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Hot answers tagged itos-lemma

16

A common way to use Ito's lemma is also to solve the SDEs. The most classic example (I guess) is the geometric Brownian motion: $$dX_t = \mu X_t dt + \sigma X_t dW_t$$ and this can be solved easily by applying Itô's lemma with $$f(x)=\ln(x)$$ That's the BnB example: $$f'(x)=\frac{1}{x}$$ $$f''(x)=-\frac{1}{x^2}$$ and by Itô: $$d(ln(X_t))=\... 12 My understanding is because the Ito's integration definition keeps the martingale property. With Brownian motion W(t, \omega) defined, to define stochastic integration in a Riemann–Stieltjes style:$$\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 ) $$, ... 11 The difference between the two is that the first will lead you to a discretization scheme of the process. So you will have to simulate a whole (approximate) trajectory of (meaning by that X'_{t_0},...,X'_{t_n}) up to time T (the expiry of your vanilla option) to get to X'_T which is then only an approximation of X_T. The second method is exact and ... 11 If you are given a diffusion process X_t, and a C^{1,2} transformation Y_t=f(t,X_t) of the process X_t. Then Itô's lemma gives you the SDE followed by the process Y_t in terms of dX_t, and dt and partial derivatives of f up to order 1 in time and 2 in x. If you are given the SDE followed by X_t in terms of Brownian motion, drift, and ... 11 These are all examples on Ito Formula in its general form (with quadratic variations): 10 In fact Ito and Stratonovich calculus are both mathematically equivalent. In the following paper you can e.g. see that both derivations lead to the same result, i.e. the Black-Scholes equation: Black-Scholes option pricing within Ito and Stratonovich conventions by J. Perello, J. M. Porra, M. Montero and J. Masoliver From the abstract: Options ... 10 Baxter and Rennie say it better than me, so I will summarize them. Suppose that N_t is not stochastic and f(.) is a smooth function then the Taylor expansion is$$ 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 $$and the term (dN_T)^2 and higher terms are zero. Ito showed that this is not the case in the ... 10 Let$$ 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) $$is a compound Poisson process, with V_j i.i.d. jump sizes (positive random variables) whose statistical properties are not relevant for what needs to be proven and N_t a standard Poisson process of intensity \lambda. The processes W_t, N_t and ... 8 In general, if you have a process that you can write under the form F(B_t,t) where F is \mathcal{C}^{2,1} then Itô's lemma gives you the drift term and diffusion term of dF. Then if the resulting SDE has a null drift (that's where Black Scholes PDE comes from), and you get a only local martingale. For it to be a proper martingale you can look at ... 8 1. This integral is not Ito's Integral. Indeed Y_t is a random time change with time change rate \frac{W_t}{1+W_t^2}. (Oksendal, Sixth edition,page 147) 2. Sometimes this trick is useful.Indeed we assume that we are going to solve Riemann integral !. Let$$f''(x)=\frac{-2x}{(1+x^2)^2}$$then$$f'(x)=\left(\frac{1}{1+x^2}\right)+c_1$$and$$f(x)=\...

7

$X_t$ 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) $$df(t,X_t) = \frac{... 6 If by 'solve' you mean how do we know that \ln S_t is the right change of variable, then you can go by the following (not rigorous) line of thought: Ito's fomula suggests that given an SDE$$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) + \...

6

The dynamics \begin{align*} \frac{dS_t}{S_t} =\mu dt + \sigma dW_t. \end{align*} is under the real-world measure $\mathbb{P}$. Then, \begin{align*} d\ln S_t =\Big(\mu-\frac{1}{2}\sigma^2 \Big) dt + \sigma dW_t. \end{align*} Therefore, \begin{align*} \ln S_T = \ln S_t + \Big(\mu-\frac{1}{2}\sigma^2 \Big)(T-t) + \sigma \big(W_T-W_t\big).\tag{1} \end{align*} ...

5

In quantitative finance, we sometimes find ourselves choosing a new stochastic model for what market variables are random, and how. For example, someone might decide that they like the SDE $$dS = \mu\ S\ dt + \left( \frac{S_0}{S} \right)^{\frac32} \sigma\ S\ dW$$ because they want to capture a leverage effect. Now, this SDE ...

5

I think you should see the hint as follows: $$d(W_t^{n+1})=d(f(W_t))$$ with $$f(x)=x^{n+1}$$ Apply Ito: $$d(W_t^{n+1}) = f'(W_t)dW_t + \frac{1}{2} f''(W_t) d<W>_t$$ $$d(W_t^{n+1}) = (n+1) W_t^n dW_t + \frac{1}{2} n (n+1) W_t^{n-1} dt$$ If you integrate, you get: W_{t_2}^{n+1}-W_{t_1}^{n+1}=(n+1) \int_{t_1}^{t_2} W_t^n dW_t+ \frac{1}{... 5 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*} as W 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\... 5 Your logic is fine X_t \sim \mathcal {N}(X_0+\mu t, \sigma^2 t) $$Thus, \left (\frac {X_t}{\sigma\sqrt {t}}\right)^2  indeed exhibits a non central chi-squared distribution$$ \left (\frac {X_t}{\sigma\sqrt {t}}\right)^2 \sim \chi^2\left(k=1,\lambda=\left (\frac {X_0+\mu t}{\sigma\sqrt {t}}\right)^2\right) $$whence the law of S_t := X_t^2. As ... 5 What can be shown is that the above expressions are equal in probability. First check the distribution. As any linear combination of a Gaussian is Gaussian the right hand side is Gaussian - the left hand side too. Then we need the 2 moments: The expected values - it is zero ... easy to see. Next what you did not specify is that the correlation between ... 5 You seem to use the term "volatility" to describe two very different quantities: (1) the diffusion coefficient of your SDE and (2) the standard deviation of the log-returns under your modelling assumptions. While the first may be negative, the second may not. [Interpretation 1] Consider a probability space (\Omega,\mathcal{F},\mathbb{P}) and a standard ... 4 I think there is a typo in your first equation. The running variable should be s, as in d\left( e^{\beta(t-s)} r(s) \right). Let's start with your integral. Let R_u = e^{-\beta u} r_u. Your integral becomes$$ e^{\beta t} \int_s^t d R_u \, . $$Recall that dR_u = R_{u+du} - R_u. The integral evaluates to e^{\beta t}(R_t -R_s), which simplifies ... 4 Buy copies of Brent Oksendal's "Stochastic Differential Equations An Introduction with Applications" and Thomas Bjork's "Arbitrage Theory in Continuous Time." These are well written graduate level textbooks. I can't promise it will be painless, but if you want to understand continuous time derivative pricing models these are a place to start. Another ... 4 I thought this was an interesting example to add. It concerns a "ratio model" of habit (as opposed to a "difference" model of habit). See, for example, Abel (1990, American Economic Review). Let$$ 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, ... 4 Based on Ito's isometry, \begin{align*} E_t (r^2_{t+1}) &= E_t \bigg(\int_t^{t+1} \sigma_s dW_s \int_t^{t+1} \sigma_s dW_s\bigg)\\ &= E_t \bigg(\int_t^{t+1} \sigma_{\tau}^2 \,d\tau\bigg) \\ &= E_t\bigg(\int_0^1 \sigma_{\tau+t}^2 \,d\tau\bigg) \\ &=\int_0^1 E_t\big(\sigma_{\tau+t}^2\big) \,d\tau. \end{align*} The identity \begin{align*} E_t (r^... 4 Maybe I'm missing something? Given f:\mathbb{R}^n \rightarrow \mathbb{R}^m, you can write f = (f_1,\ldots,f_m), where each f_i:\mathbb{R}^n \rightarrow \mathbb{R}. Apply Ito to each f_i separately. 3 The second method you use is correct and, actually, is completely equivalent to the first one. The reason is that the proof of Ito's lemma relies on a Taylor expansion of the second order. Notice that Wikipedia's formulation of Ito's lemma is a bit misleading, as they write dX_t = \mu_t dt + \sigma_t dB_t  but, actually, the functions $\mu$ and $\... 3 To answer the more general question that seems to be giving you trouble, Ito's lemma is the stochastic version of the chain rule of standard calculus. What is it useful for? That's like asking what the chain rule is useful for. Calculus is useful in quantitative finance, and in particular, for stochastic processes, you need to use the stochastic version ... 3 To add to the answer of TheBridge: I understand your second question in the sense if you could use Ito's lemma for all stochastic processes. This is definitely not the case: It can also be used for processes with bounded quadratic variation (e.g. Wiener process) - you should google this term or look it up in wikipedia: http://en.wikipedia.org/wiki/... 3 The standard method to manage your kind of problem (i.e. dealing with stochastic processes that are note presented or built thanks to a Brownian motion) is to use a measure change. The power of Brownian motion is that you have a lot of representation theorems (Doob-Meyer theorem, Wold theorem, etc) that allows to (thanks to a change of measure or a ... 3 The logic from Bob Jansen is correct. The problem is abuse of ideas and notation the integral symbol from the deterministic world gets sloppily applied to random variables. Unlike normal$dt$, which is always positive,$dW_t$can go 'backwards'. Thus increments of terms like$W_t dW_t\$ have a first element that goes up and down with the second element (...

3

For the first question, since by definition, \begin{align*} \varepsilon_t^{if} = e^{i \int_0^{t}f\big(\frac{1}{\xi}\langle M\rangle_s\big)\frac{dM_s}{\sqrt{\xi}} + \frac{1}{2}\int_0^t f\big(\frac{1}{\xi}\langle M\rangle_s\big)\frac{d\langle M\rangle_s}{\xi}}, \end{align*} then, \begin{align*} d\varepsilon_t^{if} = i \varepsilon_t^{if} f\Big(\frac{1}{\xi}\...

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