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3

Unfortunately, I have never really found an explanation for basic stochastic calculus which is both mathematically accessible and technically correct. I will give the usual mathematical treatment, but I concede that it is not particularly accessible. $W$ is a process on $[0,\infty)$ with the following basic properties: $W(0)=0$ $W$ is a Gaussian process. $... 3 We can compute the covariance between$W_t$and$I_tas follows: \begin{align*} covar(W_t, I_t) &=E\left(W_t \int_0^t W_s ds \right)\\ &= E\left( \int_0^t W_t W_s ds \right)\\ &= \int_0^t E\left(W_t W_s \right) ds\\ &= \int_0^t \min(t, s)\, ds\\ &=\frac{t^2}{2}. \end{align*} Alternaticely, note that \begin{align*} d(tW_t) = W_t dt + t ... 2 It seems that when we talk about Poisson random measure, we often assume that the characteristic measure\nu$is$\sigma$-finite. Then the random measure$\tilde{N}(t,dx)$is also$\sigma$-finite. Of course,$\tilde{N}(t,A)$is not a martingale if$\nu(A)=\infty$. To define the stochastic integral $$\int^T_0 f(t,x)\tilde{N}(dt,A),$$we first consider the ... 2$dX_t = (\beta-\alpha X_t) \, dt + \sigma \, dB_tForget a moment about the SDE and consider the associated ordinary differential equation $$dx_t = (\beta- \alpha x_t) \, dt \tag{1}$$ instead. If I would ask you to solve this ODE, you would (hopefully) first solve the homogeneous equation $$dx_t = -\alpha x_t \, dt$$ and find that the solution of this ... 2 As suggested by you, I put my comment as an answer. I suggest you have a look at lemma 3 (the iff part of the assertion) here . Best regards. 2 Note that \begin{align*} d(sW_s) = sdW_s + W_s ds. \end{align*} Then you can integrate on both sides. 2 As it was hinted in the comments, the result is not true unless you assume uniform convergence somewhere. Here is a simple example. LetX_n = X$have non-degenerate Bernoulli distribution and$Y_{un} = X \mathbf{1}_{u\ge n}$. Then$P(Y_{un}\neq X_n)\to 0$,$u\to\infty$, and$e^{itY_{un}} \to 1$,$n\to\infty$, in all senses. However,$e^{itX_n} = e^{itX}\...

2

You have to prove the convergence of the integral $$\int_{-\infty}^{+\infty}\left|e^{i\xi \sqrt tu}\right|e^{-u^2/2}\mathrm du$$ for each complex number $\xi$. Since we can write $\xi=a+ib$ where $a$ and $b$ are real numbers, it suffices to prove the convergence of the integral $$\int_{-\infty}^{+\infty}e^{b\sqrt tu}e^{-u^2/2}\mathrm du$$ for each real ...

1

By the stationarity of the increments, this is equivalent to $$\min_{s \in [0,t]} \mathbb{P}(|X_s| \leq \epsilon)>0$$ for all $t>0$. Without any additional assumptions this does, in general, not hold true. Consider for example $X_t := t$ (i.e. a deterministic drift process), then $$\mathbb{P}(|X_t| \leq \epsilon) = 0$$ for all $t>\epsilon$. ...

1

You'll want to have a look at "A Signed Measure on Path Space Related to Wiener Measure" by K. Hochberg http://www.jstor.org/stable/2243147 Hocberg constructs a Markovian signed measure (of unbounded variation) on the space of continuous paths, associated with the operator $Lu:={\partial^4u\over\partial x^4}$. The transition density $p$ is the fundamental ...

1

Roughly and intuitively speaking, a self financing strategy is (for me) a trading strategy which requires no extra cost during the trading except for the initial capital. Suppose that we are in a continuous time case. Let $(\Omega,\mathcal{F},P)$ be a prob space with filtration $\mathbb{F}$, $S$ (a semimartingle) be the discounted price process of the stock, ...

1

Although Brownian paths are not differentiable point wise, we may interpret their time derivative in a distributional sense to get a generalized stochastic process called white noise. We denote it by $$\eta (t,\omega )=\overset{\centerdot }{\mathop{B}}\,(t,\omega )$$ We also use the notation $$d\eta=dB_t$$ The term white noise arises from the spectral ...

1

Your idea is fine, however, there are a few mistakes. Here is another solution. Note that, \begin{align*} u_t &= \int_{t-1}^{t} \int_{0}^{s}e^{-\kappa(t-s)-c(s-r)}dW(r) \,ds \\ &= \int_{0}^{t-1} \int_{t-1}^{t}e^{-\kappa(t-s)-c(s-r)}ds\,dW(r) + \int_{t-1}^{t} \int_{r}^{t}e^{-\kappa(t-s)-c(s-r)}ds\,dW(r)\\ &= e^{-\kappa t} \int_{t-1}^t e^{(\kappa ...

1

Here is my calculation, which may be wrong or not, I am not quite sure... \begin{align} Var(u_t) &= E(u_t^2) = E\left(\int_{t-1}^{t} \int_{0}^{s}e^{-\kappa(t-s)-c(s-r)}dW(r) \,ds\right)^2 \\ &= E\left(\int_{0}^{t} \int_{r}^{t}e^{-\kappa(t-s)-c(s-r)}\,ds\, dW(r) \right)^2 \\ &= \int_{0}^{t} \left(\int_{r}^{t}e^{-\kappa(t-s)-c(s-r)}ds\right)^2\, ... 1 You already know that [Z,Z]_t = \int_0^t \sigma^2(s) \,ds for any Itô processdZ_t = b(t) \, dt + \sigma(t) \, dW_t.$$Moreover, the quadratic covariation is defined via the polariation formula, i.e.$$[X,Y]_t = \frac{1}{2} [X+Y,X+Y]_t - [X,X]_t-[Y,Y]_t. \tag{1}$$Since both (X_t)_{t \geq 0} and (Y_t)_{t \geq 0} are Itô processes, we have$$[X,X]...

1

It looks like there is a missing Brownian term in your expression for $X(t)$ If so then the only term that will appear in the quadratic variation is the product of the two random terms as these are effectively order $dt$, with all order terms in the product now higher order Hence the resulting quadratic variation will be $\Delta_x \Delta_y \cdot dt$ For ...

1

I have a couple of suggestions. I'm a big fan of Osborne and Rubinstien's A Course in Game Theory, but that may not have enough decision theory. You might also try The Economics of Risk and Time By Christian Gollier. More hardcore is work by Peter Fishburn.

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Well, simple it is: $$E \left[ \int_0^1 X_t^2 dt\right] = \int_0^1 E\left[X_t^2\right] dt = \int_0^1 \int_0^t E\left[u(s)^2\right]ds\, dt \\= \int_0^1 \int_s^1 E\left[u(s)^2\right]dt\, ds = \int_0^1 E\left[(1-s)u(s)^2\right] ds = \int_0^1 E\left[(1-t)u(t)^2\right] dt.$$ Hence your equality follows, taking into account that $E\left[X_1^2\right] = \... 1 The answer is: Yes, we can. It's not important that we talk about real Hilbert spaces. So, let's replace$\mathbb R$by$\mathbb K\in\left\{\mathbb C,\mathbb R\right\}$in the question. Let$(t,x)\in[0,\infty)\times H(e_n)_{n\in\mathbb N}$and$(f_n)_{n\in\mathbb N}$be orthonormal bases of$K$and$H$, respectively Note that$\$\tilde\Phi:=\Phi_tQ^\...

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