# Tag Info

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

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Let $$dS_t = \mu S_t dt + \sigma S_t dW_t + S_{t^-} dJ_t$$ where $$J_t = \sum_{j=1}^{N_t} (V_j - 1)$$ is a compound Poisson process, with $V_j$ i.i.d. jump sizes (positive random variables) whose statistical properties are not relevant for what needs to be proven and $N_t$ a standard Poisson process of intensity $\lambda$. The processes $W_t$, $N_t$ and ...

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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: $$\int_0^t f(t, \omega) d W(t, \omega) = \lim_{\| \Delta_n\| \to 0 } \sum_{i=1}^{n} f(\tau_i,\omega) \left ( W(t_i, \omega) - W(t_{i-1}, \omega) \right )$$ , ...

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Just a few notes How to make sense of $\text dW_t$ is the entire point of stochastic calculus. It's far beyond the scope of any answer here. You should read some introductory lecture notes/books on stochastic calculus. You could start here. The idea: Riemann-Stieltjes integrals are of the form $\int_0^t f(s)\mathrm{d}g(s)$ and are well-defined if $f$ is ...

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I will provide some references such that you can see where the different processes are used. These papers typically motivate their models and show which effect the single paramaters have and what asset price dynamics the model intends to capture. Geometric Brownian motion The geometric Brownian motion is the easiest model for exponential growth with ...

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Itô's Lemma The standard version of Itô's Lemma applies to a single Itô process $\text{d}X_t=\mu(t,X_t)\mathrm{d}t+\sigma(t,X_t)\mathrm dW_t$. Then, $$\mathrm{d}f(t,X_t) = \left(f_t+\mu(t,X_t)f_x + \frac{1}{2}\sigma(t,X_t)^2f_{xx}\right)\mathrm{d}t+\sigma(t,X_t)f_x\mathrm dW_t.$$ Let $\text{d}Y_t=m(t,Y_t)\mathrm{d}t+s(t,Y_t)\mathrm dW_t^{(2)}$ be a second ...

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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 financial ...

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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 $tW_t$: $d \left(tW_t\right)=tdW_t+W_tdt$ $W_tdt=d \... 9 As usual with those kind of integrals, another way to reach the result is to: Express$W_s$in integral form as$\int_0^s dW_uUse Fubini theorem to change the integration bounds of the resulting double integral More specifically, \begin{align} \int_0^t W_s ds &= \int_0^t \int_0^s dW_u ds \\ &= \int_0^t \int_u^t ds dW_u \\ ... 9 IfM$and$N$are independent (your references appear to make this assumption), then$M+N$is also a Poisson process. So, using the polarization identity: $$dMdN = 2^{-1}\left[(d(M+N))^2 - (dM)^2 - (dN)^2\right]$$ $$= 2^{-1}\left[d(M+N) - dM - dN \right] = 0$$ (A proof of$(dX)^2 = dX$for a Poisson process$X$is available here.) 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) $$df(t,X_t) = \frac{... 8$$ \frac{1}{2} \frac{\partial^2 f}{\partial S^2} dS^2 \approx \frac{1}{2} \sigma^2 S^2\frac{\partial^2 f}{\partial S^2} dt$$(for small dt, ignoring (dt)^2 terms ) \sigma is embedded in dS = \mu S dt + \sigma S dW and$$ dS^2 = \mu^2 S^2 dt^2 + 2\mu \sigma S^2 dt dW + \sigma^2 S^2 dt \approx \sigma^2 S^2 dt$$You picked up 1/2\Gamma \sigma^2 from ... 7 If by 'solve' you mean how do we know that \ln S_t is the right change of variable, then you can go by the following (not rigorous) line of thought: Ito's fomula suggests that given an SDE$$dX_t = \mu(X_t,t)dt+\sigma(X_t,t)dW_t$$and a function f(x,t): the SDE for the process Y_t=f(X_t,t) will satisfy$$dY_t = [f_t(X_t,t) + f_x(X_t,t)\mu(X_t,t) + \... 7 First, for Ito processes and Brownian motion. Ito process is a continuous-time trajectory with random evolution, so non-smooth and very kinky - also has a fractal look: no matter how much you'd zoom in, it will look similar. Ito process consists in fact of two parts: the drift part (deterministic evolution) and the diffusion part (where all the kinkiness and ... 7 Let $$Z_t = \exp(-X_t)$$ with $$X_t = \sigma(T-t)W_t+\sigma\int_0^tW_sds+\int_0^Tf(0,u)du+\int_0^t\int_s^T\alpha(s,u)du ds$$ and$W_t$a standard Brownian motion, along with the usual assumptions. We can write$X_t=f(t,W_t)and apply Itô's lemma to get: $$dX_t = \frac{\partial f}{\partial t}(t,W_t) dt + \frac{\partial f}{\partial W_t} (t,W_t) dW_t + \... 7 Ito Lemma (as 'Taylor expansion'): For X an Ito process and f = f(t, x) ∈ C^{1,2}(\mathbb{R}^2) a deterministic function, the stochastic process$$Y_t = f(t,X_t)$$is an Ito process and we have$$df (t,X_t) = \partial_tf(t,X_t)\,dt + \partial_xf(t,X_t)\,dX_t + \frac{1}{2} \partial_{xx}^2f(t,X_t)(dX_t)^2. $$Note: Functions$$g(t,x)= \partial_tf(t,x), ... 7 What a great question! I've had a go at it below, I'd say I'm about 75% sure of the result I've got to but I'd love feedback from others. I'm going to use the definition of the Ito integral, \begin{align} \int^t_0 \sigma_s dW_s = \lim_{n \to \infty} \sum_{i=1}^n \sigma_{t_{i-1}} \bigl( W_{t_i} - W_{t_{i-1}} \bigr) \end{align} where t_n = t. Then, using the ... 7 This is impossible unless you are very intelligent with good memory-retention skills and already mathemathically proficient in the field of analysis and statistics (and no, a single course in basic probability theory does not suffice). And even if you are, two weeks is an extremely short amount of time. However, assuming these criteria are satisfied, I would ... 7 Hints: First show (using Ito Lemma) that \exp(iuX_t) = 1 + iu \int_0^t\exp(iuX_s) h(s) d W_s -2^{-1}u^2 \int_0^t\exp(iuX_s) h(s)^2ds$$Then show (by taking expectations):$$ E[\exp(iuX_t)] = 1 - 2^{-1}u^2 \int_0^tE[\exp(iuX_s)] h(s)^2ds $$Finally note that this is an ODE in unknown variable x(t):=E[\exp(iuX_t)], x(0)=1:$$ x'(t) =- 2^{-1}u^2h(t)^2 ... 6 Let \begin{align*} X_t = W(t)W_*(t) - \frac{1}{2}\int_0^t\big(W_*(u)^2+ W(u)^2\big)du. \end{align*} Then, \begin{align*} dX_t &= W(t) dW_*(t) + W_*(t) dW(t) -\frac{1}{2}\left(W_*(t)^2+ W(t)^2\right)dt, \end{align*} asW$and$W_*are independent. Consequently, \begin{align*} X_t = \int_0^t \big[W(s) dW_*(s) + W_*(s) dW(s)\big] -\frac{1}{2}\int_0^t\... 6 I thought this was an interesting example to add. It concerns a "ratio model" of habit (as opposed to a "difference" model of habit). See, for example, Abel (1990, American Economic Review). Let $$x_t = \lambda \int_{-\infty}^t e^{-\lambda(t-s)} c_s ds.$$ (For context,x_t$is a log habit index that is given by a geometric average of past consumption, ... 6 You are "deriving" with respect to$t$(the time index in your stochastic process).$f(t,x) = x^2$so$f(t,W_t) = W_t^2$. And Ito's lemma tells you$W_b^2 - W_a^2 = \int_{t=a}^b d(W_t^2) = \int_{t=a}^b df(t,W_t) = \int_{t=a}^b 2W_t dW_t + \int_{t=a}^b dt$for all$0 \le a <= b$. PS: Actually you are not deriving. The differential notation is just a ... 6 In stochastic calculus, only stochastic integrals are defined. The differential form is just a notation. That is, $$dF=g(t)dW_t$$ is just another expression for the integral $$F=\int_0^t g(s) dW_s.$$ See, for example, in this book or this book, all Ito's lemmas are expressed in integral forms. For your question, note that$F$is not a function of$t$and$...

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Quantiles are preserved under monotonic transformations, hence the quantile for $Y$ is simply the exponential of the quantile of $X$, no need for corrections whatsoever (see here for instance). Put otherwise, let $q$ denote the quantile $\alpha$ of $X$ i.e. $$\Bbb{P}(X \leq q) = \alpha$$ then \begin{align} \Bbb{P}( X \leq q ) &= \Bbb{P}( \underbrace{\...

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Answer Assuming the Poisson process $N_t$ is independent from the Brownian motions $(W_{1,t},W_{2,t})$, you'll have \begin{align} df(X_{1,t},X_{2,t}) &= \frac{\partial f}{\partial X_{1,t}} dX_{1,t}^c + \frac{\partial f}{\partial X_{2,t}} dX_{2,t}^c + \dots \\ &+ \frac{1}{2} \frac{\partial^2 f}{\partial X_{1,t}^2 } d\langle X_{1,t} \rangle_t^c + \...

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Consider OP's general formula $f(g(t),X_t)$. In case of ambiguity, let us claim that $f=f(t,x)$ is defined with variables $t$ and $x$, $g=g(s)$ is defined with the variable $s$, and $h=h(u,x)=f(g(u),x)$ is defined with variables $u$ and $x$. Then Ito's formula states that $${\rm d}h(u,X_u)=\frac{\partial h}{\partial u}(u,X_u)\,{\rm d}u+\frac{\partial h}{\... 6 write down Ito's lemma for the function X:$$dX=\frac{\partial X}{\partial Y}dY+\frac{1}{2}\frac{\partial^2 X}{\partial Y^2}(dY)^2+\frac{\partial X}{\partial c}dc+\frac{1}{2}\frac{\partial^2 X}{\partial c^2}(dc)^2+\frac{\partial^2 X}{\partial Y \partial c}dYdc+\frac{\partial^2 X}{\partial c \partial Y}dcdY$$Using the following: \frac{\partial X}{\... 6 Using Fubini's argument, assuming that f is deterministic$$E(I_t^2) = E\left(\int_0^t f(s) W_s ds\int_0^t f(u) W_u du\right)=\int_0^t\int_0^t{f(s)f(u)min(s,u)duds} If $f$ is continuous(even piece wise) you can prove that $I_t$ is normally distributed.

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