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4

First, simulate the paths of the process $(X_t)$ defined by $$X_t=\sigma W_t+\left(r-\tfrac12\sigma^2\right)t,$$ using Euler's scheme $X^\varepsilon_0=0$ and$$X_{n+1}^{\varepsilon}=X_n^{\varepsilon}+\sigma\sqrt\varepsilon Z_n+\left(r-\tfrac12\sigma^2\right)\varepsilon,$$ for every positive $\varepsilon$ and every $n$, where the process $(Z_n)$ is i.i.d. ...

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It is exactly the other way round: We define $(1)$ to be the same thing as $(2)$; it's just a more convenient way to write $(2)$. That this notation makes sense can be seen by integrating both sides of $(1)$ from $0$ to $T$: $$\underbrace{\int_0^T dX_t}_{X_T-X_0 = X_T-x} = \int_0^T b(t,X_t) \, dt + \int_0^T \sigma(t,X_t) \,dW_t.$$ A similar situation pops ...

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Let $Y_t=\displaystyle\int_0^t\mathrm e^{au}\mathrm dW_u$ then, for every $t$, $X_t=b\mathrm e^{-at}Y_t+z(t)$ where $z(\ )$ is deterministic hence $$\mathrm{cov}(X_t,X_s)=(b\mathrm e^{-at})(b\mathrm e^{-as})\mathrm{cov}(Y_t,Y_s),$$ and, for every $t$, $Y_t=\displaystyle\int_0^\infty g_t(u)dW_u$ where $g_t$ is deterministic since $g_t(u)=\mathrm e^{au}\mathbf ... 1 Note that $$X_t := - \int_0^t H_s \, dB_s - \frac{1}{2} \int_0^t H_s^2 \, ds$$ defines an Itô process. Therefore we may apply Itô's formula to$f(x) := e^x: \begin{align*} f(X_t)-f(X_0) &= \int_0^t f'(X_s) \, dX_s + \frac{1}{2} \int_0^t f''(X_s) \, d\langle X \rangle_s \\ \Leftrightarrow M_t - 1 &= - \int_0^t M_s H_s \, dB_s - \frac{1}{2} ... 1 Complementing the answer above to make explicit use of Ito's isometry as you requested. The appropriate version of ito's isometry to use in this case is the following: \mathbb{E} \int_0^T f(r)dW(r) \int_0^T g(r) dW(r) = \mathbb{E} \int_0^T f(s)g(s)ds,$$where in your case f(r) = e^{ar} \mathbf{1}_{(0,s)}(r), and f(r) = e^{ar} \mathbf{1}_{(0,t)}(r). I ... 1 Note that \mathrm d\langle V,W\rangle_t=\sigma\rho\sqrt{V_t}\mathrm dt hence, for every t\gt0,$$\rho=\frac1{\sigma t}\int_0^t\frac{\mathrm d\langle V,W\rangle_s}{\sqrt{V_s}}.$$Likewise, for every t\gt0,$$\sqrt{1-\rho^2}=\frac1{\sigma t}\int_0^t\frac{\mathrm d\langle V,Z\rangle_s}{\sqrt{V_s}}.\$ Edit: Numerically, the integral involving the ...

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