Timeline for Does the definition of conditional probability involve events from more than one random experiment?
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| Sep 11 at 13:04 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Aug 12 at 3:52 | history | edited | User1865345 | CC BY-SA 4.0 |
edited tags; edited title
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| Aug 12 at 0:04 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jul 17, 2018 at 19:03 | answer | added | kjetil b halvorsen♦ | timeline score: 1 | |
| Sep 8, 2017 at 18:44 | comment | added | whuber♦ | Yes, I think you have it exactly right. The very act of writing a conditional probability implies $A$ and $B$ are events in the same space. Although you can always put events from different spaces $A\subset\Omega_A$ and $B\subset\Omega_B$ into a common space (simply form their product $\Omega_A\times\Omega_B$ and identify $A$ with $A\times\Omega_B$ and $B$ with $\Omega_A\times B$), this isn't worth much, because then $\Pr(A\mid B) = \Pr(A)$. | |
| Sep 8, 2017 at 18:31 | comment | added | Sanyo Mn | @whuber: yes I mean a probability space. After reading your comments I think as follows: The events should be in the same probability space, but whether to define a flip coin and a role of a dice in the same probability space or not depends on how you define it according to your needs. Do you agree? | |
| Sep 8, 2017 at 18:18 | comment | added | whuber♦ | @Henry I don't understand how that responds to or clarifies my comment, but it does seem like the basis of a good answer. | |
| Sep 8, 2017 at 18:15 | comment | added | Laksan Nathan | @Henry thanks for adding this. exactly the idea I've had in mind | |
| Sep 8, 2017 at 18:14 | comment | added | Henry | @whuber: yes but $A$ might be heads on the flip of a coin while $B$ might be a total of $7$ on the roll of two dice. The probability space would then have to contain both the coin flip and the dice roll | |
| Sep 8, 2017 at 18:11 | comment | added | whuber♦ | I wonder what you mean by "random experiment." Note that for "$A\cap B$" to make sense, $A$ and $B$ must be subsets of a common set. @Inathan Recall that events are defined to be measurable sets in a probability space. I believe the question might concern whether $A$ and $B$ must be events in the same probability space. | |
| Sep 8, 2017 at 18:10 | comment | added | Laksan Nathan | It is not about experiments here. It can be any two events A and B. So 'yes' to your second question. 'no' to the first. | |
| Sep 8, 2017 at 17:54 | history | asked | Sanyo Mn | CC BY-SA 3.0 |