Aug
5
comment How does the value of an asset (valued in two different currencies) change when the exchange rate changes?
OK, so I'm in the wrong place. Thanks, good to know. Is there another stack exchange more suitable to theoretical questions like this?
Aug
5
comment How does the value of an asset (valued in two different currencies) change when the exchange rate changes?
Can someone enlighten me on why this question is being downvoted? I really thought hard about it before asking, and put time into presenting the question in a way which I thought was accessible and coherent.
Aug
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revised How does the value of an asset (valued in two different currencies) change when the exchange rate changes?
corrected dollar signs into pounds
Aug
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awarded Editor
Aug
5
comment How does the value of an asset (valued in two different currencies) change when the exchange rate changes?
I've now changed my answer based on the idea that algorithmic traders will operate faster than any other participants.
Aug
5
revised How does the value of an asset (valued in two different currencies) change when the exchange rate changes?
Answer is totally changed from previous.
Aug
5
comment How does the value of an asset (valued in two different currencies) change when the exchange rate changes?
Hi Chris, I've now posted an answer to this see below. My answer was a "negative answer" in that I claim you just can't calculate x. However, what you're saying about automated trades is now making me second think it. If algorithmic trading operates so fast, then shouldn't it be possible to determine "x" just by looking at the order book and reading off the price where the two prices would match at the new exchange rate after all the buy/sells (of equal volume) have been executed?
Aug
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answered How does the value of an asset (valued in two different currencies) change when the exchange rate changes?
Aug
5
comment How does the value of an asset (valued in two different currencies) change when the exchange rate changes?
Yes, of course but I think this is getting off topic. Gold is just an example.
Aug
5
comment How does the value of an asset (valued in two different currencies) change when the exchange rate changes?
You're tying "value" of the asset to a particular currency, then adjusting to the other currency. This makes sense for a product which is made in a particular country, say US. It will cost the manufacturer $1, regardless of how US-CAN exchange rate changes (assuming they have no operations/suppliers from Canada). But what about when the market decides the price. I thought gold was illustrative. Chris Degnen further clarified that this can be traded on LSE and NYSE. You say, price quotes reach the value. What then is the "value" of gold? Its value on the LSE? Or its value on the NYSE?
Aug
5
comment How does the value of an asset (valued in two different currencies) change when the exchange rate changes?
So gold on the LSE is different to gold on the NYSE? Is this a standard way to define an asset? I thought it was the same asset, i.e. gold, but its value changed depending on where it was traded. In any case, I don't think you've completely answered my question. You are right costs often prevent the arbitrage op. But I was trying to understand in an idealised world. In any case, even with these costs, if overnight one currency collapses then arbitrage ops would exist i.e., above the spread. Wouldn't this put supply-demand pressures on both "assets" as you say.
Aug
4
asked How does the value of an asset (valued in two different currencies) change when the exchange rate changes?
Jul
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awarded Popular Question
Jun
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awarded Editor
Feb
18
accepted Is $\hat{\phi}^{-}\hat{\phi}^{+}$ a well defined observable in the Quantum field theory of a scalar field?
Feb
4
comment Is $\hat{\phi}^{-}\hat{\phi}^{+}$ a well defined observable in the Quantum field theory of a scalar field?
Sorry, I don't see how an observable which doesn't commute point-wise on spacelike separation could commute when it is smeared. e.g., $[\hat{\mathcal{O}}(x),\hat{\mathcal{O}}(y)]=\Delta(x-y)$. $\hat{\mathcal{O}}(f)=\int f(x)\hat{\mathcal{O}}(x) dx$ Suppose Supp f is spacelike to Supp f' then, $[\hat{\mathcal{O}}(f),\hat{\mathcal{O}}(f')]=\int f(x)f'(y)\Delta(x-y) dxdy$.
Feb
4
comment Is $\hat{\phi}^{-}\hat{\phi}^{+}$ a well defined observable in the Quantum field theory of a scalar field?
@V.Moretti, perhaps then it is instructive to rephrase my question in terms of the measurement of charge. Let me assume the measurement corresponds to your definition above, take two \Omega$ regions spacelike separated, is it possible (even if exponentially improbable) to signal by measuring charge in this way? Is this how one proves that one must consider a measurement of charge as being over the whole space? i.e.,does one use a no-signaling argument to prove that the operator is non-local-- or is there another way to see that non-microcausality implies the measurement is non-local.
Feb
4
comment Is $\hat{\phi}^{-}\hat{\phi}^{+}$ a well defined observable in the Quantum field theory of a scalar field?
However, there are differences when one considers coherent states, $|\Psi\rangle$, then the terms $\langle \Psi| E^+(x)E^+(x)|\Psi\rangle$ and $\langle \Psi| E^-(x)E^-(x)|\Psi\rangle$ give additional contributions which distinguish these kinds of detectors.
Feb
4
comment Is $\hat{\phi}^{-}\hat{\phi}^{+}$ a well defined observable in the Quantum field theory of a scalar field?
This is an interesting point. Glauber understood that a detector which works on an absorption principle never measures the vacuum fluctuations. You are saying that an absorbing detector is somehow equivalent to a renormalized detector which absorbs and emits).
Feb
4
comment Is $\hat{\phi}^{-}\hat{\phi}^{+}$ a well defined observable in the Quantum field theory of a scalar field?
Also nice point about being able to express $\hat{\mathcal{O}}$ as superpositions of the microcausality-obeying fields. I realised the same thing but was confused about what it meant :) So your interpretation here is that the Glauber detector must describe a detection which is "intrinsically extended", do you have some understanding of how this reconciles with Glauber's original argument that the detector is absorbing particles at the position, x? It seems, superficially, to be a local operation.
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