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Challenge for logic nuts
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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 
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7/23/2014 7:08:14 PM Posted: 2 years ago Darn. Forgot that logic symbols don't show up. Here's the link: http://alexanderpruss.blogspot.com.br...

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7/23/2014 7:27:56 PM Posted: 2 years ago At 7/23/2014 7:04:59 PM, Toviyah wrote: F That. 
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7/23/2014 8:24:39 PM Posted: 2 years ago I would, BUT I CAN'T WORK UNDER THESE CONDITIONS. *throws paper at wall*

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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: 404 coherent debate topic not found. Please restart the debate with clear resolution. Uphold MarxistLeninistMaoistSargonistn7ism. 
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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: Did I just read right, Well now you know how I feel when you get all technical' LMAO 
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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% 
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7/31/2014 8:57:20 PM Posted: 2 years ago At 7/23/2014 7:04:59 PM, Toviyah wrote: Yes. 
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8/1/2014 9:46:48 AM Posted: 2 years ago At 7/23/2014 7:04:59 PM, Toviyah wrote: 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 
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8/1/2014 10:10:52 AM Posted: 2 years ago At 7/23/2014 7:04:59 PM, Toviyah wrote: It's basically saying that if there's a nonzero 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). 
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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: ....so that huge, impressivelooking chunk of logickese is nothing more than an elaborate version of pascal's wager!? _____ 
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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: That's exactly what it is. Toviyah cited the source for it above: [http://alexanderpruss.blogspot.com.br...] 