Dear George,

You see the problem with the natural approach, as you mentioned, specifically was that when we merge the boubly-indexed sequence into a singly-indexed one with the help of function the new sequence stops being decreasing. But with the definition that I gave for we can do the same process just fine for the corresponding functions in order to construct the new sequence that we need for the set intersection. The reason of course is that there is no decreasing requirement for the sequences that appear in the definition of .

]]>Hi ST,

Thanks for your comment. I’ll have a reread through the post, when I have some time, and consider how well your suggested changes work out.

Excellent post, interesting stuff. I have a suggestion on the arranging of proof 3 (Capacitability Theorem). As you correctly mention after it, the problem with the decreasing sequences can be dealt with if we are more strict and demand the choice of to depend on . In order to model this we make the following definitions:\\

.\\

The rest of the proof can easily be transformed analogously with what you do, i think these approach makes it a bit tidier, what do you think.

]]>Maybe I read past the line “*some* mathematical model of quantum coin tossing” a bit quickly. No doubt there does exist such models, but it probably requires some interaction between the coins so that the measurements do not commute.

]]>Hi J.P.,

Hmm, do you have a reference for that? It sounds like you are supposing that measuring the state of the different coins are non-commuting measurements. That’s not what I was assuming in this post. I was just assuming that they behave according to Bose-Einstein statistics. E.g., measuring the polarization of different photons, which are commuting observables (or, measuring the spins of any identical set of bosons)

]]>When two classical coins are tossed, and the first coin measured to be heads, and the second coin measured to be tails, measuring the first coin again gives heads with probability 1.

When two quantum coins are tossed, and the first coin measured to be heads, and the second coin measured to be tails, measuring the first coin again gives tails with non-zero probability.

]]>Hi Hurgudurgu,

I have seen similar statements before in various books & papers. Nothing that I can recall precisely right now, but I think that it is quite well known that sufficiently “nice” strong Markov martingales satisfy the representation property. If I remember any references, I will come back here. Probably googling the relevant terms would get something, although probably not the exact conditions stated here.

Thanks

George

Thank you for your beautiful blog! I would be interested in the following statement you made: “It should be clear that this argument is quite general, and applies to any continuous martingale which is Markov with twice continuously differentiable transition densities”. Do you have a reference for this, some book/paper where this is proven? Thanks

]]>Even though I jokingly credit my aunt for my writing talent, I know that it is a ability I have fostered from childhood. Though my mother is a writer, I also started out young.

I’ve always had a way with words, according to my favorite professor . I was always so excited in English when we had to do a research assignment .

Now, I help current learners achieve the grades that have always come easily to me. It is my way of giving back to communities because I understand the troubles they must overcome to graduate.

Leonard Rossi – Academic Writer – http://www.asadeaguia.net Confederation

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