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Note that \begin{align*} dJ_c(s) = dB_s -cJ_c(s) ds. \end{align*} Moreover, $d\langle J_c, J_c\rangle_t = dt$. Then, by Ito's lemma, \begin{align*} \int_0^1J_c(s) dJ_c(s) &= \int_0^1 d\left(J_c^2(s) -s \right)\\ &= J_c^2(1) - 1. \end{align*} Since \begin{align*} J_c(1) &= \int_0^1 e^{-c(1-s)}dB_s \end{align*} is normal with zero mean and the ...

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Finally, I found it wrong. The correct way is dB_x = mvnrnd(zeros(K, 1), Omega_xx/n, n); % n-by-m matrix B_x = cumsum(dB_x,1); % n-by-m matrix B_x = [zeros(1,K); B_x]; J_c = exp(C*[1:n]/n)'.*cumsum(exp(-C*[1:n]/n).*[zeros(K,1) dB_x(1:n-1,:)'])'; dJ_c=[J_c(1,:); diff(J_c)]; % m-by-n matrix int_J = J_c'*dJ_c;

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Define $X_t = -\int_0^t\beta_s\,dW_s$ and $Y_t = -\frac{1}{2}\int_0^t\beta_s^2\,ds$. Then $Z_t = e^{X_t+Y_t}$. Even though you did not mention it I am guessing that there is a condition on $(\beta_s)_{s\geq 0}$ that makes $X$ a local martingale. Since Brownian motion has continuous paths, $X$ is continuous as well, which implies $Z$ is continuous as well. ...

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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. $... 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 chi-squared random variable with k degrees of freedom is given as $$\chi^2=\sum\limits_{i=1}^{k}{{{X}_{i}}^{2}}$$ where the$X_i$'s are all independent and have$N(0, 1)$distributions. Also recall that I claimed that$\chi^2$has a gamma distribution with parameters$r=k=2$and$\alpha=\frac{1}{2}$, thus $$M_{\chi^{\,2}}(t)=\left(\frac{1}{1-2t}\right)^{\... 1 I think that if f(t,x)= e^ {\beta t}, we have$$df/dt=\beta f(t,x), df/dx=e^ {\beta t}, d^2f/dx^2=0$$So using Ito formula we can receive the differential of d (e^ {\beta t}R(t)) -$$d (e^ {\beta t}R(t)) = df(t,R(t))= \beta e^ {\beta t}R(t)dt+e^ {\beta t}dR(t)=\beta e^ {\beta t}R(t)dt+e^ {\beta t}adt-\beta e^ {\beta t} R(t) dt +e^ {\beta t} \sigma dW(... 1 Calculate the ito differential: $$dN=\beta N(t)dt-\alpha e^{\beta t}\sin(\alpha W(t))dW-\alpha^2N(t)dt.$$ It follows that if$\beta=\alpha^2$, then$N(t)$is a martingale. Thus surely you know how to calculate$E[N(t)]$and also$\cos(\alpha W(t))=N(t)e^{-\beta t}, so: $$E[N(t)]=E[e^{\beta t}\cos(\alpha W(t))]=e^{\beta t}E[\cos(\alpha W(t))]$$ 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 I assume you want to prove it using Ito calculus, otherwise it is just moments of a gaussian random variable. By Ito:W^{n}_t=\int_{0}^t n W^{n-1}_s dW_s+\frac{n(n-1)}{2}\int_{0}^tW^{n-2}_sds $$using induction hypothesis (to get the local martingale part being a true martingale), I have:$$\mathbb{E}[W^{n}_t]=\frac{n(n-1)}{2}\int_{0}^t\mathbb{E}[W^{n-2}... 0 Question 1 (Girsanov's theorem) LetW_t$be a Brownian motion under the physical measure$\mathbb{P}$. Define $$L_t := \exp \left\{-\int_{0}^{t} X_s dW_s - \frac{1}{2} \int_{0}^{t} X_s^2ds \right\},$$ and define an equivalent martingale measure$\mathbb{Q}$by setting$d\mathbb{Q}/d\mathbb{P} = L_t$, then$B_t = W_t + \int_{0}^{t} X_s ds$is a standard ... 2$dX_t = (\beta-\alpha X_t) \, dt + \sigma \, dB_t$Forget 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 ... 1 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] = \...

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