# Tag Info

## Hot answers tagged brownian-motion

51

This type of integral has appeared so many times and in so many places; for example, here, here and here. Basically, for each sample $\omega$, we can treat $\int_0^t W_s ds$ as a Riemann integral. Moreover, note that \begin{align*} d(tW_t) = W_t dt + tdW_t. \end{align*} Therefore, \begin{align*} \int_0^t W_s ds &= tW_t -\int_0^t sdW_s \tag{1}\\ &= \...

19

Using the Ito Formula The general approach that often works for these kinds of question is to search for functions such that their Ito differential contains the terms that we are interested in. In your case, we are looking for a function $f(t, x)$ such that $f_t(t, x) = t x$. Let \begin{equation} f(t, x) = \frac{1}{2} t^2 x \end{equation} with \begin{...

18

Here is a short list (to be edited and improved - community wiki) : Standard brownian motion (also called Wiener process) for which: $d\, W_t \sim \mathcal N(0, \sqrt{d t})$ Geometric brownian motion, used in the Black-Scholes model (1973): $d\,X_t = \mu X_t\,dt + \sigma X_t\,dW_t$ Constant elasticity of variance ("CEV") model (1975): $d\,X_t=\mu X_t dt + \... 18 Brownian motion is simply the limit of a scaled (discrete-time) random walk and thus a natural candidate to use. It is very intuitive and arguably one of the simplest and best understood time-continuous stochastic processes. Also, don't forget that you obtain many more stochastic processes as functions of a (time-changed) Brownian motion. In many books on ... 16 Baxter and Rennie say it better than me, so I will summarize them. Suppose that$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 ... 14 Martingales + Markovian Here is the motivation. Conditional expectations are martingales by the tower property of conditional expectations (an easy exercise to show). Suppose$r=0$, by the risk neutral pricing theorem$E^\star\left[h(X_T)\bigg|\mathscr{F}_t,\,X_t=x\right]$is the price of any derivative security with$X$as the underlying asset and payoff ... 13 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 ... 12 Yes, you need Cholesky factorization. You can find the general idea here: http://www.goddardconsulting.ca/option-pricing-monte-carlo-basket.html Plus the implementation in MATLAB here: http://www.goddardconsulting.ca/matlab-monte-carlo-assetpaths-corr.html The code in general should be easily translatable. The only difficulty is the Cholesky factorization ... 12 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 ... 12 Quadratic variation and variance are two different concepts. Let$X $be an Ito process and$t\geq 0$. Variance of$X_t$is a deterministic quantity where as quadratic variation at time$t $that you denoted by$[X,X]_t $is a random variable. What is confusing you is the fact that when$X $is a martingale then$X^2_t-[X,X]_tis a martingale thus you ... 12 Another approach consists in using the Fubini theorem to write that \begin{align} \int_0^T u W_u du &= \int_0^T \int_0^u u\, dW_v\, du \tag{W_u = \int_0^u dW_v} \\ &= \int_0^T \int_v^T u\, du\, dW_v \tag{Fubini}\\ &= \frac{1}{2}\int_0^T (T^2 - v^2) dW_v \end{align} This is an Itô integral. Since the integrand ... 11 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 thatS_0is 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: E[d(S_t/... 11 Here's my favorite example of an intraday strategy on S&P500 futures that at least used to work: Intraday Share Price Volatility and Leveraged ETF Rebalancing I pull it out whenever people start talking about market efficiency. The strategy is very simple: if S&P500 futures are up or down more than 2% on the day with two hours left until close, ... 11 By construction, the Itô integral, I_t=\int_0^t X_s\text{d}W_s, is a martingale if \int_0^t \mathbb{E}[X_s^2]\text{d}s<\infty. The martingale property, \mathbb{E}_s[I_t]=I_s implies \mathbb{E}[I_t]=I_0=0. Because W_s\overset{d}{=}\sqrt{s}Z, where Z\sim N(0,1), we indeed have \begin{align*} \int_0^t\mathbb{E}\left[\frac{1}{(1+W_s^2)^2}\right]\... 11 You need to rotate them so we can find some orthogonal axes. A simple way to think about this is by remembering that we can decompose the second of two brownian motions into a sum of the first brownian and an independent component, using the expression \begin{align} W_{t,2} = \rho_{12} W_{t,1} + \sqrt{1-\rho_{12}^2} \tilde{W}_{t,2} \end{align} where \tilde{... 11 Besides @StackG's splendid answer, I would like to offer an answer that is based on the notion that the multivariate Brownian motion is of course multivariate normally distributed, and on its moment generating function. We know that \mathbb{E}\left(W_{i,t}W_{j,t}\right)=\rho_{i,j}t i.e. an N-dimensional vector X of correlated Brownian motions has ... 10 Just to add to the already nice answers, the result can also be obtained using the (stochastic) Fubini theorem. \begin{align} \int_0^t W_s ds &= \int_0^t \int_0^s dW_u\, ds \tag{W_s=\int_0^s dW_u}\\ &= \int_0^t \int_u^t ds\,dW_u \tag{Fubini} \\ &= \int_0^t (t-u) dW_u \tag{\int_u^t ds = t-u } \end{align} And we fall back on the same equation ... 10 Physical objects move according to simple smooth curves that can be represented by low order polynomials: a straight line, a parabola, an ellipse, etc. Financial market prices move in a completely different way, as can be seen by looking at any graph of stock prices, interest rates etc. in a newspaper: there are constant, erratic fluctuations, sometimes in ... 9 The Feynman-Kac theorem primarily makes sense in a pricing context. If you know that some function solves the Feynman-Kac equation you can represent it's soluation as an Expectation with respect to the process. (confer this document) On the other hand a pricing function solves the FK-PDE. Thus often one would try solving the PDE to get a closed form ... 9 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 ... 9 First you need to correct the formula to: W_t^2 = \rho W_t^1 + \sqrt{1-\rho^2} Z_t, $$where Z_t is a BM independent of W_t^1 If you calculate the variance and the covariance, then you see that it is true:$$ V[W_t^1] = t $$and$$ V[W_t^2] = \rho^2 V[W_t^1] + (1-\rho^2) V[Z_t] = \rho^2 t + (1-\rho^2) t = t, $$which is the desired variance. For the ... 9 Here is the general approach you can follow to generate two correlated random variables. Let's suppose, X and Y are two random variable, such that:$$X \sim N(\mu_1, \sigma_1^2)Y \sim N(\mu_2, \sigma_2^2)$$and$$cor(X,Y)=\rho$$Now consider: y=bx + e_i, where x (=\frac{X-\mu_1}{\sigma_1}) and y (=\frac{Y-\mu_2}{\sigma_2}) both follow ... 9 We can obtain a closed-form expression for price correlation given (log) return correlation when the two stocks follow geometric Brownian motion:$$S_1(t) = S_1(0)e^{(\mu_1- \frac{1}{2} \sigma_1^2)t}e^{\sigma_1Z_1(t)},\\ S_2(t) = S_2(0)e^{(\mu_2- \frac{1}{2} \sigma_2^2)t}e^{\sigma_2Z_2(t)},$$where \text{corr}(Z_1(t),Z_2(t)) = E[Z_1(t)Z_2(t)]=\rho t. ... 9 It is indeed Riemann integrable, so you don't need stochastic integration. For a given path, you can interpret the integral in the Riemann sense. For a given t, the paths are random, so it is a random variable. You can also express it as an Ito’s process. To see the connection, just apply ito's lemma to tW_t: d \left(tW_t\right)=tdW_t+W_tdt W_tdt=d \... 9 I provide a solution in three steps. The first step carefully outlines how to split up the expectation and what new measures are used. This first step does not require any special model assumption and holds in a very general framework. I derive a formula for the option price that resembles the standard Black-Scholes formula. In a second step, I assume that ... 8 I would calculate it this way, \mathbb{E}[(W_s+W_t−2W_0)^2] = \mathbb{E}\left[\left((W_s-W_0)+(W_t-W_0)\right)^2\right]\\ \hspace{4cm}=\mathbb{E}[(W_s-W_0)^2]+\mathbb{E}[(W_t-W_0)^2]+2\mathbb{E}[(W_s-W_0)(W_t-W_0)] \\ \hspace{4cm}=s+t+2\mathbb{E}[W_sW_t]\\ \hspace{4cm}=s+t+2\min(s,t) 8 X_t being a stochastic process, one cannot use ordinary calculus to express the differential of a (sufficiently well-behaved) function f of t and X_t. Instead one should turn to Itô's lemma, one of the key results of stochastic calculus, which stipulates (assuming X_t is here a continuous, square integrable stochastic process)$$ df(t,X_t) = \frac{... 8 if you talk about correlation then: compute expectation: $$\mathbb{E}(W_t)=0\text{ and }\mathbb{E}(\int_0^tW_d ds)=0$$ variance: $$\text{Var}(W_t)=t\text{ and }\text{Var}(\int_0^tW_s ds)=\frac{t^3}{3}$$ covariance: $$\mathbb{E}(W_t\int_{0}^tW_sds)=\int_{0}^t\mathbb{E}(W_tW_s)ds=\int_0^tsds=\frac{t^2}{2}$$ then you get:\$\text{Corr}(W_t,\int_0^tW_s ds)= \...

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