# Tag Info

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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}\\ &= \...

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The way you do it in the first place is a discretization of the Geometric Brownian Motion (GBM) process. This method is most useful when you want to compute the path between $S_0$ and $S_t$, i.e. you want to know all the intermediary points $S_i$ for $0 \leq i \leq t$. The second equation is a closed form solution for the GBM given $S_0$. A simple ...

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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 $$f(t, x) = \frac{1}{2} t^2 x$$ with \begin{...

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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): $d\,X_t=\mu X_t dt + \... 17 Brownian motion is simply the limit of a scaled (discrete-time) random walk and thus a natural candidate to use. It is very intuitive and arguably one of the simplest and best understood time-continuous stochastic processes. Also, don't forget that you obtain many more stochastic processes as functions of a (time-changed) Brownian motion. In many books on ... 16 Baxter and Rennie say it better than me, so I will summarize them. Suppose that$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 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, ... 10 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/... 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 ... 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 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)},$$where \text{corr}(Z_1(t),Z_2(t)) = E[Z_1(t)Z_2(t)]=\rho t. ... 9 Physical objects move according to simple smooth curves that can be represented by low order polynomials: a straight line, a parabola, an ellipse, etc. Financial market prices move in a completely different way, as can be seen by looking at any graph of stock prices, interest rates etc. in a newspaper: there are constant, erratic fluctuations, sometimes in ... 9 I provide a solution in three steps. The first step carefully outlines how to split up the expectation and what new measures are used. This first step does not require any special model assumption and holds in a very general framework. I derive a formula for the option price that resembles the standard Black-Scholes formula. In a second step, I assume that ... 8 I would calculate it this way, \mathbb{E}[(W_s+W_t−2W_0)^2] = \mathbb{E}\left[\left((W_s-W_0)+(W_t-W_0)\right)^2\right]\\ \hspace{4cm}=\mathbb{E}[(W_s-W_0)^2]+\mathbb{E}[(W_t-W_0)^2]+2\mathbb{E}[(W_s-W_0)(W_t-W_0)] \\ \hspace{4cm}=s+t+2\mathbb{E}[W_sW_t]\\ \hspace{4cm}=s+t+2\min(s,t) 8 To provide a straight forward answer: It is not a good model. It never was, it never will be. Until we all do not come up with a better model that provides better modeling accuracy while it is equally intuitive and makes similarly simplifying assumptions the BS model with its geometric brownian motion component is here to stay. It actually does not matter ... 8 X_t being a stochastic process, one cannot use ordinary calculus to express the differential of a (sufficiently well-behaved) function f of t and X_t. Instead one should turn to Itô's lemma, one of the key results of stochastic calculus, which stipulates (assuming X_t is here a continuous, square integrable stochastic process)$$ df(t,X_t) = \frac{... 8 Yes, the term Brownian Bridge seems to be used loosely. I assume you are talking about continuously monitored barriers by the way, since you mention the probability of the barrier being crossed in between the path time points. If that's the case then "naive" Monte Carlo simulation will have what is called "simulation bias". That's exactly because the ... 8 @Ivan's comment regarding the covariances is the key. Consider an equally spaced partition\Pi_n = \left\{ t_0 = 0, t_1 = \Delta_n, \ldots, t_n = t \right\}$of the interval$[0, t]$, where$t_i = i \Delta_n$and$\Delta_n = t / n$so that X_t = \lim_{n \rightarrow \infty} X_n, \qquad X_n = \sum_{i = 1}^n W_{t_i} \left( t_i - t_{i - 1} \... 8 It is indeed Riemann integrable, so you don't need stochastic integration. For a given path, you can interpret the integral in the Riemann sense. For a given t, the paths are random, so it is a random variable. You can also express it as an Ito’s process. To see the connection, just apply ito's lemma to$tW_t$:$d \left(tW_t\right)=tdW_t+W_tdtW_tdt=d \...

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The means are equal Suppose $f$ is analytic so that we can give it a Taylor series that works everywhere such that $f(x) = \sum a_n x^n$, and then let us let this be bounded too. To show that the expectations are the same consider taking the expectation of $U_t$ \mathbb{E}(U_t) = \mathbb{E}\left(\int_0^t f(W_s) \,\mathrm{d}s\right)\\ \end{...

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