Total Posts:12|Showing Posts:1-12
Jump to topic:

Challenge for logic nuts

Toviyah
Posts: 88
Add as Friend
Challenge to a Debate
Send a Message
7/23/2014 7:04:59 PM
Posted: 2 years ago
Decode this for me:

Proof of Theorem: Say that an event E is continuous provided that for any 0X04;xX04;P(E), there is an event FX38;E with P(F)=x. By Axiom 5, without loss of generality {XW12;A} and {YW12;A} are continuous for any (Borel measurable) A. (Proof: If necessary, enrich the probability space that X is defined on to introduce a random variable U uniformly distributed on [0,1] and independent of X. The enrichment will not change any gamble orderings by Axiom 5. Then if 0X04;xX04;P(XW12;A), just choose aW12;[0,1] such that aP(XW12;A)=x and let F={XW12;A&UX04;a}. Ditto for Y.)

Now, given an event A and a random variable X, let AX be the random variable equal to X on A and equal to zero outside of A. Let A={X=W22;W34;} and B={Y=W22;W34;}. Define the random variables X1 and Y1 on [0,1] with uniform distribution by X1(x)=W22;W34; if xX04;P(A) and X1(x)=0 otherwise, and Y1(x)=W22;W34; if xX04;P(B) and Y1(x)=0 otherwise. Since P(A)X05;P(B) by (7), it follows that X1(x)X04;Y1(x) everywhere and so X1X04;Y1 by Axiom 3. But AX and BY are probabilistically equivalent to X1 and Y1 respectively, so by Axiom 5 we have AXX04;BY. If we can show that AcX<BcY then the conclusion of our Theorem will follow from the second part of Axiom 4.

Let X2=AcX and Y2=BcY. Then P(X2=+W34;)<P(Y2=+W34;), X2* and Y2* have finite expected values and X2 and Y2 never have the value W22;W34;. We must show that X2X04;Y2. Let C={X2=+W34;}. By subdivisibility, let D be a subset of {Y2=+W34;} with P(D)=P(C). Then CX2 and DY2 are probabilistically equivalent, so CX2X04;DY2 by Axiom 5. Let X3=CcX2 and Y3=DcY3. Observe that X3 is everywhere finite. Furthermore P(Y3=+W34;)=P(Y2=+W34;)W22;P(X2=+W34;)>0.

Choose a finite N sufficiently large that NP(Y3=+W34;)>E(X3)W22;E(Y3*) (the finiteness of the right hand side follows from our integrability assumptions). Let Y4 be a random variable that agrees with Y3 everywhere where Y3 is finite, but equals N where Y3 is infinite. Then E(Y4)=NP(Y3=+W34;)+E(Y3*)>E(X3). Thus, Y4>X3 by Axiom 2. But Y3 is greater than or equal to Y4 everywhere, so Y3X05;Y4. By Axiom 1 it follows that Y3>X3. but DY2X05;CX2 and X2=CX2+X3 and Y2=DY2+Y3, so by Axiom 4 we have Y2>X2, which was what we wanted to prove.

Thanks
Envisage
Posts: 3,646
Add as Friend
Challenge to a Debate
Send a Message
7/23/2014 7:27:56 PM
Posted: 2 years ago
At 7/23/2014 7:04:59 PM, Toviyah wrote:
Decode this for me:

Proof of Theorem: Say that an event E is continuous provided that for any 0X04;xX04;P(E), there is an event FX38;E with P(F)=x. By Axiom 5, without loss of generality {XW12;A} and {YW12;A} are continuous for any (Borel measurable) A. (Proof: If necessary, enrich the probability space that X is defined on to introduce a random variable U uniformly distributed on [0,1] and independent of X. The enrichment will not change any gamble orderings by Axiom 5. Then if 0X04;xX04;P(XW12;A), just choose aW12;[0,1] such that aP(XW12;A)=x and let F={XW12;A&UX04;a}. Ditto for Y.)

Now, given an event A and a random variable X, let AX be the random variable equal to X on A and equal to zero outside of A. Let A={X=W22;W34;} and B={Y=W22;W34;}. Define the random variables X1 and Y1 on [0,1] with uniform distribution by X1(x)=W22;W34; if xX04;P(A) and X1(x)=0 otherwise, and Y1(x)=W22;W34; if xX04;P(B) and Y1(x)=0 otherwise. Since P(A)X05;P(B) by (7), it follows that X1(x)X04;Y1(x) everywhere and so X1X04;Y1 by Axiom 3. But AX and BY are probabilistically equivalent to X1 and Y1 respectively, so by Axiom 5 we have AXX04;BY. If we can show that AcX<BcY then the conclusion of our Theorem will follow from the second part of Axiom 4.

Let X2=AcX and Y2=BcY. Then P(X2=+W34;)<P(Y2=+W34;), X2* and Y2* have finite expected values and X2 and Y2 never have the value W22;W34;. We must show that X2X04;Y2. Let C={X2=+W34;}. By subdivisibility, let D be a subset of {Y2=+W34;} with P(D)=P(C). Then CX2 and DY2 are probabilistically equivalent, so CX2X04;DY2 by Axiom 5. Let X3=CcX2 and Y3=DcY3. Observe that X3 is everywhere finite. Furthermore P(Y3=+W34;)=P(Y2=+W34;)W22;P(X2=+W34;)>0.

Choose a finite N sufficiently large that NP(Y3=+W34;)>E(X3)W22;E(Y3*) (the finiteness of the right hand side follows from our integrability assumptions). Let Y4 be a random variable that agrees with Y3 everywhere where Y3 is finite, but equals N where Y3 is infinite. Then E(Y4)=NP(Y3=+W34;)+E(Y3*)>E(X3). Thus, Y4>X3 by Axiom 2. But Y3 is greater than or equal to Y4 everywhere, so Y3X05;Y4. By Axiom 1 it follows that Y3>X3. but DY2X05;CX2 and X2=CX2+X3 and Y2=DY2+Y3, so by Axiom 4 we have Y2>X2, which was what we wanted to prove.

Thanks

F That.
n7
Posts: 1,360
Add as Friend
Challenge to a Debate
Send a Message
7/23/2014 10:13:32 PM
Posted: 2 years ago
At 7/23/2014 7:27:56 PM, Envisage wrote:
At 7/23/2014 7:04:59 PM, Toviyah wrote:
Decode this for me:

Proof of Theorem: Say that an event E is continuous provided that for any 0X04;xX04;P(E), there is an event FX38;E with P(F)=x. By Axiom 5, without loss of generality {XW12;A} and {YW12;A} are continuous for any (Borel measurable) A. (Proof: If necessary, enrich the probability space that X is defined on to introduce a random variable U uniformly distributed on [0,1] and independent of X. The enrichment will not change any gamble orderings by Axiom 5. Then if 0X04;xX04;P(XW12;A), just choose aW12;[0,1] such that aP(XW12;A)=x and let F={XW12;A&UX04;a}. Ditto for Y.)

Now, given an event A and a random variable X, let AX be the random variable equal to X on A and equal to zero outside of A. Let A={X=W22;W34;} and B={Y=W22;W34;}. Define the random variables X1 and Y1 on [0,1] with uniform distribution by X1(x)=W22;W34; if xX04;P(A) and X1(x)=0 otherwise, and Y1(x)=W22;W34; if xX04;P(B) and Y1(x)=0 otherwise. Since P(A)X05;P(B) by (7), it follows that X1(x)X04;Y1(x) everywhere and so X1X04;Y1 by Axiom 3. But AX and BY are probabilistically equivalent to X1 and Y1 respectively, so by Axiom 5 we have AXX04;BY. If we can show that AcX<BcY then the conclusion of our Theorem will follow from the second part of Axiom 4.

Let X2=AcX and Y2=BcY. Then P(X2=+W34;)<P(Y2=+W34;), X2* and Y2* have finite expected values and X2 and Y2 never have the value W22;W34;. We must show that X2X04;Y2. Let C={X2=+W34;}. By subdivisibility, let D be a subset of {Y2=+W34;} with P(D)=P(C). Then CX2 and DY2 are probabilistically equivalent, so CX2X04;DY2 by Axiom 5. Let X3=CcX2 and Y3=DcY3. Observe that X3 is everywhere finite. Furthermore P(Y3=+W34;)=P(Y2=+W34;)W22;P(X2=+W34;)>0.

Choose a finite N sufficiently large that NP(Y3=+W34;)>E(X3)W22;E(Y3*) (the finiteness of the right hand side follows from our integrability assumptions). Let Y4 be a random variable that agrees with Y3 everywhere where Y3 is finite, but equals N where Y3 is infinite. Then E(Y4)=NP(Y3=+W34;)+E(Y3*)>E(X3). Thus, Y4>X3 by Axiom 2. But Y3 is greater than or equal to Y4 everywhere, so Y3X05;Y4. By Axiom 1 it follows that Y3>X3. but DY2X05;CX2 and X2=CX2+X3 and Y2=DY2+Y3, so by Axiom 4 we have Y2>X2, which was what we wanted to prove.

Thanks

F That.
404 coherent debate topic not found. Please restart the debate with clear resolution.


Uphold Marxist-Leninist-Maoist-Sargonist-n7ism.
johnlubba
Posts: 2,892
Add as Friend
Challenge to a Debate
Send a Message
7/31/2014 1:30:49 PM
Posted: 2 years ago
At 7/23/2014 7:27:56 PM, Envisage wrote:
At 7/23/2014 7:04:59 PM, Toviyah wrote:
Decode this for me:

Proof of Theorem: Say that an event E is continuous provided that for any 0X04;xX04;P(E), there is an event FX38;E with P(F)=x. By Axiom 5, without loss of generality {XW12;A} and {YW12;A} are continuous for any (Borel measurable) A. (Proof: If necessary, enrich the probability space that X is defined on to introduce a random variable U uniformly distributed on [0,1] and independent of X. The enrichment will not change any gamble orderings by Axiom 5. Then if 0X04;xX04;P(XW12;A), just choose aW12;[0,1] such that aP(XW12;A)=x and let F={XW12;A&UX04;a}. Ditto for Y.)

Now, given an event A and a random variable X, let AX be the random variable equal to X on A and equal to zero outside of A. Let A={X=W22;W34;} and B={Y=W22;W34;}. Define the random variables X1 and Y1 on [0,1] with uniform distribution by X1(x)=W22;W34; if xX04;P(A) and X1(x)=0 otherwise, and Y1(x)=W22;W34; if xX04;P(B) and Y1(x)=0 otherwise. Since P(A)X05;P(B) by (7), it follows that X1(x)X04;Y1(x) everywhere and so X1X04;Y1 by Axiom 3. But AX and BY are probabilistically equivalent to X1 and Y1 respectively, so by Axiom 5 we have AXX04;BY. If we can show that AcX<BcY then the conclusion of our Theorem will follow from the second part of Axiom 4.

Let X2=AcX and Y2=BcY. Then P(X2=+W34;)<P(Y2=+W34;), X2* and Y2* have finite expected values and X2 and Y2 never have the value W22;W34;. We must show that X2X04;Y2. Let C={X2=+W34;}. By subdivisibility, let D be a subset of {Y2=+W34;} with P(D)=P(C). Then CX2 and DY2 are probabilistically equivalent, so CX2X04;DY2 by Axiom 5. Let X3=CcX2 and Y3=DcY3. Observe that X3 is everywhere finite. Furthermore P(Y3=+W34;)=P(Y2=+W34;)W22;P(X2=+W34;)>0.

Choose a finite N sufficiently large that NP(Y3=+W34;)>E(X3)W22;E(Y3*) (the finiteness of the right hand side follows from our integrability assumptions). Let Y4 be a random variable that agrees with Y3 everywhere where Y3 is finite, but equals N where Y3 is infinite. Then E(Y4)=NP(Y3=+W34;)+E(Y3*)>E(X3). Thus, Y4>X3 by Axiom 2. But Y3 is greater than or equal to Y4 everywhere, so Y3X05;Y4. By Axiom 1 it follows that Y3>X3. but DY2X05;CX2 and X2=CX2+X3 and Y2=DY2+Y3, so by Axiom 4 we have Y2>X2, which was what we wanted to prove.

Thanks

F That.

Did I just read right, Well now you know how I feel when you get all technical'

LMAO
sadolite
Posts: 8,839
Add as Friend
Challenge to a Debate
Send a Message
7/31/2014 7:01:30 PM
Posted: 2 years ago
Logic dictates that this will never serve a useful purpose to solve it.
It's not your views that divide us, it's what you think my views should be that divides us.

If you think I will give up my rights and forsake social etiquette to make you "FEEL" better you are sadly mistaken

If liberal democrats would just stop shooting people gun violence would drop by 90%
Rusty
Posts: 2,109
Add as Friend
Challenge to a Debate
Send a Message
7/31/2014 8:57:20 PM
Posted: 2 years ago
At 7/23/2014 7:04:59 PM, Toviyah wrote:
Decode this for me:

Proof of Theorem: Say that an event E is continuous provided that for any 0X04;xX04;P(E), there is an event FX38;E with P(F)=x. By Axiom 5, without loss of generality {XW12;A} and {YW12;A} are continuous for any (Borel measurable) A. (Proof: If necessary, enrich the probability space that X is defined on to introduce a random variable U uniformly distributed on [0,1] and independent of X. The enrichment will not change any gamble orderings by Axiom 5. Then if 0X04;xX04;P(XW12;A), just choose aW12;[0,1] such that aP(XW12;A)=x and let F={XW12;A&UX04;a}. Ditto for Y.)

Now, given an event A and a random variable X, let AX be the random variable equal to X on A and equal to zero outside of A. Let A={X=W22;W34;} and B={Y=W22;W34;}. Define the random variables X1 and Y1 on [0,1] with uniform distribution by X1(x)=W22;W34; if xX04;P(A) and X1(x)=0 otherwise, and Y1(x)=W22;W34; if xX04;P(B) and Y1(x)=0 otherwise. Since P(A)X05;P(B) by (7), it follows that X1(x)X04;Y1(x) everywhere and so X1X04;Y1 by Axiom 3. But AX and BY are probabilistically equivalent to X1 and Y1 respectively, so by Axiom 5 we have AXX04;BY. If we can show that AcX<BcY then the conclusion of our Theorem will follow from the second part of Axiom 4.

Let X2=AcX and Y2=BcY. Then P(X2=+W34;)<P(Y2=+W34;), X2* and Y2* have finite expected values and X2 and Y2 never have the value W22;W34;. We must show that X2X04;Y2. Let C={X2=+W34;}. By subdivisibility, let D be a subset of {Y2=+W34;} with P(D)=P(C). Then CX2 and DY2 are probabilistically equivalent, so CX2X04;DY2 by Axiom 5. Let X3=CcX2 and Y3=DcY3. Observe that X3 is everywhere finite. Furthermore P(Y3=+W34;)=P(Y2=+W34;)W22;P(X2=+W34;)>0.

Choose a finite N sufficiently large that NP(Y3=+W34;)>E(X3)W22;E(Y3*) (the finiteness of the right hand side follows from our integrability assumptions). Let Y4 be a random variable that agrees with Y3 everywhere where Y3 is finite, but equals N where Y3 is infinite. Then E(Y4)=NP(Y3=+W34;)+E(Y3*)>E(X3). Thus, Y4>X3 by Axiom 2. But Y3 is greater than or equal to Y4 everywhere, so Y3X05;Y4. By Axiom 1 it follows that Y3>X3. but DY2X05;CX2 and X2=CX2+X3 and Y2=DY2+Y3, so by Axiom 4 we have Y2>X2, which was what we wanted to prove.

Thanks

Yes.
Sidewalker
Posts: 3,713
Add as Friend
Challenge to a Debate
Send a Message
8/1/2014 9:46:48 AM
Posted: 2 years ago
At 7/23/2014 7:04:59 PM, Toviyah wrote:
Decode this for me:

Proof of Theorem: Say that an event E is continuous provided that for any 0X04;xX04;P(E), there is an event FX38;E with P(F)=x. By Axiom 5, without loss of generality {XW12;A} and {YW12;A} are continuous for any (Borel measurable) A. (Proof: If necessary, enrich the probability space that X is defined on to introduce a random variable U uniformly distributed on [0,1] and independent of X. The enrichment will not change any gamble orderings by Axiom 5. Then if 0X04;xX04;P(XW12;A), just choose aW12;[0,1] such that aP(XW12;A)=x and let F={XW12;A&UX04;a}. Ditto for Y.)

Now, given an event A and a random variable X, let AX be the random variable equal to X on A and equal to zero outside of A. Let A={X=W22;W34;} and B={Y=W22;W34;}. Define the random variables X1 and Y1 on [0,1] with uniform distribution by X1(x)=W22;W34; if xX04;P(A) and X1(x)=0 otherwise, and Y1(x)=W22;W34; if xX04;P(B) and Y1(x)=0 otherwise. Since P(A)X05;P(B) by (7), it follows that X1(x)X04;Y1(x) everywhere and so X1X04;Y1 by Axiom 3. But AX and BY are probabilistically equivalent to X1 and Y1 respectively, so by Axiom 5 we have AXX04;BY. If we can show that AcX<BcY then the conclusion of our Theorem will follow from the second part of Axiom 4.

Let X2=AcX and Y2=BcY. Then P(X2=+W34;)<P(Y2=+W34;), X2* and Y2* have finite expected values and X2 and Y2 never have the value W22;W34;. We must show that X2X04;Y2. Let C={X2=+W34;}. By subdivisibility, let D be a subset of {Y2=+W34;} with P(D)=P(C). Then CX2 and DY2 are probabilistically equivalent, so CX2X04;DY2 by Axiom 5. Let X3=CcX2 and Y3=DcY3. Observe that X3 is everywhere finite. Furthermore P(Y3=+W34;)=P(Y2=+W34;)W22;P(X2=+W34;)>0.

Choose a finite N sufficiently large that NP(Y3=+W34;)>E(X3)W22;E(Y3*) (the finiteness of the right hand side follows from our integrability assumptions). Let Y4 be a random variable that agrees with Y3 everywhere where Y3 is finite, but equals N where Y3 is infinite. Then E(Y4)=NP(Y3=+W34;)+E(Y3*)>E(X3). Thus, Y4>X3 by Axiom 2. But Y3 is greater than or equal to Y4 everywhere, so Y3X05;Y4. By Axiom 1 it follows that Y3>X3. but DY2X05;CX2 and X2=CX2+X3 and Y2=DY2+Y3, so by Axiom 4 we have Y2>X2, which was what we wanted to prove.

Thanks

42
"It is one of the commonest of mistakes to consider that the limit of our power of perception is also the limit of all there is to perceive." " C. W. Leadbeater
Enji
Posts: 1,022
Add as Friend
Challenge to a Debate
Send a Message
8/1/2014 10:10:52 AM
Posted: 2 years ago
At 7/23/2014 7:04:59 PM, Toviyah wrote:
Decode this for me:

Proof of Theorem: Say that an event E is continuous provided that for any 0X04;xX04;P(E), there is an event FX38;E with P(F)=x. By Axiom 5, without loss of generality {XW12;A} and {YW12;A} are continuous for any (Borel measurable) A. (Proof: If necessary, enrich the probability space that X is defined on to introduce a random variable U uniformly distributed on [0,1] and independent of X. The enrichment will not change any gamble orderings by Axiom 5. Then if 0X04;xX04;P(XW12;A), just choose aW12;[0,1] such that aP(XW12;A)=x and let F={XW12;A&UX04;a}. Ditto for Y.)

Now, given an event A and a random variable X, let AX be the random variable equal to X on A and equal to zero outside of A. Let A={X=W22;W34;} and B={Y=W22;W34;}. Define the random variables X1 and Y1 on [0,1] with uniform distribution by X1(x)=W22;W34; if xX04;P(A) and X1(x)=0 otherwise, and Y1(x)=W22;W34; if xX04;P(B) and Y1(x)=0 otherwise. Since P(A)X05;P(B) by (7), it follows that X1(x)X04;Y1(x) everywhere and so X1X04;Y1 by Axiom 3. But AX and BY are probabilistically equivalent to X1 and Y1 respectively, so by Axiom 5 we have AXX04;BY. If we can show that AcX<BcY then the conclusion of our Theorem will follow from the second part of Axiom 4.

Let X2=AcX and Y2=BcY. Then P(X2=+W34;)<P(Y2=+W34;), X2* and Y2* have finite expected values and X2 and Y2 never have the value W22;W34;. We must show that X2X04;Y2. Let C={X2=+W34;}. By subdivisibility, let D be a subset of {Y2=+W34;} with P(D)=P(C). Then CX2 and DY2 are probabilistically equivalent, so CX2X04;DY2 by Axiom 5. Let X3=CcX2 and Y3=DcY3. Observe that X3 is everywhere finite. Furthermore P(Y3=+W34;)=P(Y2=+W34;)W22;P(X2=+W34;)>0.

Choose a finite N sufficiently large that NP(Y3=+W34;)>E(X3)W22;E(Y3*) (the finiteness of the right hand side follows from our integrability assumptions). Let Y4 be a random variable that agrees with Y3 everywhere where Y3 is finite, but equals N where Y3 is infinite. Then E(Y4)=NP(Y3=+W34;)+E(Y3*)>E(X3). Thus, Y4>X3 by Axiom 2. But Y3 is greater than or equal to Y4 everywhere, so Y3X05;Y4. By Axiom 1 it follows that Y3>X3. but DY2X05;CX2 and X2=CX2+X3 and Y2=DY2+Y3, so by Axiom 4 we have Y2>X2, which was what we wanted to prove.

Thanks

It's basically saying that if there's a non-zero probability for an infinite payoff with only the risk of a finite loss, then the reward from taking the gamble is worth the risk. Similarly, if both options offer a chance at an infinite payoff but there's a greater probability of an infinite payoff from taking the gamble and no higher risk of a negative infinite payoff then you should take the bet. So therefore it's better to choose to believe in God because there's a better chance of an infinite payoff (heaven).
UchihaMadara
Posts: 1,049
Add as Friend
Challenge to a Debate
Send a Message
9/8/2014 6:57:27 PM
Posted: 2 years ago
At 8/1/2014 10:10:52 AM, Enji wrote:
At 7/23/2014 7:04:59 PM, Toviyah wrote:
Decode this for me:

Proof of Theorem: Say that an event E is continuous provided that for any 0X04;xX04;P(E), there is an event FX38;E with P(F)=x. By Axiom 5, without loss of generality {XW12;A} and {YW12;A} are continuous for any (Borel measurable) A. (Proof: If necessary, enrich the probability space that X is defined on to introduce a random variable U uniformly distributed on [0,1] and independent of X. The enrichment will not change any gamble orderings by Axiom 5. Then if 0X04;xX04;P(XW12;A), just choose aW12;[0,1] such that aP(XW12;A)=x and let F={XW12;A&UX04;a}. Ditto for Y.)

Now, given an event A and a random variable X, let AX be the random variable equal to X on A and equal to zero outside of A. Let A={X=W22;W34;} and B={Y=W22;W34;}. Define the random variables X1 and Y1 on [0,1] with uniform distribution by X1(x)=W22;W34; if xX04;P(A) and X1(x)=0 otherwise, and Y1(x)=W22;W34; if xX04;P(B) and Y1(x)=0 otherwise. Since P(A)X05;P(B) by (7), it follows that X1(x)X04;Y1(x) everywhere and so X1X04;Y1 by Axiom 3. But AX and BY are probabilistically equivalent to X1 and Y1 respectively, so by Axiom 5 we have AXX04;BY. If we can show that AcX<BcY then the conclusion of our Theorem will follow from the second part of Axiom 4.

Let X2=AcX and Y2=BcY. Then P(X2=+W34;)<P(Y2=+W34;), X2* and Y2* have finite expected values and X2 and Y2 never have the value W22;W34;. We must show that X2X04;Y2. Let C={X2=+W34;}. By subdivisibility, let D be a subset of {Y2=+W34;} with P(D)=P(C). Then CX2 and DY2 are probabilistically equivalent, so CX2X04;DY2 by Axiom 5. Let X3=CcX2 and Y3=DcY3. Observe that X3 is everywhere finite. Furthermore P(Y3=+W34;)=P(Y2=+W34;)W22;P(X2=+W34;)>0.

Choose a finite N sufficiently large that NP(Y3=+W34;)>E(X3)W22;E(Y3*) (the finiteness of the right hand side follows from our integrability assumptions). Let Y4 be a random variable that agrees with Y3 everywhere where Y3 is finite, but equals N where Y3 is infinite. Then E(Y4)=NP(Y3=+W34;)+E(Y3*)>E(X3). Thus, Y4>X3 by Axiom 2. But Y3 is greater than or equal to Y4 everywhere, so Y3X05;Y4. By Axiom 1 it follows that Y3>X3. but DY2X05;CX2 and X2=CX2+X3 and Y2=DY2+Y3, so by Axiom 4 we have Y2>X2, which was what we wanted to prove.

Thanks

It's basically saying that if there's a non-zero probability for an infinite payoff with only the risk of a finite loss, then the reward from taking the gamble is worth the risk. Similarly, if both options offer a chance at an infinite payoff but there's a greater probability of an infinite payoff from taking the gamble and no higher risk of a negative infinite payoff then you should take the bet. So therefore it's better to choose to believe in God because there's a better chance of an infinite payoff (heaven).

....so that huge, impressive-looking chunk of logickese is nothing more than an elaborate version of pascal's wager!? -_____-
Enji
Posts: 1,022
Add as Friend
Challenge to a Debate
Send a Message
9/8/2014 7:23:21 PM
Posted: 2 years ago
At 9/8/2014 6:57:27 PM, UchihaMadara wrote:
At 8/1/2014 10:10:52 AM, Enji wrote:
At 7/23/2014 7:04:59 PM, Toviyah wrote:
Decode this for me:

Proof of Theorem: Say that an event E is continuous provided that for any 0X04;xX04;P(E), there is an event FX38;E with P(F)=x. By Axiom 5, without loss of generality {XW12;A} and {YW12;A} are continuous for any (Borel measurable) A. (Proof: If necessary, enrich the probability space that X is defined on to introduce a random variable U uniformly distributed on [0,1] and independent of X. The enrichment will not change any gamble orderings by Axiom 5. Then if 0X04;xX04;P(XW12;A), just choose aW12;[0,1] such that aP(XW12;A)=x and let F={XW12;A&UX04;a}. Ditto for Y.)

Now, given an event A and a random variable X, let AX be the random variable equal to X on A and equal to zero outside of A. Let A={X=W22;W34;} and B={Y=W22;W34;}. Define the random variables X1 and Y1 on [0,1] with uniform distribution by X1(x)=W22;W34; if xX04;P(A) and X1(x)=0 otherwise, and Y1(x)=W22;W34; if xX04;P(B) and Y1(x)=0 otherwise. Since P(A)X05;P(B) by (7), it follows that X1(x)X04;Y1(x) everywhere and so X1X04;Y1 by Axiom 3. But AX and BY are probabilistically equivalent to X1 and Y1 respectively, so by Axiom 5 we have AXX04;BY. If we can show that AcX<BcY then the conclusion of our Theorem will follow from the second part of Axiom 4.

Let X2=AcX and Y2=BcY. Then P(X2=+W34;)<P(Y2=+W34;), X2* and Y2* have finite expected values and X2 and Y2 never have the value W22;W34;. We must show that X2X04;Y2. Let C={X2=+W34;}. By subdivisibility, let D be a subset of {Y2=+W34;} with P(D)=P(C). Then CX2 and DY2 are probabilistically equivalent, so CX2X04;DY2 by Axiom 5. Let X3=CcX2 and Y3=DcY3. Observe that X3 is everywhere finite. Furthermore P(Y3=+W34;)=P(Y2=+W34;)W22;P(X2=+W34;)>0.

Choose a finite N sufficiently large that NP(Y3=+W34;)>E(X3)W22;E(Y3*) (the finiteness of the right hand side follows from our integrability assumptions). Let Y4 be a random variable that agrees with Y3 everywhere where Y3 is finite, but equals N where Y3 is infinite. Then E(Y4)=NP(Y3=+W34;)+E(Y3*)>E(X3). Thus, Y4>X3 by Axiom 2. But Y3 is greater than or equal to Y4 everywhere, so Y3X05;Y4. By Axiom 1 it follows that Y3>X3. but DY2X05;CX2 and X2=CX2+X3 and Y2=DY2+Y3, so by Axiom 4 we have Y2>X2, which was what we wanted to prove.

Thanks

It's basically saying that if there's a non-zero probability for an infinite payoff with only the risk of a finite loss, then the reward from taking the gamble is worth the risk. Similarly, if both options offer a chance at an infinite payoff but there's a greater probability of an infinite payoff from taking the gamble and no higher risk of a negative infinite payoff then you should take the bet. So therefore it's better to choose to believe in God because there's a better chance of an infinite payoff (heaven).

....so that huge, impressive-looking chunk of logickese is nothing more than an elaborate version of pascal's wager!? -_____-

That's exactly what it is. Toviyah cited the source for it above: [http://alexanderpruss.blogspot.com.br...]