# 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 $$... 6 We know that (\tilde{W}_t) := (-W_t) is also a Wiener process so$$ 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 + ... 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

The convexity of the exponential function of the stochastic variable $W$ makes its expectation greater than the exponentiation of the expectation of $W$. This is an example of Jensen's inequality, $E[e^{\sigma W}]> e^{\sigma E[W]}=1$. $\sigma$ can be interpreted as the magnitude of the convexity of the exponential function. This can be seen by Taylor ...

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 ...

4

Orthogonality and independence are different concepts. The concepts are the same for Wiener processes because in the context of normal random variables, independence is equivalent to orthogonality (i.e. uncorrelatedness) Independence is the standard definition for probability. Let $\mathcal{F}, \mathcal{G}$ be the sigma algebras generated by two ...

4

For a basic introduction, the three chapters in Hull's Options, Futures, and Other Derivatives on Binomial Trees, Wiener Processes and Ito's Lemma, and The Black-Scholes-Merton Model helped me start to understand the basic concepts within a broader context. After that, Shreve's two books seems to be pretty popular (see here and here). He explains things ...

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 $$\lambda_i(s) = ... 4 You have typo "vol^2", but it should be "vol". Its$$\sqrt{\sigma^2T}=\sigma\sqrt{T}$$4 If at first you don't have a model at all, then geometric Brownian motion is not bad. As others before me said: log-returns are normally distributed in this model. This is debatable and there are times and markets where this is not true. There is more than enough research about this. But why is a model based on Brownian motion not that bad? The reason is ... 3 Consider an (arithmetic) Ornstein-Uhlenbeck process as a model of the asset price X_t:$$dX_t = \kappa(\mu-S_t)dt + \sigma dW_t$$where \mu is the mean-reversion level, \sigma is a volatility parameter, W_t is Brownian motion, and \kappa is the reversion speed. An Ornstein-Uhlenbeck process will revert to the mean infinitely often if \kappa ... 3 EQ1 is uni-variate case. EQ2 is multivariate case, in which you have to use correlated X_t. His way of doing is making Y_t independent so that you can simulate freely. He does so by finding PC on \Delta. Alternatively, you could generate correlated X_t in your simulation. To benchmark your model / code, you should first test and reproduce a given ... 3 A key property of Brownian motion is independent increments. So if x-1 > y, then$$ \mathbb{E}[\Delta W_x \Delta W_y] = 0 because the time intervals [x-1,x] and [y-1,y] do not overlap. If they do overlap, i.e. x-1 \leq y < x, then \begin{align} \mathbb{E}[\Delta W_x \Delta W_y] =&\ \mathbb{E}[(W_x - W_{x-1}) (W_y-W_{y-1})] \\ =&\ ... 3 Bond Price Dynamics I do not know the source of the bond dynamics you show above but seeing how we are dealing with an affine model there is a very elegant way to derive those. Due to the model being affine the bond price is given byP(t,T)=A(t,T)e^{-r(t)B(t,T)}$$you can find the exact formulas for A(t,T) and B(t,T) in this document (or just read ... 3 I will try to answer this a bit differently. The rigorous answer: because Ito calculus tells us that we need the second order term. Look at$$ S_t = S_0\exp(\mu t + \sigma B_t). $$Assume that S_0 is known and fixed and look at by Ito's formula$$ d(S_t/S_0) = \mu dt + \sigma B_t + \frac{\sigma^2}{2} dt. $$Then with some abuse of notation:$$ ...

3

This is a good shorter reference: http://www.impan.pl/CZM/tankov.pdf. Cont and Tankov have also written a longer book about modelling with Levy processes that I think is really good. There's going to be a strong connection between the sequence of jump times and the Levy measure $\nu$. In a single unit of time, $\nu(dx)$ is a measure (not necessarily a ...

3

Autocorrelation is the correlation of a series with itself. Suppose $X = {X_1, X_2, X_3, ...}$ is your time series. Then the autocorrelation between $X_t$ amd $X_s$ is: $$\frac{E[(X_t-\mu_t)(X_s-\mu_s)]}{\sigma_t \sigma_s}$$ This can be simplified quite a lot if the series you have is stationary (a common assumption), in which case the autocorrelation ...

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

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 ...

3

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 ...

2

To complement @SRKX comment ,i'll try to explain the "simple mathematical proof" beetween both formula : I assume you know the geometric or arithmetic brownian motion : Geometric: \begin{equation*} dS = \mu S dt + \sigma Sdz \end{equation*} Arithmetic : \begin{equation*} dS = \mu dt + \sigma dz \end{equation*} Then another important stochastic tool you ...

2

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 ...

2

I think this question might be asking for the central limit theorem. If we consider a process W which varies as a series of independent random steps, then the Central Limit Theorem tells us that after many steps, the value of W will be normally distributed.

2

This interesting question provides excellent links to Dynamic Nelson-Siegel Term Structure Models for interest rates for No Arbitrage and exposes key formulation in an interesting way. Appendix in p37 of ssrn link says $\lambda$ is market price of diffusion risk. However, in the DNS model the $\lambda$ is eigenvalues of $\kappa$, which then part of ...

2

$$S_t = S_0\exp((r-\frac{\sigma^2}{2})t+\sigma W_t)$$ is not yet a martingale for it is not dirftless. From a probabilistic point of vew the "drift adjustment" comes into play so that the expected value of $S_t$ will be $e^{rt}$ rathern than $e^{(r+0.5\sigma^2)t}$. For the expected value of a log-normaly distributed variable with mean $\mu$ and vol ...

2

You are asking an interesting question. Firstly, a Submartingale has increasing or equal expectation (not decreasing). Secondly, the process $dX_t=X_tdW_t$ is a true martingale (not strictly local), since its solution (by Ito): $$X_t=X_0e^{W_t-\frac{t}{2}}$$ has $E(X_t)=X_0$ constant expectation ($e^{-\frac{t}{2}}E(e^{W_t})=1, W_t\sim N(0,t)$). The ...

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]$$Then, getting back to your simulation, we can say that your time step is very big (1 year) and you simulate ... 2 You have$$\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 ...

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