# Tag Info

## Hot answers tagged stochastic-calculus

26

A stationary process is one where the mean and variance don't change over time. This is technically "second order stationarity" or "weak stationarity", but it is also commonly the meaning when seen in literature. In first order stationarity, the distribution of $(X_{t+1}, ..., X_{t+k})$ is the same as $(X_{1}, ..., X_{k})$ for all values of $(t, k)$. ...

18

This is pure speculation: MFE's are really tailored toward valuation models (how can we develop a model to price x swap, etc.). You don't entirely have to worry about those details in order to trade them: you're just quoted a price based on these models. But if you go in-house at a bank and are working as a product quant (structured products, etc.), then ...

12

My understanding is because the Ito's integration definition keeps the martingale property. With Brownian motion $W(t, \omega)$ defined, to define stochastic integration in a Riemann–Stieltjes style: $$\int_0^t f(t, \omega) d W(t, \omega) = \lim_{\| \Delta_n\| \to 0 } \sum_{i=1}^{n} f(\tau_i,\omega) \left ( W(t_i, \omega) - W(t_{i-1}, \omega) \right )$$ , ...

10

Baxter and Rennie say it better than me, so I will summarize them. Suppose that $N_t$ is not stochastic and $f(.)$ is a smooth function then the Taylor expansion is $$df(N_t) = f'(N_t)dN_t + \frac{1}{2}f''(N_t)(dN_t)^2 + \frac{1}{3!} f'''(N_t)(dN_t)^3 + \ldots$$ and the term $(dN_T)^2$ and higher terms are zero. Ito showed that this is not the case in the ...

10

In fact Ito and Stratonovich calculus are both mathematically equivalent. In the following paper you can e.g. see that both derivations lead to the same result, i.e. the Black-Scholes equation: Black-Scholes option pricing within Ito and Stratonovich conventions by J. Perello, J. M. Porra, M. Montero and J. Masoliver From the abstract: Options ...

10

These are all examples on Ito Formula in its general form (with quadratic variations):

9

There are many numerical approaches to solving stochastic integrals such as the above. Assuming that there is no closed form slight-of-hand, the easiest approach is the Monte Carlo approach. I would recommend referring to Glasserman's excellent "Monte Carlo Methods in Financial Engineering" If you are not familiar with MC, think of it as evaluating ...

9

I think this question has no easy answer but I'll give it a shot anyway (beware: oversimplification ahead!). The main idea of the Malliavin calculus is to be able to differentiate stochastic processes like Brownian motion (or more general martingales with bounded quadratic variation), which are not differentiable in the traditional sense (because of their ...

8

(1) You analytically solve a stochastic differential equation (SDE) using Ito's lemma. Your second equation (the discretized one) is how you could model one path over one step. To find the solution, you would model many of these paths over many steps and then take the expectation (i.e., Monte Carlo methods). The solution to the SDE models all of these paths ...

8

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 $E[W_pW_qW_r] = 0$.

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 $$... 7 A very excellent discussion of stationarity as it relates to trading can be found in Sherry's (Sherrys'?) Mathematics of Technical Analysis (poorly organized, but very useful book). As he puts it, if the price changes of a stock, etc., are stationary over a time period, the underlying rules generating the price changes are effectively unchanged. The ... 7 A process is defined here and is simply a collection of random variables indexed (in general) by time. Otherwise I know the concept stated by Shane under the name of "weak stationarity", strong stationary processes are those that have probability laws that do not evolve through time. More formally let X_t be a given process, then let's call P_X the ... 7 This is the separable differential equation for simple continuous compounding! See this very accessible article for a step-by-step derivation (esp. under continuous compounding): http://plus.maths.org/content/have-we-caught-your-interest 7 The classic argument using risk-neutral pricing is to assume that discounted stock prices are \tilde{P}-martingales where \tilde{P} is the risk-neutral probability measure. Then, you know that$$\frac{S_t}{(1+r)^t}=\tilde{E}[\frac{S_T}{(1+r)^T} | \mathcal{F}_t]$$by definition of a martingale process. As the discounts are non-stochastic, you can ... 7 The part where you say that$$\frac{dS_t}{S_t} = d\ln(S_t)$$is wrong, because S is a stochastic variable. This is exactly what Itô tells you with his formula that you apply right do compute your dZ. The difference comes from the quadratic variation of the process S which you express as (dS)^2. If you don't add this term when the variable are ... 7 It's a lemma! Ito's Lemma gives the change of coordinates rule for stochastic calculus. The multiplication rule is a shorthand way of expressing it. 7 Stochastics are usually applied in the field of derivatives pricing. In this setting the task is to price a derivative such that it fits into the landscape of tradable instruments (no-arbitrage). We work using the risk-neutral measure - usually denoted by Q. The measure is derived from other traded instruments. In risk analysis (e.g. calculate the VaR, ES ... 7 A stochastic differential equation is nothing more than a short-hand notation for a corresponding integral equation. So the initial SDE you provided actually means$$ \int_0^t d S_u = \int_0^t \mu(S_u, u) du + \int_0^t\sigma(S_u, u) dW_u$$This is how the SDE is defined (see e.g. here). The reason is that you cannot differentiate a Brownian motion. It does ... 6 The model for the stock is the Bachelier model with the solution$$ S(t) = S(0) + \sigma W(t) $$Thus the law of the stock S(t) is Gaussian with mean S(0) and variance \sigma^2 t. For average process Z(T) is thus the average of linear Brownian motion, we can rewrite this as$$ Z(T) = \frac{1}{T} \int_0^T S(0) + \sigma W(t) dt = S(0) + ...

6

I like Richard's answer, but I think we can compute the mean and the variance of $\int_0^T W_t dt$ by ourselves using Ito's lemma. Let $f(W_t, t) = t W_t$. $$d( t W_t ) = W_t dt + t dW_t .$$ Integrating both sides, and re-arranging the terms, we get $$\int_0^T W_t dt = T W_T - \int_0^T t dW_t \, .$$ We'll be using Ito's isometry formula $\mathbb{E} ... 6 If$\alpha(t)$is of finite variation, then the product rule is the same as in ordinary calculus: $$d(\alpha(t)X_t) = \alpha(t) dX_t + X_t d\alpha(t).$$ If you had$X_t$and$Y_t$as processes, you would get $$d(X_t Y_t) = X_t dY_t + Y_t dX_t + d [X,Y]_t.$$ If$Y$has finite variation, the last quadratic covariation term is zero. The second equation ... 6 Okay so I'll take Jase answer and format it properly so that it answers your question and it will be useful for users in the future. For clarity, let me restate the dynamics of the Modified Ornstein-Uhlenbeck model using the more common notation: $$dS_t = \theta (\mu-S_t)dt + \sigma S_t dW_t$$ This blog post provides a closed form solution: $$S_t = S_0 ... 6 If you consider X_1 a random variable which is normally distributed with mean \mu and variance \sigma^2 them S_1 = \exp(X_1) is log-normally distributed with mean \exp(\mu + \sigma^2/2) and variance (\exp(\sigma^2)-1)\exp(2\mu+\sigma^2). This follows from the definitions of the normal distribution and the log-normal distribution and deriving the ... 6 The convexity of the exponential function of the stochastic variable W makes its expectation greater than the exponentiation of the expectation of W. This is an example of Jensen's inequality, E[e^{\sigma W}]> e^{\sigma E[W]}=1. \sigma can be interpreted as the magnitude of the convexity of the exponential function. This can be seen by Taylor ... 6 Shreve's theorem also called "Girsanov II" indeed represents a special case of the general "Girsanov I" from Wiki above, with$$Y_t:=W_t,X_t:=-\int_0^t\Theta_udW_u$$We can show:$$[Y,X]=-\int_0^t\Theta_udu$$by using general Stochastic Calculus rules (e.g. p.37, 6.6 here): ... 6 The trick is to start with the highest power, rewrite it as something you know (a third order moment) and then work backwards on the remaining terms. By that I mean you can complete the cube as follows:$$E[W_t^3 - 3tW_t|\mathcal{F}_s] = E[(W_t-W_s)^3 - C -3tW_t|\mathcal{F}_s]$$where you'll need to find C such that the equality holds (i.e. C=W_s^3 + ... 6 You derivation here is flawed because you are deriving with respect to two processes and you do not take into account that the variable W_t is stochastic and hence S_t is as well. So, to derive S_t from dS_t, you have to apply Ito's Lemma, see this question for details. This is the "classic" way you see it. If you want to do it the other way ... 6 In the integral$$\int_0^t S_u dW^{*}_u \, ,$dW^{*}_u \equiv W^{*}_{u+du} - W^{*}_u$is independent from the integrand$S_u$. So,$\mathbb{E}\left[ \int_0^t S_u dW^{*}_u\middle\vert \mathcal{F}_0\right] = \int_0^t \mathbb{E}\left[S_u \middle\vert \mathcal{F}_0\right]\mathbb{E}\left[dW^{*}_u\middle\vert \mathcal{F}_0\right] = 0\$, since ...

6

As @SRKX suggested, full answers are provided below. (a). Using Ito's lemma \begin{align*} d\left(W_t^3\right) &= 3W_t^2 dW_t +3W_t dt. \end{align*} Integrating on both sides, we obtain that \begin{align*} W_t^3 = 3\int_0^t W_s^2 dW_s +3\int_0^t W_s ds. \end{align*} Consequently, \begin{align*} \int_0^t W_s^2 dW_s = \frac{1}{3}W_t^3 -\int_0^t W_s ...

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