Discussion of article "Triangular arbitrage" - page 6
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I couldn't understand this terminology using the search engine.
Ok, in brief. even if arbitrage only on the fact, there may be a situation when the skew is not on two currencies, but on three or four or more (partial execution, etc.).
If you arbitrage by limits, then the situation when several arbitrage situations are triggered at once is not uncommon.
In this case, there may be an optimal way to resolve the skew (deviation) that is more efficient than the initial one. Besides, during the time of triggering and analysis the situation may change dramatically and the optimal chain may change to another one.
If you arbitrage on the exchanger, the arbitrage situation triggering means the portfolio deviation from the base portfolio, which should be resolved. In the general case, the portfolio deviation can be anything, including non-zero for any asset in the portfolio. I am interested in the general optimal solution of the problem.
OK, to summarise. even if arbitrage is based on fact only, there may be a situation where the skew is not on two currencies, but on three or four or more (partial execution, etc.).
Without an example, I can't seem to get a handle on what was meant.
Marketplaces are entered. Let, for example, EURUSD buy by 1 lot. So buy is less desirable for EURUSD than if it were zero lots. Therefore, EURUSD_Ask markup (up to infinity) as an edge of a directed graph.
It is extremely problematic to align positions perfectly. we can say that there is almost always a directional position.
To be more precise, this type of arbitrage is absolutely impossible on the Forex market. The reason has already been given: If we buy GBP for EUR, we will not be able to close this position by selling GBP for USD (Sell GBPUSD). We will have two positions Sell EURGBP and Sell GBPUSD. Unfortunately, in MT there is no possibility to make such an operation, despite the fact that we have GBP and our deposit in USD.
Without an example, I can't seem to understand what was meant.
Buy 1 EURUSD,
chain EUR - GBP - USD
Sell 1 EURGBP with limit (ideally FOK) executes 0.3,
update the data, for EUR --> USD the best chain is EUR - AUD - USD.
Sell 0.7 EURAUD executes 0.3.
So we have a divergence for EUR, USD, GBP, AUD, which we need to close in an optimal way. Everything is quite simple here, because there is only one negative deviation (USD). Plus we should keep in mind that the cost of switching from one currency to another may differ depending on the lot.
To be more precise, this type of arbitrage is absolutely impossible on the Forex market.
You're not thinking straight.
fxsaber:
Therefore, EURUSD_Ask marcapim (up to infinity) as an edge of a directed graph.
Can you explain this phrase? I don't understand.
You're the one who didn't think it through.
And you answered without thinking.
I didn't read the wiki, but here's someone else's translation.
1) Bank A sells $5,000,000 to Bank B for euros, getting €4,085,500. ($5,000,000,000 × 0.8171 €/$ = €4,085,500)
2) Bank A then sells this received €4,085,500 to another Bank B for pounds, receiving £3,430,311. (€4,085,500 ÷ 1.1910 €/£ = £3,430,311)
3) Last Bank A sells £3,430,311 to Bank D back for dollars, receiving $5,025,406. (£3,430,311 × 1.4650 $/£ = $5,025,406)
is understandable and reasonable. However, the quotes here are inverted, but it is understandable, if dollars are sold for euros at 0.8171, it means buying euros for dollars Buy EURUSD 1/0.8171 = 1.2238
All the other options are all drawn-out attempts to interpret triangular arbitrage.
You didn't think it through.
Practising arbitration...
Practising arbitrage...
Maybe something similar to the strategy "Martingale", the same concept with different interpretation?
In any case, it is impossible to do what is in the quote on the Forex market. That's exactly what I was talking about. I don't know any other options.
To be more precise, this type of arbitrage on the forex market is absolutely impossible.
Well, impossible so impossible )