# Tag Info

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 ...

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 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 ...

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 ... 4 I think all they are doing is integrating and estimating $$P(|W_t| \leq 2) = \int_{-2}^{2} \frac{d}{dr} P(W_t \leq r) dr$$ so $$P(|W_t| \leq 2) \leq 4 \sup \limits_{r \in [-2,2]} \frac{d}{dr}P(W_t \leq r)$$ The normal density is maximal at zero and we are done. 4 This is wrong! Notice that$dX_t=\mu(t,X_t)dt + \sigma(t,X_t)dW$is a shorthand for $$\int_0^tdX_s = \int_0^t \mu(s,X_s)ds + \int_0^t\sigma(s,X_s)dW_s$$ Integrating: $$X_t-X_0 = \int_0^t \mu(s,X_s)ds + \int_0^t\sigma(s,X_s)dW_s \text{ (eq.1)}$$ If we take expectations, remembering that$\mathbb{E}[\int_0^t\sigma(s,X_s)dW_s]=0$, we have ... 4 The first process $$B_{t+dt} = B_t + Z$$ where$Z$is independent of$(B_s)_{s \le t}$and follows a Gaussian distribution with mean$0$and varince$dt$is a standard Brownian motion (thus the variance of$B_t$is$t$). For the second process let us recall the definition from your link: $$E[B^H_t B^H_s] = \frac12 ( t^{2H} + s^{2H} - |t-s|^{2H}),$$ thus ... 4 The first process is a BM. The second does not exist in continuous time. The variance goes down too slowly with dt and the process blows up at the limit. You can break the (0,1) interval into 1, 100, 1000, 1000000 steps and see that happening. Variance of a martingale has to scale with dt: if it is too fast then the process dies, if it is too slow then ... 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 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 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 $$... 3 Just a bit of illustration added to @John's answer. Look at log prices \log(P_t), assume that you know P_0 then$$ \log(P_t) = \log(P_0) + r_1 + \cdots r_t $$where r_i = \log(P_i)-\log(P_{i-1}) are the log returns. By modelling the log-returns (which as already said take values on the whole real line which is a nice property for modelling) we model ... 3 EDIT: My reasoning below seems to be wrong. The process as you write it tends to infinity if a is big enough and positive and if \lambda_0 is positive. I would not call this process non-meanreverting OU. It is just an Ito process of a simple form. If we remove the stochastic part then we get$$ d\lambda_t = a \lambda_t dt with the solution (if ... 3 Not sure about the correctness of the first approach, but second approach uses 1 /\sqrt k to scale the variance of the total sum by k. So the difference of two processes (say W_t and W_{t+\Delta t}) generated by the random walk would have a variation of \Delta t, which satisfies one of conditions needed to get a Wiener's process. 3 To solve for U_t, we can proceed as follows. First, note that \begin{align*} d\left(e^{(\theta + \frac{1}{2}\xi^2)t - \xi W_t} U_t \right) &= e^{(\theta + \frac{1}{2}\xi^2)t - \xi W_t} U_t \left((\theta+\xi^2) dt -\xi dW_t\right) \\ &\qquad+ e^{(\theta + \frac{1}{2}\xi^2)t - \xi W_t} dU_t -\xi^2e^{(\theta + \frac{1}{2}\xi^2)t - \xi W_t} U_t dt\\ ... 3 If you want to address interesting problems that are interesting for financial mathematics, I do not believe you have the good list. Pricing. For instance, most of explicit formulas for pricing that are not available yet will never be. In this direction, you should have a look at simulation techniques. See for instance Nonlinear Option Pricing. Interesting ... 3 Your notations are really hard to follow as you define \mathbb{P} twice at the beginning. The notation \mathbb{P} = \mathbb{\hat{P}} and \mathbb{P} =\mathbb{\tilde{P}} is not meaningful as the probability measure \mathbb{P} is already fixed and used for the real world probability measure. I think that this is the reason why you are getting confused. ... 3 Given efficient markets, asset prices should be unpredictable in the sense that any upcoming returns are uncorrelated with current or past returns. Hence for traded assets the price should follow something more similar to a GBM than an O-U process. However, many financial metrics are not prices; for example interest rates or volatility. O-U processes may ... 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 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 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 ...

Only top voted, non community-wiki answers of a minimum length are eligible