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## Hot answers tagged mathematics

10

I think there are a lot of different ways to specify this problem. For simplicity, consider independent Garch processes $$r_{1,t} \sim N\left(0,\sigma_{1,t}^{2}\right)$$ $$\sigma_{1,t}^{2} = \beta_{1,1}+\beta_{1,2}\varepsilon_{1,t-1}^{2}+\beta_{1,3}\sigma_{1,t-1}^{2}$$ and $$r_{2,t} \sim N\left(0,\sigma_{2,t}^{2}\right)$$ $$\sigma_{2,t}^{2} = ... 8 I think he was jokingly suggesting to breed top PhD candidates in pure mathematics. I often heard complaints that a lot of PhDs in Mathematical disciplines lack a rigorous base in pure Mathematics. Obviously the hedge fund manager was not suggesting that the proof of the hypothesis will be in any way relevant to trading or financial pricing applications. On ... 8 Edit: Freddy's answer is good -- we wrote concurrently. He rightly points out that QF is a broad field, and that it is among other things a community. Here, I describe a practical, down-to-earth path for getting your feet wet in one key piece of it -- software and model development for derivatives analysis, starting with vanilla options. Your best bet ... 8 At the top of this list I still recommend you to seek employment in order to learn from others in QF space. Could you possible work in a quant team within an investment bank where you currently reside? Start to reach out to the quant finance community so you are connected once you decide to locate to where you can practice this discipline.reach out to ... 5 Regarding your first question. You can try to argue in the following manner. Using the fact that E_{Q^t}[P(t,T)|\mathcal{F}_s]P(s,t)=P(s,T), then:$$p(s)=\lim_{T\to\infty}P(s,T)^{\frac{1}{T}}=\lim_{T\to\infty}(P(s,t)E_{Q^t}[P(t,T)|\mathcal{F}_s])^{\frac{1}{T}}=\lim_{T\to\infty}E_{Q^t}[P(t,T)|\mathcal{F}_s]^{\frac{1}{T}}$$as ... 4 I can offer you two explanations, one more economical, and the other mathematical. The one based on economics is based on no arbitrage (and probably what you're looking for): You are aware of the "Second" FTAP, which says roughly, that there is precisely one equivalent true/local/\sigma martingale measure if and only if the market is complete, i.e. all ... 4 The intuitive reason is that an arbitrary payoff in L^0_+ can always throw away wealth to be in L^\infty_+. Strange economically, but it's a common trick whenever a duality argument is going to be used. A bit more formally: Since L^\infty_+ \subset L^0_+, it's clear that Condition 1 implies Condition 2. If Condition 1 doesn't hold, take a ... 4 First question: Looking at the paper, we see that the authors assume, for some \epsilon > 0,$$ \underset{n}{\limsup} P \left(| \widehat{h}(y^n) - \widehat{h}(y) | > \epsilon \right) > \epsilon, $$and from this they wish to deduce (*), or (3.13) in their notation, for some \epsilon. I claim that if \left \{ \widehat{h}(y^n) : n \in ... 4 The other answers are useful and sensible. I have worked full time in equity research for nearly two decades, so very much a "qualitative" rather than a quantitative approach. However, all the firms for which I have worked had quants and because of my casual interest in the area I've spent a lot of time talking to quant teams over the years, often over a ... 3 Expanding a bit on chrisaycock's answer, and noting in particular from the abstract In mathematical finance, solutions to obstacle problem for the elliptic Heston operator correspond to value functions for perpetual American-style options on the underlying asset. we can see that this would be used to price those few rare cases of perpetual options. ... 3 From this abstract: The Heston stochastic volatility process is a degenerate diffusion process where the degeneracy in the diffusion coefficient is proportional to the square root of the distance to the boundary of the half-plane. The generator of this process with killing, called the elliptic Heston operator, is a second-order, degenerate-elliptic ... 3 I'll throw this in as an "application of RMT" ... EDHEC and FTSE use RMT to decide the optimal number of principal components in their covariance estimation procedure for which they use PCA (Principal Component Analysis). For details look here or here in Appendix C section 4 for details. 3 I think one of the best (and very current) articles about how to break into QF (for any kind of background) is: "On becoming a Quant" by Mark Joshi For your special background in mathematics see this excerpt from section 9: The main challenge for a pure mathematician is to be able to get one’s hands dirty and learning to be more focussed on getting ... 3 The desired decomposition: Suppose that x + y + \theta \cdot S \in A(z), with z = x + y. Then \theta \cdot S has the property that \theta \cdot S > -x - y, by definition of admissibility. It follows that \frac{x}{x+y} \theta \cdot S > -x, and therefore x + \frac{x}{x+y} \theta \cdot S \in A(x). Similar reasoning implies that y + ... 3 Constructing the right supermartingale is the key step. Consider the process$$ X_t := \underset{Q \in \mathbb{P}}{\text{esssup}} \ E_Q \left[ H | \mathcal{F}_t \right].  This process $X$ is well-defined, since $\underset{Q \in \mathbb{P}}{\sup} E_Q [H] < \infty$. One can show that $X_t$ is a supermartingale under every $Q$ in $\mathbb{P}$. This ...

2

Regarding your first question, let me reformulate $p(s)$: $p(s)=(\frac{P(s,T)}{P(s,t)})^\frac{1}{T}= \frac{e^{-\frac{(T-s)R(s,T)}{T}}}{e^{⁻\frac{(t-s)R(s,t)}{T}}}.$ Now, because $R(s,t)$ is finite, the term $e^{⁻\frac{t-s}{T}R(s,t)}$ converges to 1 for $T\to\infty$. On the other hand, the term $e^-\frac{(T-s)R(s,T)}{T}$ converges to ...

2

If you are able to derive quantities in the LMM with unconditional expectation as functions of $L(0,T)$ and $\lambda(s,T)$ for $0 \leq s \leq T$ then expression of the time $t$ conditional expectation is exactly the same function in $L(t,T)$ and $\lambda(s,T)$ for $t \leq s \leq T$ - just replace $0$ by $t$. This is due to the fact that the model is ...

2

I didn't read the papers you linked but I can understand that lie groups may be used much as there are used in quantum field theories to build up gauge theories for interaction of particles. The purpose is to have a model that is invariant according to a given transformation group. This introduce interaction terms in the equations. I can imagine that the ...

1

Although I don't think that this is a question that fits in here, I will give you a reference. You might want to have a look at the so called greeks, you find a first overview here: http://en.wikipedia.org/wiki/Greeks_(finance)

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A detailed description of the Hurst Exponent can be found here. A further (rather short search of Google) turned up this site claiming to provide an Excel Workbook with, among other things, Hurst Exponent estimation.

1

Riemann Hypothesis is a very import conjecture in mathematics, but it also an extremely hard problem, top mathematicians have worked on it for over 100 years and could not solve it. Moreover, one cannot start to really think about it without proper understanding of the problem; it might take years to understand what is going on even for people with strong ...

1

One possibility worth exploring is to use the support vector machine learning tool on the Metatrader 5 platform. Firstly, if you're not familiar with it, Metatrader 5 is a platform developed for users to implement algorithmic trading in forex and CFD markets (I'm not sure if the platform can be extended to stocks and other markets). It is typically used for ...

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