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## Hot answers tagged stochastic-processes

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 $$... 7 We know that (\tilde{W}_t) := (-W_t) is also a Wiener process so$$ E[W_pW_qW_r] = E[\tilde{W}_p\tilde{W}_q\tilde{W}_r] = (-1)^3E[W_pW_qW_r] $$and that implies that E[W_pW_qW_r] = 0. 6 The trick is to start with the highest power, rewrite it as something you know (a third order moment) and then work backwards on the remaining terms. By that I mean you can complete the cube as follows:$$E[W_t^3 - 3tW_t|\mathcal{F}_s] = E[(W_t-W_s)^3 - C -3tW_t|\mathcal{F}_s]$$where you'll need to find C such that the equality holds (i.e. C=W_s^3 + ... 6 If at first you don't have a model at all, then geometric Brownian motion is not bad. As others before me said: log-returns are normally distributed in this model. This is debatable and there are times and markets where this is not true. There is more than enough research about this. But why is a model based on Brownian motion not that bad? The reason is ... 6 To provide a straight forward answer: It is not a good model. It never was, it never will be. Until we all do not come up with a better model that provides better modeling accuracy while it is equally intuitive and makes similarly simplifying assumptions the BS model with its geometric brownian motion component is here to stay. It actually does not matter ... 6 Basically, prices usually have a unit root, while returns can be assumed to be stationary. This is also called order of integration, a unit root means integrated of order 1, I(1), while stationary is order 0, I(0). Time series that are stationary have a lot of convenient properties for analysis. When a time series is non-stationary, then that means the ... 6 I will assume a white noise is a process (\varepsilon_t) with zero mean, no autocorrelation and constant variance \sigma^2 > 0 while a random walk is a process (x_t) defined by$$ x_{t+1} = x_t + \varepsilon_{t+1} $$where \varepsilon is a white noise. 1) No since Var(x_{t+1}) = Var(x_t) + Var(\varepsilon_{t+1}) is stricly increasing while ... 5 Brownian motion - because it is simple, and results in intuitive closed form solutions, and it's not a terrible description of asset prices, especially when employed in high-frequency event time. Geometric - because the returns compound, and equities cannot go below zero due to the fact that they are limited liability corporations There are many, many ... 5 if Y=1 the stock price doesn't change since it's a percentage change not an absolute, so we have to subtract one when drift compensating. See my book Concepts etc for my discussion. 5 Mean reversion speed \kappa is better interpreted with the concept of half-life, which can be calculated from \text{HL} = \ln(2) / \kappa. For example, if the mean reversion coefficient is \kappa = 1.5, then the half-life of the process is \ln(2) / 1.5 = 0.46209812 years, or about 6 months. Let's assume that the current interest rate is 1% and the ... 4 You have typo "vol^2", but it should be "vol". Its$$\sqrt{\sigma^2T}=\sigma\sqrt{T}$$4 Perhaps overly simplistic and repeating the pt above, but when doing statistics, ideally we want to compare like with like. Returns can be comparable with each other. Prices on the other hand always depend on the previous price. 4 For a martingale dX=a(X,t)\,dt+b(X,t) dW(t) where a and b are not constant, your tree will not recombine in general . This is the main issue. See for instance: Florescu, I. and F. G. Viens (2008, March). Stochastic volatility: Option pricing using a multinomial recombining tree. Applied Mathematical Finance 15 (2), 151-181. It deals with the case ... 4 In this case it is just the notion that your payoff function should not explode at some point - made mathematically rigorous. Have a look at the following picture from wikipedia: Intuitively the Lipschitz condition (or Lipschitz continuity) ensures that your payoff function always remains entirely outside the white cone, so it cannot e.g. become ... 4 Write X_t = A_t B_t with A_t = e^{(\lambda - \eta)t} and B_t = \left(\frac{\eta}{\lambda} \right)^{N_t}. Then dX_t = A_t dB_t + B_t dA_t by the product rule of calculus. There are no second order terms since both A_t and B_t are finite variation (i.e. \langle A_t, B_t\rangle= 0). Next, dA_t = (\lambda - \eta)A_t dt, and dB_t = B_t \cdot ... 3 Basically, Black-Scholes is an "industry standard" formula. It is widely used by practitioners and usually augmented with extra specifications or intuition. It has a closed form solution, which is rare in option pricing models. It is also relative to simple to understand. Otherwise, you usually need to rely on Monte Carlo simulation or some other way. And ... 3 Just a bit of illustration added to @John's answer. Look at log prices \log(P_t), assume that you know P_0 then$$ \log(P_t) = \log(P_0) + r_1 + \cdots r_t $$where r_i = \log(P_i)-\log(P_{i-1}) are the log returns. By modelling the log-returns (which as already said take values on the whole real line which is a nice property for modelling) we model ... 3 I would say Take log of first equation to get rid of dependence on x_t Apply Kalman filter equations to estimate parameters I believe Conrad and Kaul (1988) J of Business do exactly what you describe. 3 There is a shortcut around the Forward Equation when you are looking for the stationary distribution. Let me write$$ dX = \mu(X)dt +\sigma(X)dW $$for$$ \mu(x)=b(1-x)-ax\ \text{ and }\ \sigma^2(x)=x(1-x) $$The Forward Equation indeed states that the stationary distribution p(x) will be satisfied for \partial p/\partial t = 0, therefore$$ ...

3

Let's consider the following example: the process is initialized randomly with $\pm1$ and then stays there forever. Seems stationary to me, but it would never cross its mean.

3

The more phenomenological definitions in his books are probably more helpful. Whether one uses the fractal dimension, Hurst coefficient, or exponential coefficient alpha, there is a value that corresponds to pure Brownian motion, a regime relative to this value that corresponds to persistence of motion, and the opposite regime that corresponds to ...

3

For a Brownian motion, if you wait $dt$, the variance will grow linearly with (proportionally to) $dt$. For a fractional Brownian motion, it will grow with a power law of $dt$, in fact in $dt^{H}$, where $H$ is the Hurst exponent. See wikipedia for more details. It means the fBM will somehow keep memory of the past. When $H$ is lower than 1/2, it will mean ...

3

EDIT: My reasoning below seems to be wrong. The process as you write it tends to infinity if $a$ is big enough and positive and if $\lambda_0$ is positive. I would not call this process non-meanreverting OU. It is just an Ito process of a simple form. If we remove the stochastic part then we get $$d\lambda_t = a \lambda_t dt$$ with the solution (if ...

2

First of all, a filtration $( \mathscr{F}_t )_{t \geq 0 }$ is a "set" of sigma algebras indexed usually by time t that are increasing. That is, for every $t>0$, $\mathscr{F}_t$ is a sigma algebra and $\mathscr{F}_t \subseteq \mathscr{F}_T$ for all $0\leq t \leq T$. The canonical example, is the filtration generated by a process, say Brownian Motion $W$: ...

2

You have $$\widetilde{W}_t=W_t+\int\Theta(u)du$$ which is in general not a Brownian motion, because it has a drift component. But 5.3.1 states $$M_t=M_0+\int \Gamma(u)dW_u\tag{5.3.1}$$ , which holds only for a Brownian motion $W$ (and $M_t$ martingale). So one cannot trivially replace $W_t$ and $W_t+\int\Theta(u)du=\widetilde{W}_t$ in 5.3.2 aswell by ...

2

You should look at confidence interval. Normally, your confidence interval size is proportional to the standard deviation, looking something like: with probability $p$ your value will be in the interval: $$[\bar{S} - k*StdDev, \bar{S} + k*StdDev]$$ Then, getting back to your simulation, we can say that your time step is very big (1 year) and you simulate ...

2

I think you are on the right track here. You made a sign error in the first line, unfortunately: $$E[W_p W_q W_r] = E[W_r W_p^2 + W_pW_q^2 - W_qW_p^2]=\\ E[(W_r-W_q)W_p^2]+E[W_pW_q^2]= E[W_pW_q^2]$$ The first term is $0$ by independence (as $p<\text{min}(r,q)$ and the square does not affect independence). To take care of the second term we do the ...

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