# Tag Info

20

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

13

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 ...

13

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 ...

11

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

9

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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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 ...

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) $$df(t,X_t) = \frac{... 8 \max(B_T,S_T)=\max(0,S_T-B_T)+B_T, so this is just a call option (with strike B_T) plus B_T. 7 If by 'solve' you mean how do we know that \ln S_t is the right change of variable, then you can go by the following (not rigorous) line of thought: Ito's fomula suggests that given an SDE$$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 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 \... 7 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_u Use 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 \\ ... 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*} 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\... 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.$$See, for example, in this book or this book, all Ito's lemmas are expressed in integral forms. For your question, note that F is not a function of t and ... 6 Quantiles are preserved under monotonic transformations, hence the quantile for Y is simply the exponential of the quantile of X, no need for corrections whatsoever (see here for instance). Put otherwise, let q denote the quantile \alpha of X i.e.$$\Bbb{P}(X \leq q) = \alphathen \begin{align} \Bbb{P}( X \leq q ) &= \Bbb{P}( \underbrace{\... 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}dcdYUsing the following: \frac{\partial X}{\... 6 As @Canardini pointed out, \begin{align*} E\big(I_t^2\big) &= 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)dsdu\\ &= \int_0^t\left(\int_0^u f(s)f(u) s ds + \int_u^t f(s)f(u) u ds \right)du\\ &= \int_0^t \int_0^u sf(s) f(u)ds du + \int_0^t \int_u^t uf(u)f(s) ds du\\ &=2\int_0^t uf(u) \... 5 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. $$(For context, x_t is a log habit index that is given by a geometric average of past consumption, ... 5 Your logic is fine$$ 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 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 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 ... 5 May I point your attention to my following paper, where I address this question in an intuitive manner on page 12: von Jouanne-Diedrich, Holger, Ito, Stratonovich and Friends (May 18, 2017).Available at SSRN: https://ssrn.com/abstract=2956257 The basic idea in the discrete case is to go through all possible branches and then square the differences. In each ... 5 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 + \... 5 Apply the Ito product rule, noting the cov of a deterministic and stochastic term is zero:\begin{align} d\left(e^{at}r_t\right)&=e^{at} dr_t+r_t de^{at} \\[6pt] &=e^{at} dr_t+r_t e^{at} d(at) \\[6pt] &=e^{at} dr_t+r_t e^{at} a dt \end{align}

5

The second solution is correct, so the solution at t is: $F_t^2=F_0^2e^{2 \sigma W_t-\sigma^2t}$ Now apply expectation to both sides. $E\left[F_t^2\right]=F_0^2 \, E\left[e^{2 \sigma W_t-\sigma^2t}\right]$ The term in the exponent is just Gaussian so let’s call it X: $X=2 \sigma W_t-\sigma^2t$ It’s mean and variance are: $E[X]=-\sigma^2t$ \$V[X]=4 \...

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