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#31
12-29-2017, 06:19 PM
 stefanos Member SOA Join Date: Mar 2015 Posts: 77

Quote:
 Originally Posted by Z3ta Yes just apply the following: $\text{\underline{Marcie's Theorem}} \\ \text{If }S=A_{1}\cup A_{2}\text{ and }A_{1}\cap A_{2}=\emptyset\text{ then }P(A_{1})=P(A_{2})=\frac{1}{2}$
This, in a sense, is true One can decompose A_1 in countably many pieces, move them around, and then reassemble them as A_2, regardless of their areas.

(A special case of the Banach-Tarski paradox)
#32
12-29-2017, 07:27 PM
 Academic Actuary Member Join Date: Sep 2009 Posts: 7,662

Quote:
 Originally Posted by Tsin I found other solutions and they are waaay more complicated. Seems like this problem is super hard. http://mathforum.org/library/drmath/view/56637.html
I did try the formula given and got .85975 so it seems quite accurate for a large n.
#33
12-29-2017, 07:41 PM
 Z3ta Member SOA Join Date: Sep 2015 Posts: 361

Quote:
 Originally Posted by stefanos This, in a sense, is true One can decompose A_1 in countably many pieces, move them around, and then reassemble them as A_2, regardless of their areas. (A special case of the Banach-Tarski paradox)
Edit: I should say it is not true because it requires going outside of the sigma algebra the probability measure is defined on... so it is not true in any sense in our framework.

Last edited by Z3ta; 12-29-2017 at 07:56 PM..
#34
12-29-2017, 08:25 PM
 stefanos Member SOA Join Date: Mar 2015 Posts: 77

Quote:
 Originally Posted by Z3ta Edit: I should say it is not true because it requires going outside of the sigma algebra the probability measure is defined on... so it is not true in any sense in our framework.
If it wasn't clear before: my intention was to alter the framework by using paradoxical (non-measurable) decompositions, which represent very legit math, but half-jokingly in this context.
#35
12-29-2017, 08:37 PM
 Z3ta Member SOA Join Date: Sep 2015 Posts: 361

Quote:
 Originally Posted by stefanos If it wasn't clear before: my intention was to alter the framework by using paradoxical (non-measurable) decompositions, which represent very legit math, but half-jokingly in this context.
You mean it’s weird to respond seriously to a post with a wink emoji that was a response to something I wrote in jest? You may have a point

 Tags binary, martingale, sequence, string