# 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 ...

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 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 ... 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 $$... 3 I would say Take log of first equation to get rid of dependence on x_t Apply Kalman filter equations to estimate parameters I believe Conrad and Kaul (1988) J of Business do exactly what you describe. 3 Just a bit of illustration added to @John's answer. Look at log prices \log(P_t), assume that you know P_0 then$$ \log(P_t) = \log(P_0) + r_1 + \cdots r_t $$where r_i = \log(P_i)-\log(P_{i-1}) are the log returns. By modelling the log-returns (which as already said take values on the whole real line which is a nice property for modelling) we model ... 3 Let's consider the following example: the process is initialized randomly with \pm1 and then stays there forever. Seems stationary to me, but it would never cross its mean. 3 The more phenomenological definitions in his books are probably more helpful. Whether one uses the fractal dimension, Hurst coefficient, or exponential coefficient alpha, there is a value that corresponds to pure Brownian motion, a regime relative to this value that corresponds to persistence of motion, and the opposite regime that corresponds to ... 3 EDIT: My reasoning below seems to be wrong. The process as you write it tends to infinity if a is big enough and positive and if \lambda_0 is positive. I would not call this process non-meanreverting OU. It is just an Ito process of a simple form. If we remove the stochastic part then we get$$ d\lambda_t = a \lambda_t dt $$with the solution (if ... 3 I think all they are doing is integrating and estimating$$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. 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\\ ... 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 If we are going to have the form \begin{align*} dr = A dt + BdW_t, \end{align*} Then both A and B are functions oft$and$r_t$, otherwise,$r_tis normal. However, note that \begin{align*} r_t = \exp\Bigg(\frac{1}{\sigma(t)}\bigg(\int_0^t \theta(s)\sigma(s) ds +\sigma(0)\ln r_0 + \int_0^t\sigma^2(s) dW_s\bigg)\Bigg). \end{align*} That is,r_t$is ... 2 Exhibiting a counter-example is straight-forward enough. For example, let$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 ...

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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

Because you can hedge. Once you have delta hedged, the pay-off is symmetric about up and down moves so drift doesn't matter. Also the delta-hedged call and the delta hedged put have to have the same value since they have the same pay-off. (Put-call parity) Yet any argument that the call should be worth more because of drift says that the put should be ...

2

Yes and No. In the absence of arbitragers, the price of the option will be different for each speculator based on their drift expectations (and each speculator has a risk in his position and will limit his ability to trade large sizes to avoid bankruptcy) and the option price will converge to priced off a supply-and-demand driven drift expectation. ...

2

I think to gain intution you have to understand that the same agents that value the stocks will value the options. And agents compensate for volatility by demanding higher expected returns. Therefore you should ask: Why are stocks priced as they are in the first place? In your example, the stock with higher volatility has much lower expected return. This ...

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