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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

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 $y$ with $$h_(x) = \sum_0^n \theta_i x_i = ...


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 ...


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

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

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

If you allow $X_t$ to be two dimensional then a model with a stock price $X_t^1$ and its variance process $X_t^2$ (stochastic volatility) would fit your definition. In such cases to my knowledge we often don't have a closed form of the density of $X_T^1$ but in some cases we have a closed form of the Laplace transform. An example is the Heston model.


3

Note that you can understand the $\Delta$ as an "operator" acting on $r$. So just act on $r$ twice: $$\Delta^2 r_t = r_t - 2 r_{t-1} + r_{t-2}. $$ In fact if you write the $r$ as a vector, $r = (r_1, r_2, \ldots, r_N)$, then $\Delta$ is an $N\times N$ matrix with elements $\Delta_{i,j} = \delta_{i,j} - \delta_{i-1,j}$. The AR(2) model can be written as ...


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

As @Rustam notes, "correlation" of deterministic functions in the sense you describe is a special case of allowing $\mu$ and $\sigma$ to have a term structure of arbitrary shape. Since the latter is easy to treat, no one bothers with restricted forms of it. Now, there quite a few people who deal with models that let $\sigma$ change with $S$. I am thinking ...


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 ...


2

This book goes through exactly this problem in quite detail (with C++ codes included). I've worked through it in the past, but can't sum it up off the top of my head.


2

Since the variance gamma process can actually be expressed as the difference of two gamma processes, the parameters are quite easy to estimate. Taking the mean (rate) and variance (rate) of the positive values and negatives will give you the variables necessary to estimate the total variance gamma process parameters. They are described in a more recent ...


2

Saying that you can't analyze something as is does not make it garbage. You can't eat flour "as-is", but that doesn't mean you throw it out. In order to use "standard" analysis tools, you must first transform the series into something compatible. Some examples of such a transformation include k-th order differences or a log transformation. These ...


2

This is the answer to the first version of the question which asked whether a stationary process has an increasing variance over time. No the definition of (weakly) stationary (http://en.wikipedia.org/wiki/Stationary_process) is that the variance is the same for each point in time. In the literature it is often dealt with the covariance function. For ...


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

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 $\lambda_1(s)+\lambda_2(s)$.


2

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 by $$P(t,T)=A(t,T)e^{-r(t)B(t,T)}$$ you can find the exact formulas for $A(t,T)$ and $B(t,T)$ in this document (or just read ...


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 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 $$ ...


1

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.


1

The question is not 100% clear. If you set $X = W_t-W_s$ where $t-s = 1$ then this is equal in distribution to $W_1-W_0$ and the defining property of Brownian motion is that increments are normally distributed. In the general case $W_t-W_s$ is $N(0,t-s)$, where the second parameter is variance. If you set $dW_t = W_{t+dt}-W_t = Z \sqrt{dt}$, where ...


1

adam I still think that your question is a bit vague but perhaps the following will be of some help to you. First of all ItĂ´'s theorem is a tool. It will never give you the price by itself. While working out the concrete formula one might end up using it in one context or another. In case of a european option, a borel measurable function $h$ and $X_t$ ...


1

you hypothesize that your data is generated by the following process: $y_t=\phi_0+\sum_{k=1}^P\phi_ky_{t-k}+\varepsilon_t$, where $\phi_k$ are your autocorrelation coefficients, and $\varepsilon_T$ - random errors. Next, you estimate your $\phi_t$ using one of the methods of estimation of autoregressive processes AR(P) of order P, e.g. see AR(P), there's no ...


1

1.) Autocorrelation is the correlation of a time series against the lagged version of itself. 2). First autocorrelation is the correlation of the time series against the lag(1) version of itself. Let's look at the example below Period_Numbers = [1,2,3,4,5,6,7,8,9,10] Time_Series = [10, 20, 30, 40, 50, 60, 70, 80, 90, 100] First Autocorrelation is ...


1

Any of a wide variety of local vol models, where (from your equation) $b(\cdot,\cdot)$ is some fitted surface, are unlikely to have closed-form solutions for the terminal distribution. Indeed it's well-known that these models tend to have very unusual forward term structures of volatility. As a specific example, take $b(\cdot,\cdot)$ to be an approximation ...


1

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 ...


1

You can check the wikipedia page to find out "the the basic model assumptions" for the a stationary random process, and I assume "the correct reasoning on relationships" are the model that describe a random process. But intuitively speaking, if the data are sampled from a stationary random process, then you can predict the future by deductively extrapolate ...


1

In log space the explicit solution for the density of the first passage time is the Inverse Gaussian Distribution. See, e.g., http://www.springerreference.com/docs/html/chapterdbid/205395.html or the Wikipedia page for the distribution. The only thing that should matter is the interval from the initial state to the threshold, and that is the parameter "a" ...



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