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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?

Display Name
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  • If $A$ and $B$ are mutually exclusive then they are indepedent if and only if $P(A)=0$ or $P(B)=0$. Now you can answer all your questions using this. – Kavi Rama Murthy Sep 23 '20 at 07:29
  • You cannot prove it because it is wrong. – Kavi Rama Murthy Sep 23 '20 at 07:36
  • If two events A and B are mutually exclusive, then knowing whether or not A has happened will usually affect the probabilities of B happening. If A happened, then you know that B cannot happen. This means that A and B are not independent, unless B was never going to happen regardless of A. Independence means $P(A\cap B)=P(A)P(B)$, and mutually exclusive means $P(A\cap B)=0$. The only way for both to hold is if $P(A)=0$ or $P(B)=0$. – Jaap Scherphuis Sep 23 '20 at 08:29
  • At the risk of tackiness: https://math.stackexchange.com/questions/3836556/can-two-independent-events-be-disjoint – user2661923 Sep 23 '20 at 08:41

2 Answers2

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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.

user2661923
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  • @NotAZoomedImage What do you think dependent means? If you tell us that maybe we can explain it better for you. (BTW, it does not really have anything to do with sampling spaces, which seems to be where you are misunderstanding things.) – Jaap Scherphuis Sep 23 '20 at 11:55
  • Unless I am mistaken, your understanding is wrong. Consider for example, Bayes Theorem, which applies to any two events, regardless of whether they are independent. Make the simplifying assumption that $p(A) \neq 0 \neq p(B).$ Then, per Bayes Theorem, $p(B) \times p(A|B) = p(AB).$ The very essence of characterizing two events as independent is when and only when $p(A|B) = p(A).$ This is (in effect) considered merely a special case of Bayes Theorem. Under this special case, $p(B) \times p(A) = p(AB).$ I emphasize the simplifying assumption that $p(A) \neq 0 \neq p(B).$ – user2661923 Sep 23 '20 at 12:12
  • @JaapScherphuis Note further, that given the simplifying assumption that $p(A) \neq 0 \neq p(B)$, since $p(B) \times p(A|B) = p(AB) = p(A) \times p(B|A)$, it is easy to demonstrate that $[p(A|B) = p(A)] \iff [p(B|A) = p(B)]$. – user2661923 Sep 23 '20 at 12:17
  • @NotAZoomedImage See section 1 - Statement of Theorem in https://en.wikipedia.org/wiki/Bayes%27_theorem – user2661923 Sep 23 '20 at 12:24
  • @NotAZoomedImage See also Ethan Bolker's answer which makes it easy to get a bird's eye view of the subject. – user2661923 Sep 23 '20 at 12:28
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    For the first case, the proposed example might be as follows.

    In throwing a die, $A$ is an event in which we get 7 and $B$ is an event in which we get 0. Both events are mutex and independent. Correct?

     – Display Name Sep 24 '20 at 07:55
  • @NotAZoomedImage Very creative and absolutely correct. This is why, in questions of this nature, it is often necessary to make the simplifying assumption that $p(A) \neq 0 \neq p(B)$. I overlooked that detail. I am going to immediately edit my answer accordingly. – user2661923 Sep 24 '20 at 08:20
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Short answer in English to complement the long excellent one from @user2661923.

Two events are mutually exclusive when they cannot happen simultaneously.

Two events are independent when information about whether one occurred has no effect on your estimate of the probability of the other.

Then it's clear that mutually exclusive events cannot be independent. With these meanings you can analyze the other three cases too.

Ethan Bolker
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  • This is untrue. It's not that mutually exclusive events can't be independent, but that if mutually exclusive events are independent, then one of them "almost never" happens (i.e. happens with probability zero). – Rivers McForge Sep 24 '20 at 08:31
  • @RiversMcForge Yes, events with probability $0$ are independent of everything and mutually independent of everything. But making that correct technical observation might interfere with intuitive understanding. – Ethan Bolker Sep 24 '20 at 13:02