tossing a coin twice. $A=\{HH\}$ is an event in which the head shows up twice and $B=\{TT\}$ is an event in which the tail shows up twice. Their share the same sample space $S=\{HH,HT,TH,TT\}$. As $A\cap B=\{\}$, they are mutually exclusive.

• Two events that are compared for independency must be from two sample spaces. For example,

  • Tossing a coin twice. $A=\{H\}$ is an event in which the head shows up in the first throw and $B=\{T\}$ is an event in which the tail shows up in the second throw. The sample space for the first trial is $S_1=\{H,T\}$ and the sample space for the second trial is $S_2=\{H,T\}$ As $S_1=S_2$, they are independent.

From this perspective, it seems to me that "mutually exclusiveness" and "independency" are orthogonal.

Questions

If they are not orthogonal, each the following cases should have at least one example.

Could you give me one example (or more) for each of the following?

• Two events that are mutually exclusive and independent.
• Two events that are mutually exclusive and dependent.
• Two events that are "not mutually exclusive" but independent.
• Two events that are "not mutually exclusive" but dependent.

Attempt

I am attempting to follow the comment by kavi rama murthy

If $A$ and $B$ are mutually exclusive then they are independent if and only if $P(A)=0$ or $P(B)=0$.

Consider tossing a coin twice.

• I define two events that are mutually exclusive and one of them has zero probability.

  • $A=\{\star T\}$ is an event in which the first throw is star and the second throw is tail. As $\star$ is not possible then $p(A)=0$.

  • $B=\{HT\}$ is an event in which the first throw is head and the second throw is tail. It is clear that $p(B)\not=0$.

  • Their sample space is $S=\{HH,TH,HT,TT\}$. As $A\cap B=\{\}$, they are mutually exclusive.

• Now I have to interpret the created events above by finding two sample spaces to check whether or not they are independent. Unfortunately, these two events cannot be interpreted by considering two sample space in mind unless I redefine the events as follows.

  • $A_1=\{\star\}$ and $A_2=\{T\}$. First throw is $\star$ and 2nd throw is tail.
  • $B_1=\{H\}$ and $B_2=\{T\}$. First throw is head and 2nd throw is tail.
  • $S_1=\{H,T\}$ and $S_2=\{H,T\}$. As $S_1=S_2$, $A_1$ is independent of $A_2$, $B_1$ is independent of $B_2$. However we cannot make any dependency check for $A_1$ and $B_1$ or for $A_2$ and $B_2$.

From this attempt, hopefully you understand my confusion. The two events $A$ and $B$ that I defined to check for "mutually exclusiveness" cannot be directly reused to check "independency". Or is it because of wrongly chosen example?

Answer 1

Per OP's request.

First of all, as I understand the term orthogonal, as it is being used here, it is being used as a synonym for the phrase mutually exclusive.

Suppose that you have two events: $A$ and $B$.

To facilitate the analysis, make the simplifying assumption that
$p(A) \neq 0 \neq p(B).$

Then the two events either are or are not mutually exclusive.
Further, the two events either are or are not independent.
That gives 4 possibilities.

Case 1 $A$ and $B$ are both mutually exclusive and independent.

As has been discussed, this case is impossible, so no example can be provided.

Case 2 $A$ and $B$ are mutually exclusive but are not independent.

Example:
$A$ is the event of rolling a 1 on a die.
$B$ is the event of rolling a 6 on a die.

It is impossible for both events to simultaneously occur. Therefore, they are mutually exclusive.

$p(A) = (1/6)$ and $p(A|B) = 0$.
Since $p(A) \neq p(A|B)$, the two events are not independent.
Therefore, the two events are dependent.

Case 3 $A$ and $B$ are not mutually exclusive but are independent.

Example:
$A$ is the event of rolling a [1 or 2] on a die.
$B$ is the event of rolling [an even number] on a die.

It is definitely possible for both events to simultaneously occur. This is illustrated by rolling a [2]. Therefore, the two events are not mutually exclusive.

$p(A) = (1/3).$
Assume event $B$ occurred.
Then, either a [2, 4, or 6] was rolled. Under this scenario, chance of $A$ occurring given that $B$ occurred is still (1/3).
Therefore, $p(A|B) = (1/3) = p(A).$
Therefore, these two events are independent.

Note that the exact same conclusion will inevitably be drawn if the primary focus here is event $B$, rather than event $A$.

$p(B) = (1/2).$
Assume event $A$ occurred.
Then, either a [1, or 2] was rolled. Under this scenario, chance of $B$ occurring given that $A$ occurred is still (1/2).
Therefore, $p(B|A) = (1/2) = p(B).$
Therefore, these two events are independent.

Case 4 $A$ and $B$ are not mutually exclusive and are not independent.

Example:
$A$ is the event of rolling a [1 or 2] on a die.
$B$ is the event of rolling [1, 2, or 3] on a die.

It is definitely possible for both events to simultaneously occur. This is illustrated by rolling a [2]. Therefore, the two events are not mutually exclusive.

$p(A) = (1/3).$
Assume event $B$ occurred.
Then, either a [1, 2, 3] was rolled. Under this scenario, chance of $A$ occurring given that $B$ occurred is (2/3).
Therefore, $p(A|B) = (2/3)$ and $p(A) = (1/3).$
Therefore, these two events are not independent.
Therefore, these two events are dependent.

Note that the exact same conclusion will inevitably be drawn if the primary focus here is event $B$, rather than event $A$.

$p(B) = (1/2).$
Assume event $A$ occurred.
Then, either a [1, or 2] was rolled. Under this scenario, chance of $B$ occurring given that $A$ occurred is (1).

That is, if event $A$ occurred, then it is absolutely certain that event $B$ also occurred.
Therefore, $p(B|A) = (1)$ and $p(B) = (1/2).$
Therefore, these two events are not independent.
Therefore, these two events are dependent.