# Tag Info

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] $$and that implies that E[W_pW_qW_r] = 0. 7 I will assume a white noise is a process (\varepsilon_t) with zero mean, no autocorrelation and constant variance \sigma^2 > 0 while a random walk is a process (x_t) defined by$$ x_{t+1} = x_t + \varepsilon_{t+1} $$where \varepsilon is a white noise. 1) No since Var(x_{t+1}) = Var(x_t) + Var(\varepsilon_{t+1}) is stricly increasing while ... 6 The trick is to start with the highest power, rewrite it as something you know (a third order moment) and then work backwards on the remaining terms. By that I mean you can complete the cube as follows:$$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 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 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 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} & ... 4 Milstein Scheme This scheme is described in Glasserman (2003) and in Kloeden and Platen (1992) for general processes.Hence, for simplicity, we can assume that the Stochastic Process is driven by the SDE \begin{align} &dX_t=\Xi(t,X_t)dt+\Sigma(t,X_t)dW_t\\ \end{align} Milstein discretization is, \begin{align} dX_{t+\Delta ... 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 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

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 ...

2

There is a qualitative shift in the shape of the density. When V is small it is monotone decaying. When V is large it looks more like a Gaussian. Another reason he uses two schemes is that he wants match two moments of the density. When V is small, the moment matching equations for the quadratic Gaussian are unsolvable. When V is large they are unsolvable ...

2

To answer this I sum up a paragraph of "Interest rate models - An Introduction" by A.Cairns: For $i=1,\ldots,d$ consider the OU-processes $$dX^i_t = -\frac 12 \alpha X^i_t dt + \sqrt{\alpha} dW^i_t.$$ Looking at the squared radius $R_t = \sum_{i=1}^d (X^i_t)^2$ (in $\mathbb{R}^d$) of this process we get by Ito: dR_t = \sum_{i=1}^d (2 X^i_t dX^i_t) + d ... 2 Exhibiting a counter-example is straight-forward enough. For example, let B_{t}(\omega) be a Brownian motion and \mathcal{T}(\omega) a stopping time on (\Omega,\mathbb{P}) with a continuous distribution. Then with ... 2 The above question was a typo due to the author -- the expression should be evaluated as $$E(t|\mathcal{F}_{s}^{W}) = t$$ due to the reasoning in the question. Sorry for the noise. 2 It is a Wiener integral as your integrand is a deterministic function of time. It is known that the Wiener integral is stationary gaussian process with independent increments. So z(t) \sim \mathcal N\left(0, \int_0^te^{-2k(t-s) }~ds\right) and (z(t)-z(s)) \amalg z(u), \ \forall u,s,t \in \mathbb R_+ \text{ such that }u\leq s, s\leq t  or alternatively ... 2 \begin{align*} E\Big(W_t^3-3tW_t \mid \mathcal{F}_s\Big) &= E\Big((W_t-W_s+W_s)^3-3t(W_t-W_s+W_s) \mid \mathcal{F}_s\Big) \\ &=E\Big((W_t-W_s)^3+W_s^3+3(W_t-W_s)^2W_s + 3 (W_t-W_s)W_s^2\\ &\qquad \qquad -3t(W_t-W_s)-3tW_s \mid \mathcal{F}_s\Big) \\ &=E\Big((W_t-W_s)^3\Big) + W_s^3+3W_sE\Big((W_t-W_s)^2\Big)\\ &\qquad \qquad + 3W_s^2 ... 2 I am not absolutely sure what you mean by diffusion-jump but if you mean jump-diffusion. Here are some references: Chapter 15, Concepts and Practice of Mathematical Finance, Joshi Cont and Tankov, Financial Modelling with Jump Processes Using Monte Carlo Simulation and Importance Sampling to Rapidly Obtain Jump-Diffusion Prices of Continuous Barrier ... 2 The concept of 'mean reversion' is tricky in continuous time. Most people would call 'mean reverting' a process where the drift pulls back towards a long run mean, and I assume that this is what you also mean. Something like the drift of an OU process. However, in continuous time the 'pull' can be generated by the volatility. For example the process dX_t ...

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