# Tag Info

17

The way you do it in the first place is a discretization of the Geometric Brownian Motion (GBM) process. This method is most useful when you want to compute the path between $S_0$ and $S_t$, i.e. you want to know all the intermediary points $S_i$ for $0 \leq i \leq t$. The second equation is a closed form solution for the GBM given $S_0$. A simple ...

15

From what I remember, there is no real relation between Markov and Martingale, and my intuition was confirmed by this post. Basically, it says that you can say neither of the following: If A is Markov, then A is a martingale. If A is a martingale, then A is Markov. further down the post, you can find two counter examples: $dX_t = a dt + \sigma dW_t$ is ...

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I will defer to others answering the parts of your question concerning the relationship between Markov processes and martingales (@SRKX has already given a good explanation of the relationship) and concerning statistical testing. Broadly, however, it is not possible to "prove" either assumption, but only to fail to reject them. A Non-Random Walk Down Wall ...

9

These moving strategies are also known as trend-following. If returns have positive autocorrelation, hurst exponent > 0.5 that would be good for these strategies.

8

"Treshold Garch" or T-Garch models are designed to capture this asymmetry. See this exposition by U. Chicago's Ruey Tsay who has a terrific text on time-series models in "Analysis of Financial Time Series". You can use the structure of the T-Garch models to simulate data with this property. There is a package called fGarch that creates APARCH models. A ...

8

I can clarify 100% that $(dw)^2$= $dt$ and recommend you to accept it as a fact. Like any other differential, this differential is defined in terms of its integral: $$\int_{t_{0}}^{t_{1}}(dW)^{2}\equiv\lim_{n\rightarrow\infty}\sum_{k=0}^{n-1}[W(t_{k+1})-W(t_{k})]^{2}$$ Where $t_{k}=t_{0}+k(t_{1}-t_{0})/n$. Since $$... 7 These patterns are of course well-known enough to have been "priced in" to the financial markets. Jump diffusions are a classic way to capture the phenomenon, and often have closed-form option pricing formulas associated with them. The implied option skew, for example, gets a lot flatter when you use a JD model. Jump diffusions are often combined with ... 7 We know that (\tilde{W}_t) := (-W_t) is also a Wiener process so$$ E[W_pW_qW_r] = E[\tilde{W}_p\tilde{W}_q\tilde{W}_r] = (-1)^3E[W_pW_qW_r] $$and that implies that E[W_pW_qW_r] = 0. 6 I have low frequency data (daily) from which I want to construct high frequency data, going though all the lower frequency sampling points. Bad idea in my opinion. I don't really know why you really want to do this (what's are you going to do with the generated data). If it's for backtesting purposes, it's a really bad idea as there are so many ... 6 The model for the stock is the Bachelier model with the solution$$ S(t) = S(0) + \sigma W(t) $$Thus the law of the stock S(t) is Gaussian with mean S(0) and variance \sigma^2 t. For average process Z(T) is thus the average of linear Brownian motion, we can rewrite this as$$ Z(T) = \frac{1}{T} \int_0^T S(0) + \sigma W(t) dt = S(0) + ...

6

I like Richard's answer, but I think we can compute the mean and the variance of $\int_0^T W_t dt$ by ourselves using Ito's lemma. Let $f(W_t, t) = t W_t$. $$d( t W_t ) = W_t dt + t dW_t .$$ Integrating both sides, and re-arranging the terms, we get $$\int_0^T W_t dt = T W_T - \int_0^T t dW_t \, .$$ We'll be using Ito's isometry formula \mathbb{E} ... 6 Okay so I'll take Jase answer and format it properly so that it answers your question and it will be useful for users in the future. For clarity, let me restate the dynamics of the Modified Ornstein-Uhlenbeck model using the more common notation: $$dS_t = \theta (\mu-S_t)dt + \sigma S_t dW_t$$ This blog post provides a closed form solution: $$S_t = S_0 ... 6 The trick is to start with the highest power, rewrite it as something you know (a third order moment) and then work backwards on the remaining terms. By that I mean you can complete the cube as follows:$$E[W_t^3 - 3tW_t|\mathcal{F}_s] = E[(W_t-W_s)^3 - C -3tW_t|\mathcal{F}_s]$$where you'll need to find C such that the equality holds (i.e. C=W_s^3 + ... 6 Basically, prices usually have a unit root, while returns can be assumed to be stationary. This is also called order of integration, a unit root means integrated of order 1, I(1), while stationary is order 0, I(0). Time series that are stationary have a lot of convenient properties for analysis. When a time series is non-stationary, then that means the ... 5 To complement @SRKX comment ,i'll try to explain the "simple mathematical proof" beetween both formula : I assume you know the geometric or arithmetic brownian motion : Geometric: \begin{equation*} dS = \mu S dt + \sigma Sdz \end{equation*} Arithmetic : \begin{equation*} dS = \mu dt + \sigma dz \end{equation*} Then another important stochastic tool you ... 5 In general, if you have a process that you can write under the form F(B_t,t) where F is \mathcal{C}^{2,1} then Itô's lemma gives you the drift term and diffusion term of dF. Then if the resulting SDE has a null drift (that's where Black Scholes PDE comes from), and you get a only local martingale. For it to be a proper martingale you can look at ... 5 The best I have seen so far is William Wheaton's work in this area. I don't know how much is described in his papers but he and Torto created a system that combined factor models for things like local and national price indexes with specific economics of commercial real estate ventures (such as balloon payments on construction milestones and the like). The ... 5 Apparently yes, (I haven't verified the math but have no reason to doubt it). For this simple case you can find a closed form in the following paper: Jeff A. BILMES: What HMM can do The closed form is given on part 4.4 of the paper but the whole thing is worth reading as it clearly shows the main properties of these models. You can also note that ... 5 For completeness, let's restate that the discrete case goes like this:$$\Delta S_t = S_{t+\Delta t}- S_t = \mu S_t \Delta t + \sigma \sqrt{\Delta t} Z_t with Z_t \sim \mathcal{N}(0,1). What you are doing in your case (although there is a typo in your formula) is to use the exact solution of the SDE to model the move between two points of S. ... 5 The convexity of the exponential function of the stochastic variable W makes its expectation greater than the exponentiation of the expectation of W. This is an example of Jensen's inequality, E[e^{\sigma W}]> e^{\sigma E[W]}=1. \sigma can be interpreted as the magnitude of the convexity of the exponential function. This can be seen by Taylor ... 5 For a basic introduction, the three chapters in Hull's Options, Futures, and Other Derivatives on Binomial Trees, Wiener Processes and Ito's Lemma, and The Black-Scholes-Merton Model helped me start to understand the basic concepts within a broader context. After that, Shreve's two books seems to be pretty popular (see here and here). He explains things ... 5 An AR(1), once the time series and lags are aligned and everything is set-up, is in fact a standard regression problem. Let's look, for simplicity sake, at a "standard" regression problem. I will try to draw some conclusions from there. Let's say we want to run a linear regression where we want to approximate y withh_(x) = \sum_0^n \theta_i x_i = ... 5 $$\textbf{Preface}$$ I am assuming log normal asset but this is not clear from the question? Or rather I have misinterpreted the question! Well as I see it from a a purely mathematical exercise $$d\left(\dfrac{S_t}{M_t}\right) =\frac{1}{M_t}dS_t - \frac{S_t}{M_t^2}dM_t +O(dt^2)$$ using Ito's lemma. Then we can sub in the original processes yields ... 5 If at first you don't have a model at all, then geometric Brownian motion is not bad. As others before me said: log-returns are normally distributed in this model. This is debatable and there are times and markets where this is not true. There is more than enough research about this. But why is a model based on Brownian motion not that bad? The reason is ... 5 Brownian motion - because it is simple, and results in intuitive closed form solutions, and it's not a terrible description of asset prices, especially when employed in high-frequency event time. Geometric - because the returns compound, and equities cannot go below zero due to the fact that they are limited liability corporations There are many, many ... 5 To provide a straight forward answer: It is not a good model. It never was, it never will be. Until we all do not come up with a better model that provides better modeling accuracy while it is equally intuitive and makes similarly simplifying assumptions the BS model with its geometric brownian motion component is here to stay. It actually does not matter ... 5 ifY=1$the stock price doesn't change since it's a percentage change not an absolute, so we have to subtract one when drift compensating. See my book Concepts etc for my discussion. 4 I think a simple solution is to try to construct a Brownian motion$W_t$through known points (e.g.,$W_0 = W_1 = 0$); it is also known as a Brownian Bridge [ http://en.wikipedia.org/wiki/Brownian_bridge ]. See also question 3 in http://www.math.nyu.edu/faculty/goodman/teaching/StochCalc2012/assignments/assignment4.pdf . 4 Check out these resources: The book Levy Processes in finance. This paper basically enabling you to use any distribution for asset prices: Option Valuation Using the Fast Fourier Transform 4 Note: There is a typo in your third equations. Instead of$S(u)$it should be$S(t_{i})$and in place of$S(t)$there should be$S(t_{i+1})$. In fact, given$S(t_{i})\$ we have that $$S(t_{i+1}) = S(t_{i}) \exp\left( (\mu - \frac{1}{2} \sigma^2) (t_{i+1} - t_{i}) + \sigma (W(t_{i+1}) - W(t_{i})) \right)$$ is the exact solution of the SDE. Hence, the ...

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