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## Hot answers tagged stochastic-processes

19

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 ...

16

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 ...

15

Here is a short list (to be edited and improved - community wiki) : Standard brownian motion (also called Wiener process) for which: $d\, W_t \sim \mathcal N(0, \sqrt{d t})$ Geometric brownian motion, used in the Black-Scholes model (1973): $d\,X_t = \mu X_t\,dt + \sigma X_t\,dW_t$ Constant elasticity of variance ("CEV") model (1975): $d\,X_t=\mu X_t dt + ... 13 I will defer to others answering the parts of your question concerning the relationship between Markov processes and martingales (@SRKX has already given a good explanation of the relationship) and concerning statistical testing. Broadly, however, it is not possible to "prove" either assumption, but only to fail to reject them. A Non-Random Walk Down Wall ... 10 These moving strategies are also known as trend-following. If returns have positive autocorrelation, hurst exponent > 0.5 that would be good for these strategies. 8 In general, if you have a process that you can write under the form$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$. Then if the resulting SDE has a null drift (that's where Black Scholes PDE comes from), and you get a only local martingale. For it to be a proper martingale you can look at ... 8 "Treshold Garch" or T-Garch models are designed to capture this asymmetry. See this exposition by U. Chicago's Ruey Tsay who has a terrific text on time-series models in "Analysis of Financial Time Series". You can use the structure of the T-Garch models to simulate data with this property. There is a package called fGarch that creates APARCH models. A T-... 8 We know that$(\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$E[W_pW_qW_r] = 0$. 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 $$W(t_{k+1})-W(t_{k})=\sqrt{t_{k+1}-t_{k}}\... 8 1. This integral is not Ito's Integral. Indeed Y_t is a random time change with time change rate \frac{W_t}{1+W_t^2}. (Oksendal, Sixth edition,page 147) 2. Sometimes this trick is useful.Indeed we assume that we are going to solve Riemann integral !. Let$$f''(x)=\frac{-2x}{(1+x^2)^2}$$then$$f'(x)=\left(\frac{1}{1+x^2}\right)+c_1$$and$$f(x)=\... 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 I will assume a white noise is a process$(\varepsilon_t)$with zero mean, no autocorrelation and constant variance$\sigma^2 > 0$while a random walk is a process$(x_t)$defined by $$x_{t+1} = x_t + \varepsilon_{t+1}$$ where$\varepsilon$is a white noise. 1) No since$Var(x_{t+1}) = Var(x_t) + Var(\varepsilon_{t+1})$is stricly increasing while ... 7 Stochastics are usually applied in the field of derivatives pricing. In this setting the task is to price a derivative such that it fits into the landscape of tradable instruments (no-arbitrage). We work using the risk-neutral measure - usually denoted by$Q$. The measure is derived from other traded instruments. In risk analysis (e.g. calculate the VaR, ES ... 7$X_t$being a stochastic process, one cannot use ordinary calculus to express the differential of a (sufficiently well-behaved) function$f$of$t$and$X_t$. Instead one should turn to Itô's lemma, one of the key results of stochastic calculus, which stipulates (assuming$X_t$is here a continuous, square integrable stochastic process) $$df(t,X_t) = \frac{... 6 I have low frequency data (daily) from which I want to construct high frequency data, going though all the lower frequency sampling points. Bad idea in my opinion. I don't really know why you really want to do this (what's are you going to do with the generated data). If it's for backtesting purposes, it's a really bad idea as there are so many mechanisms ... 6 The best I have seen so far is William Wheaton's work in this area. I don't know how much is described in his papers but he and Torto created a system that combined factor models for things like local and national price indexes with specific economics of commercial real estate ventures (such as balloon payments on construction milestones and the like). The ... 6 To complement @SRKX comment ,i'll try to explain the "simple mathematical proof" beetween both formula : I assume you know the geometric or arithmetic brownian motion : Geometric: \begin{equation*} dS = \mu S dt + \sigma Sdz \end{equation*} Arithmetic : \begin{equation*} dS = \mu dt + \sigma dz \end{equation*} Then another important stochastic tool you ... 6 The model for the stock is the Bachelier model with the solution$$ 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) + \frac{\sigma}{... 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: $$S_t = S_0 \... 6 The convexity of the exponential function of the stochastic variable W makes its expectation greater than the exponentiation of the expectation of W. This is an example of Jensen's inequality, E[e^{\sigma W}]> e^{\sigma E[W]}=1. \sigma can be interpreted as the magnitude of the convexity of the exponential function. This can be seen by Taylor ... 6 If at first you don't have a model at all, then geometric Brownian motion is not bad. As others before me said: log-returns are normally distributed in this model. This is debatable and there are times and markets where this is not true. There is more than enough research about this. But why is a model based on Brownian motion not that bad? The reason is ... 6 To provide a straight forward answer: It is not a good model. It never was, it never will be. Until we all do not come up with a better model that provides better modeling accuracy while it is equally intuitive and makes similarly simplifying assumptions the BS model with its geometric brownian motion component is here to stay. It actually does not matter ... 6 The trick is to start with the highest power, rewrite it as something you know (a third order moment) and then work backwards on the remaining terms. By that I mean you can complete the cube as follows:$$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 ... 6 You know that E\left[\int_{0}^{s}W_udu\right]=E\left[\int_{0}^{t}W_vdv\right]=0. By definition \begin{align} & Cov\left(\int_{0}^{s}W_u\,du\,\,,\,\int_{0}^{t}W_v\,dv\right)=E\left[\int_{0}^{s}W_u\,du\int_{0}^{t}W_v\,dv\right]-0 \end{align} then \begin{align} & Cov\left(\int_{0}^{s}W_u\,du\,\,,\,\int_{0}^{t}W_v\,dv\right)=\int_{0}^{s}\int_{0}^{t}E\,... 6 The dynamics \begin{align*} \frac{dS_t}{S_t} =\mu dt + \sigma dW_t. \end{align*} is under the real-world measure \mathbb{P}. Then, \begin{align*} d\ln S_t =\Big(\mu-\frac{1}{2}\sigma^2 \Big) dt + \sigma dW_t. \end{align*} Therefore, \begin{align*} \ln S_T = \ln S_t + \Big(\mu-\frac{1}{2}\sigma^2 \Big)(T-t) + \sigma \big(W_T-W_t\big).\tag{1} \end{align*} ... 6 By definition, the payoff of a log-contract of maturity T writes \phi(S_T) = \ln\left(\frac{S_T}{S_0}\right) $$Let \Pi_t denote the t-value of such a contingent claim. We are interested in the price at t=0, best known as the option premium. Theory tells us that the latter premium can be computed as$$ \Pi_0 = e^{-rT} E^{\mathbb{Q}} \left[ \phi(...

6

Let $$Y_t = \int_0^t N_u du$$ where $(N_t)_{t \geq 0}$ figures a Poisson process with intensity $\lambda$. Using the stochastic Fubini theorem we have that: \begin{align} Y_T &= \int_0^T N_t dt \\ &= \int_0^T \int_0^t dN_u dt \\ &\color{lightgray}{= \int_0^T \int_0^T \mathbf{1}\{u \in [0,t]\} dN_u\ dt} \\ &\color{lightgray}{= \int_0^T \...

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 ...

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