# 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] $$and that implies that E[W_pW_qW_r] = 0. 6 The trick is to start with the highest power, rewrite it as something you know (a third order moment) and then work backwards on the remaining terms. By that I mean you can complete the cube as follows:$$E[W_t^3 - 3tW_t|\mathcal{F}_s] = E[(W_t-W_s)^3 - C -3tW_t|\mathcal{F}_s]$$where you'll need to find C such that the equality holds (i.e. C=W_s^3 + ... 6 Basically, prices usually have a unit root, while returns can be assumed to be stationary. This is also called order of integration, a unit root means integrated of order 1, I(1), while stationary is order 0, I(0). Time series that are stationary have a lot of convenient properties for analysis. When a time series is non-stationary, then that means the ... 5$$ \textbf{Preface} $$I am assuming log normal asset but this is not clear from the question? Or rather I have misinterpreted the question! Well as I see it from a a purely mathematical exercise$$ d\left(\dfrac{S_t}{M_t}\right) =\frac{1}{M_t}dS_t - \frac{S_t}{M_t^2}dM_t +O(dt^2) using Ito's lemma. Then we can sub in the original processes yields ... 5 An AR(1), once the time series and lags are aligned and everything is set-up, is in fact a standard regression problem. Let's look, for simplicity sake, at a "standard" regression problem. I will try to draw some conclusions from there. Let's say we want to run a linear regression where we want to approximate y withh_(x) = \sum_0^n \theta_i x_i = ...

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 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 ...

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.

4

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 ...

4

Note that $$P(X_i >s)= \exp\Big(-\int_0^s \lambda_i(u) du \Big),$$ for $i=1, 2$. Then, $$P(\min(X_1, X_2) >s) = P((X_1>s)\cap (X_2>s)) = P(X_1>s)P(X_2>s) = \exp\Big(-\int_0^s (\lambda_1(u)+\lambda_2(u)) du \Big).$$ That is, the hazard function for $\min(X_1, X_2)$ is $\lambda_1(s)+\lambda_2(s)$. Alternatively, note that $$\lambda_i(s) = ... 4 There is a lot of ways to understand why stationarity allows to apply usual time series analysis. Here is one more. Very often, the theoretical justification of what you do in time series need to be able to identify the mean formula and the expectation:$$\frac{1}{N}\sum_{n=1}^N X_n \underset{N\rightarrow +\infty}{\longrightarrow} \mathbb{E} X, $$where the ... 4 You have typo "vol^2", but it should be "vol". Its$$\sqrt{\sigma^2T}=\sigma\sqrt{T}$$4 Perhaps overly simplistic and repeating the pt above, but when doing statistics, ideally we want to compare like with like. Returns can be comparable with each other. Prices on the other hand always depend on the previous price. 3 Consider an (arithmetic) Ornstein-Uhlenbeck process as a model of the asset price X_t:$$dX_t = \kappa(\mu-S_t)dt + \sigma dW_t$$where \mu is the mean-reversion level, \sigma is a volatility parameter, W_t is Brownian motion, and \kappa is the reversion speed. An Ornstein-Uhlenbeck process will revert to the mean infinitely often if \kappa ... 3 In general SDE's are defined on a probability space which consists of a triplet (\Omega, P, B): the space \Omega, a probability measure P, and a sigma algebra B. In short, the sigma algebra consist on the set of all events that we can assign probability to. For SDE's driven by Brownian Motion this probability space is the so called Wiener space, ... 3 EQ1 is uni-variate case. EQ2 is multivariate case, in which you have to use correlated X_t. His way of doing is making Y_t independent so that you can simulate freely. He does so by finding PC on \Delta. Alternatively, you could generate correlated X_t in your simulation. To benchmark your model / code, you should first test and reproduce a given ... 3 Just a bit of illustration added to @John's answer. Look at log prices \log(P_t), assume that you know P_0 then$$ \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 ...

2

You can write it as $$\left(\begin{array}{c}dY_t\\ dX_t\end{array} \right) = \left(\begin{array}{cc}\alpha(X_t, Y_t)& \beta(X_t,Y_t)\\ 1 & 0\end{array} \right)\cdot \left(\begin{array}{c}dW_t\\ dZ_t\end{array} \right)$$ and check Platen's conditions (Lipschitz?) as Richard pointed out on the matrix perhaps? If it is $$... 2 You should look at confidence interval. Normally, your confidence interval size is proportional to the standard deviation, looking something like: with probability p your value will be in the interval:$$[\bar{S} - k*StdDev, \bar{S} + k*StdDev]$$Then, getting back to your simulation, we can say that your time step is very big (1 year) and you simulate ... 2 First of all, a filtration ( \mathscr{F}_t )_{t \geq 0 } is a "set" of sigma algebras indexed usually by time t that are increasing. That is, for every t>0, \mathscr{F}_t is a sigma algebra and \mathscr{F}_t \subseteq \mathscr{F}_T for all 0\leq t \leq T. The canonical example, is the filtration generated by a process, say Brownian Motion W: ... 2 You have$$\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}$$, which holds only for a Brownian motion W (and M_t martingale). So one cannot trivially replace W_t and W_t+\int\Theta(u)du=\widetilde{W}_t in 5.3.2 aswell by ... 2 Basically, Black-Scholes is an "industry standard" formula. It is widely used by practitioners and usually augmented with extra specifications or intuition. It has a closed form solution, which is rare in option pricing models. It is also relative to simple to understand. Otherwise, you usually need to rely on Monte Carlo simulation or some other way. And ... 2 I think you are on the right track here. You made a sign error in the first line, unfortunately:$$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

\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 ...

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