Jointly Exhaustive

http://www.stat.cmu.edu/~cshalizi/36-220/lecture-4.pdf

It says that two events are jointly exhaustive if one or the other of them
must happen.

I only have had high school probability so I have no idea what all the symbols really mean, don't bother explaining that part.

But I don't understand what it means if "one or the other of them must happen"?

So if I have some Record Sizes

Record: 30, 46, 70

Would they be jointly exhaustive? Clearly being 30 means I cannot be 46 and so that is mutually exhaustive, but I could have 69 and that isn't included in the set (not math set) and so it wouldn't be jointly exhaustive
 Mentor Here's a simple example: a light switch that is either ON or OFF. The two events (light is on, light is off) are mutually exclusive, and jointly exhaustive - the switch must be in one of the two positions.

 Quote by Mark44 jointly exhaustive - the switch must be in one of the two positions.
What was wrong about my original question with the records?

I think (and I am) having the wrong definitions of jointly exhaustive here. I am thinking that okay, if I can come up with another item in that "concept" or genus that's not included, then it is NOT jointly exhaustive.

Jointly Exhaustive

 Quote by flyingpig http://www.stat.cmu.edu/~cshalizi/36-220/lecture-4.pdf It says that two events are jointly exhaustive if one or the other of them must happen. I only have had high school probability so I have no idea what all the symbols really mean, don't bother explaining that part. But I don't understand what it means if "one or the other of them must happen"? So if I have some Record Sizes Record: 30, 46, 70 Would they be jointly exhaustive? Clearly being 30 means I cannot be 46 and so that is mutually exhaustive, but I could have 69 and that isn't included in the set (not math set) and so it wouldn't be jointly exhaustive
In mathematical lingo, the statement "one or the other must happen" means that P(A OR B) = 1 if you have two events. If you have more events you just use something like P(A OR B OR C) = 1 and so on. (This uses the definition "must happen" means "must happen with complete certainty").
 Recognitions: Gold Member Homework Help An event A is just a subset of the sample space S. All he is saying is that if the union of a collection of sets = S they are jointly exhaustive. The sets might overlap, but if so they don't they form a partition of the sample space. Example: S = {1,2,3,4,5,6,7,8,9,10} represents the sample space for a ten position spinner. A = {1,2,3,4} B = {3,4,5,6} C = {5,6,7,8} D={7,8,9,10}. These sets are jointly exhaust S because if you spin the spinner, one (at least) of the events must happen. Since they overlap two of them might happen at the same outcome. Now consider M = {1,3,5,7,9} and N = {2,4,6,8,10}. These also exhaust S. Since they don't overlap they form a partition of S (into the evens and odds). The sets {1,2,3} {6,7,8} {9,10} aren't jointly exhaustive since the spinner might hit 4 which isn't one of these events.
 The "S" in that pdf threw me off...that's about it. I remember the phi thing means empty EDIT: @Kurt, oh

 Example: S = {1,2,3,4,5,6,7,8,9,10} represents the sample space for a ten position spinner. A = {1,2,3,4} B = {3,4,5,6} C = {5,6,7,8} D={7,8,9,10}.
So if it lands on 6, then B and C happens, but not A and D. It doesn't matter that A and D happen, but at least one, B and C happened. It is jointly exhaustive?

 Now consider M = {1,3,5,7,9} and N = {2,4,6,8,10}. These also exhaust S. Since they don't overlap they form a partition of S (into the evens and odds).
M and N are then (follow my logic here now) mutually exclusive, but they are not jointly exhaustive? Was that what you meant when you said "might" in? You said they exhaust S, but they have no intersection.

 Quote by LCKurtz An event A is just a subset of the sample space S. All he is saying is that if the union of a collection of sets = S they are jointly exhaustive. The sets might overlap, but if they don't they form a partition of the sample space.
 Quote by LCKurtz The sets {1,2,3} {6,7,8} {9,10} aren't jointly exhaustive since the spinner might hit 4 which isn't one of these events.

But all of those sets are mutually exclusive.

Would it be correct to draw the following conclusion?

If n events are mutually exclusive, then they could be jointly exhaustive. If n events are NOT mutually exclusive, then they can never be jointly exhaustive as in {1,2,3} {6,7,8} {9,10} in S: {1,2,3,4,5,6,7,8,9,10}

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 Quote by LCKurtz An event A is just a subset of the sample space S. All he is saying is that if the union of a collection of sets = S they are jointly exhaustive. The sets might overlap, but if so they don't they form a partition of the sample space. Example: S = {1,2,3,4,5,6,7,8,9,10} represents the sample space for a ten position spinner. A = {1,2,3,4} B = {3,4,5,6} C = {5,6,7,8} D={7,8,9,10}. These sets are jointly exhaust S because if you spin the spinner, one (at least) of the events must happen. Since they overlap two of them might happen at the same outcome. Now consider M = {1,3,5,7,9} and N = {2,4,6,8,10}. These also exhaust S. Since they don't overlap they form a partition of S (into the evens and odds). The sets {1,2,3} {6,7,8} {9,10} aren't jointly exhaustive since the spinner might hit 4 which isn't one of these events.
 Quote by flyingpig So if it lands on 6, then B and C happens, but not A and D. It doesn't matter that A and D happen, but at least one, B and C happened. It is jointly exhaustive?
Yes.

 M and N are then (follow my logic here now) mutually exclusive, but they are not jointly exhaustive?

 But all of those sets are mutually exclusive. Would it be correct to draw the following conclusion? If n events are mutually exclusive, then they could be jointly exhaustive.
Mutually exclusive and jointly exhaustive are different and independent concepts. Mutually exclusive says nothing about whether or not they are jointly exhaustive.
 If n events are NOT mutually exclusive, then they can never be jointly exhaustive as in {1,2,3} {6,7,8} {9,10} in S: {1,2,3,4,5,6,7,8,9,10}

 Quote by flyingpig if n events are not mutually exclusive, then they can never be jointly exhaustive as in {1,2,3} {6,7,8} {9,10} in s: {1,2,3,4,5,6,7,8,9,10}
s = {1, 2, ..., 10}

a = {1, 3, 4, ..., 10}
b = {1, 2, 4, ..., 10}

 Now consider M = {1,3,5,7,9} and N = {2,4,6,8,10}. These also exhaust S. Since they don't overlap they form a partition of S (into the evens and odds).
 Quote by LCKurtz An event A is just a subset of the sample space S. All he is saying is that if the union of a collection of sets = S they are jointly exhaustive. The sets might overlap, but if they don't they form a partition of the sample space.
Oh okay, because when you spin the spinner, it can be either odd or even in S. So M or N can happen and as in the first example since one of them could happen, it is jointly exhaustive.

 "form a partition of the sample space"
I took this for English...I feel it is not. What exactly does this mean?

 Quote by LCKurtz No. Read the examples again and see where that is answered.
Because none of the sets would "cover" the 4.

But isn't this consistent with my Record example?