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2

Since $L_t$ is a.s. positive and the function $f(x) = 1/x$ is $C^2$ on $(0,\infty)$ you can use Ito as you want to. You have messed up the algebra though. There should definitely be an $L^2$ on the bottom, not an $X^2$. With your edited post you need to be more careful. You need to show that $X_t$ is a.s. positive if you want to apply Ito like you usually ...

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Set $$X_{T}=\int_{0}^{T}\theta_{s}dW_{s}-\frac{1}{2}\int_{0}^{T}\theta_{s}^2ds$$ then $$dX_{T}=\theta_{T}dW_{T}-\frac{1}{2}\theta_{T}^2dT$$ and using Lemme Itô: $$d(e^{X_{T}})=e^{X_{T}}dX_{T}+\frac{1}{2}e^{X_T}d<X>_{T}$$ then ?? I don't know how to continue

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Let $Y_t=\int_0^t b_s dW_s$ in the Ito sense. If $E(|Y_t|)<\infty$, then $Y_t$ forms a martingale. Martingales have constant expectation, which must be zero, since $Y_0=0$ a.s.

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Continuity and the local martingale property should (hopefully) be clear. For the quadratic variation/covariation, look at the quantities $(dW_1)^2$, $(dW_2)^2$, $dW_1dW_2$. The first two should be $dt$, and the third $0$. (Use the fact that $dB_1dB_2=\rho_t$). Then integrate to get the required answer.

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One of the characterizations of Multidimensional Brownian Motion $B_i(t)$ is that the covariations satisfy: $$[B_i, B_j]_t = \delta_{ij}t$$

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You can determine the sde that $X_t^2$ satisfies using Ito's lemma. Then you can find the pdf by solving the forward Kolmogorov equations (Fokker-Planck).

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Ignoring for the moment the fact that $\sqrt{2t+B_t}$ will sometimes be negative, we have $F(t,x) = \frac23(2t+x)^\frac32,\qquad$ $\frac{\partial F}{\partial t}=2\sqrt{2t+x},\qquad$ $\frac{\partial f}{\partial x} = \frac{1}{2\sqrt{2t+x}}$, so $\int_0^T\sqrt{2t+B_t}\;dB_t = \frac23(2T+B_T)^\frac32 -\int_0^T2\sqrt{2t+B_t}+\frac{1}{2\sqrt{2t+B_t}}\;dt$. We ...

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I agree, what I mean is that if you want to use the formula $\frac{dF}{dx}(x,y)= f(x,x)$+ 2nd term, then the $f(x,x)$ is the full expression after the integral symbol, but divided by the $dy$ Therefore, applying this method formally, you should write $\frac{dM}{dt} = e^{-(t-u) } \frac{dS}{S dt}$ + 2nd term.

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even if would apply the classical calculus rules, then I think you did not use correctly the last formula your write. Namely, in the first term of the last equation, there is no $dy$. Hence, following your logic, you need to divide the first term in your equation by $dt$. This then gives the correct equation. Hope this helps.

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For (almost) every fixed $\omega \in \Omega$ it is a just a Riemann integral, because you are integrating a continuous function over a compact interval. So overall what you have is a random variable, and your interpretation of this variable is correct (although it is of course signed area). The resulting random variable doesn't really have a simple ...

2

I'm also studying a course where things like this appear, so I'll present an attempt but corrections are welcome. I know stochastic calculus doesn't always play by the rules. I get that: $$\mathbb{E}[W_T \cdot \int_0^Tf(s)dW_s] = \int_0^Tf(s)ds$$ If $\mathbf{P} = \{t_0 = 0, t_1,t_2,\dots,t_n = T\}$ is a tagged partition, $||\mathbf{P}|| = ... 1 Let$(X_t)_{t \geq 0}$be a weak solution of the given SDE, i.e. there exists a Brownian motion$(B_t)_{t \geq 0}$such that $$X_t = X_0 + \int_0^t b(s,X_s) \, ds + \int_0^t \sigma(s,X_s) \, dB_s.$$ Using the elementary estimate $$(a+b+c)^p \leq 3^p (a^p+b^p+c^p), \qquad a,b,c \geq 0,$$ we find $$|X_t|^p \leq 3^p |X_0|^p + 3^p \left| \int_0^t b(s,X_s) ... 0 This is a Wiener integral and so it has a Gaussian law with first two moments of the form : Mean$$m_t=E[\int_0^t \frac{u^\alpha}{t}dW_u]=0 $$Variance$$v_t=E[\int_0^t \frac{u^{2.\alpha}}{t^2}du]=\int_0^t \frac{u^{2.\alpha}}{t^2}du=...$$Check the web for properties on Wiener Integrals there is plenty. Best regards 0 It is the L2 limit of a family of variables of the form \frac{1}{t}\sum t_i^\alpha(W_{t_{i+1}}-W_{t_i}) hence it is measurable according to the filtration of W. 1 Fix \epsilon>0. Since \|H\|^2 < \infty, we can choose R>0 sufficiently large such that$$\mathbb{E} \left( \int_R^{\infty} H_t^2 \, dt \right) < \epsilon.$$By assumption, there exists H^n \in \mathcal{E} such that$$\mathbb{E} \left( \int_0^R |H_t^n-H_t|^2 \, dt \right) \leq \epsilon.$$Without loss of generality, we may assume ... 0 (too long for a comment)divide by (1-t) we find$$ \frac{1}{1-t}X_t^x = x\frac{(t-s)}{(1-s)(1-t)} +\frac{1}{1-s}X_s^x + \left(B_t-B_s +\int_s^t\frac{r}{1-r}dB_t\right) $$and$$ x\frac{(t-s)}{(1-s)(1-t)} = -x\left(\frac{1}{1-s} - \frac{1}{1-t}\right) $$thus we get$$ \frac{1}{1-t}X_t^x - \frac{1}{1-s}X_s^x + x\left(\frac{1}{1-s} - \frac{1}{1-t}\right) = ... 0 Note that, by the first equation,$X_0^x = 0$, and so the result follows from taking$s=0$in the second equation. 2 Hint: Using Itô's formula, show that$X_t := \sin(W_t) e^{t/2}$satisfies $$dX_t = \cos(W_t) e^{t/2} \, dW_t.$$ Conclude that $$\sin(W_T) = \int_0^T e^{-(T-t)/2} \cos(W_t) \, dW_t.$$ Remark: The idea is to find a determinstic function$f$such that$X_t := f(t) \sin(W_t)is a martingale. By Itô's formula, a sufficient condition is $$\frac{1}{2} ... 0 HINT:$$\int_0^t\int_0^{s}\frac{B_s'}{(1-s')^2}ds'ds=\int_0^t\int_s'^t \frac{B_s'}{(1-s')^2}ds'ds \tag 1$$and$$\int_0^t \frac{B_s'(t-s')}{(1-s')^2}ds'=\int_0^t \frac{B_s'(t-1+1-s')}{(1-s'^2)}ds' \tag 2$$Can you complete? We are given$$X_t=xt+B_t-(1-t)\int_0^t \frac{B_s}{(1-s)^2}ds \tag 3$$and asked to show that this X_t is a solution of the ... 0 We will prove this by induction, first the induction basis n=0:$$ X_t^0 \equiv 1 = H_0(W_t,t), hence the induction basis holds. Now suppose it holds for n, we will prove it for n+1: Then \begin{align*} X_t^{n+1} &= 1 + \int_0^t X_s^n d W_s \\ &= 1+ \int_0^t \sum_{k=0}^n \frac{1}{k!} H_k(W_s,s) dW_s \\ &= 1 + \sum^n_{k=0} \int_0^t ... 2 For an Itô process (X_t)_{t \geq 0} of the formX_t-X_0 = \int_0^t \sigma(s) \, dB_s + \int_0^t b(s) \, ds \tag{1}$$Itô's formula reads$$f(X_t)-f(X_0) = \int_0^t f'(X_s) \sigma(s) \, dB_s + \int_0^t \left( \frac{1}{2} f''(X_s) \sigma^2(s) + f'(X_s) b(s) \right) \, ds.$$Now: X_t := \int_0^t s \, dB_s - \frac{t^3}{6} is an Itô process. Choose ... 1 the answer to this is quite simple if you look carefully to the definition of H, you should notice that h_{i-1} is an \mathcal{F}_{t_{i-1}}-measurable random variable. From this and the following elementary properties on conditional expectation we get the result : If X is \mathcal{F}-measurable and for any bounded random variable Y, we have a.s. : ... 0 Just seen this - probably too late, but if not..: re. (1): You are in t, i.e. t is now, and V_t and S_t are real market prices observable right in this moment (t). re (2): same answer as to (1). S_t and V_t are here not symbols for random variables, but current realizations which you see now (in time t). Of course, you don't know future ... 1 Let Y_t=ln(X_t), by Ito's lemma we derive that the process Y_t follows the SDE dY_t=\sigma dW_t,\quad Y_0=0 which has solution Y_t= \sigma W_t. Because W_t has distribution N(0,t), Y_t has distribution N(0,\sigma^2t). Since X_t=e^{Y_t} and X_t has normal distribution, we conclude that X_t has log-normal distribution, so the ... 1 Integrating any locally bounded predictable process (in particular a continuous adapted process) with respect to a local martingale (resp locally square integrable martingale) results in a new process which is a local martingale (resp locally square integrable). A proof of this can be found in Jacod & Shirjaev (Chapter 1; Theorem 4.31). (It basically ... 6 From the stochastic integral analogue of integration by part, which is a corollary of Ito's Lemma,$$d(W_t\ln t) = \frac{W_t}{t}dt+(\ln t)dW_t,$$so$$I:=\int_0^t \frac{W_s}{s}ds=W_t\ln t-\int_0^t (\ln s)dW_s=\int_0^t \ln\frac{t}{s}dW_s$as$W_t\ln t\to 0$in probability, as$t\to 0^+$. Since the random variable$I$is a linear combination of independent ... -1 Here's my take on this, Let$S_t(s) = \int_0^t \mu(t,S_t,s)dt + \int_0^t \sigma(t,S_t,s)dW_t$Then,$S_t(s)$is a normally distributed random variable parametrized by it's expectation and variance. To see that$S_t(s)$is normal,$S_t(s) = \int_0^t \mu(t,S_t,s)dt + \int_0^t \sigma(t,S_t,s)dW_t$, which is equal to a constant$\int_0^t \mu(t,S_t,s)dt = ...

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