# Tag Info

## Hot answers tagged market-model

4

Art markets typically have huge transaction costs of the order of 10%, caused by buyers premium and auction fees. Therefore long holding periods are unavoidable, with long-term returns somewhere between those of bonds and equities. By its very nature, art is not easily replicated so arbitrage or derivatives are out. The rationality of agents (aka collectors) ...

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The general effect of quantitative analysis of the markets is to enforce randomness. Suppose a strategic quant finds a predictable pattern where a stock always rises on Tuesdays. His institution will commence buying the stock every Monday, and selling on Tuesday. The trading itself pushes the stock price up on Monday and down on Tuesday (in general), so if ...

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For Q1, the function $a(t)$ is the instantaneous correlation. The form given by (2) is basically the Cholesky decomposition. Of course, you may directly show, uisng Levy's characterization, that $$\widetilde{W}(t) = \int_0^t\bigg[\frac{1}{\sqrt{1-||a(t)||^2}} dZ(t) -\frac{a(t)^T}{\sqrt{1-||a(t)||^2}} dW^B(t) \bigg]$$ is a standard scalar Brownian motion ...

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Q1: $$(1)\rightarrow(2)$$ (1): $a(t)$ is the instantaneous correlation of $\rho(Z_t,W_t)$ because: $$\rho(dZ_t,dW_t)=\dfrac{Cov(dZ_t,dW_t)}{\sigma_{dZ_t}\sigma_{dW_t}}=\dfrac{E(dZ_t\cdot dW_t)}{\sqrt{dt} \sqrt{dt}}=\dfrac{\langle dZ_t, dW_t\rangle}{t}=a(t)$$ $\Rightarrow$ (2) holds as following, in the 1-dim case: $dZ_t\sim N(0,dt),$ ...

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I have asked myself the very same question when I first read the book. As far as I can tell, the "scalability" condition is only imposed for technical reasons. It simplifies the subsequent proof of the Fundemental Theorem of Asset Pricing in constrained markets. There are several papers that have shown that the theorem is valid for conic constraints. ...

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I would say the financial- and the art market is very different, only the roots of the market / auctions is the same. As the art market is unique and very illiquid, alot of the strategies from the modern financial market simply does not apply. I have been building (and still maintains) a toolbox of models, which mostly try to find trends based on multiple ...

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I think one should look at the problem from two different angles to get an answer to this. Firstly, you can look (as you said you did) look at $\hat{\epsilon}$ in terms of a disturbance like you said, meaning the returns $R_{it}$ are depending linearly on the $R_{mt}$ - the market or factor returns. Then you can figure there is some regression involved an ...

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Idiosyncratic volatility is NOT included in the regressors, so it should not be and actually cannot be part of your matrix X. Idiosyncratic volatility is the volatility (of Y) your matrix X (explanatory variables) cannot explain (i.e. remaining unexplained part), so it is the error term of your regression equation. Just compute the standard deviation of ...

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Calibrating to swaption prices would give you the right volatilities for your model, but you have to use the floating notes (or similar instruments, as swaps) in order to get the right drifts. In any case, your model have to be able to exactly replicate the floating notes prices in order to be considered a valid model, and you can feel comfortable to use it ...

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it's difficult to say that they are not popular. Some people definitely use them for live pricing. I'd say the real question is "why are they not popular in the academic literature"? One answer would simply be that most the questions that arise in their use are ones of fiddliness which do not make good papers.

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EDIT: I changed the answer to have it more on topic. Summary It boils down to Mark Joshi's answer. I wanted to add something more. Answer A probability measure $Q1$ and a numeraire $N1(t)$ are associated if all prices expressed relative to $N1$ are martingales under $Q1$: \frac{price(t)}{N1(t)} = \mathbb{E}^{Q1} \left[ \left. \frac{price(T)}{N1(T)} \, ...

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Different measures have different properties. Using a particular measure may make it easy to derive an analytic formula since a rate is driftless. When performing Monte Carlo, the sign of the drifts changes with measure which affects convergence. There is also the problem in the terminal measure that the numeraire can get very small and so some paths can ...

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