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No, you cannot simply set $$g(t,x) = \int_0^t e^{ks} \, dx$$ since the stochastic integral $\int_0^t e^{ks} \, dB_s$ is not pathwise defined. In order to find the differential of $(Y_t)_{t \geq 0}$, you have to apply Itô's formula for Itô processes. Namely, if $$dZ_t = \sigma_t \, dB_t$$ then Itô's formula states that \begin{align*} f(t,Z_t)-f(0,Z_0) ... 0 Here's a tip. Look at the integration-by-parts formula for stochastic calculus: http://en.wikipedia.org/wiki/It%C5%8D_calculus#Integration_by_parts 2 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.$$A similar situation pops ... 4 First, simulate the paths of the process (X_t) defined by$$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,$$for every positive \varepsilon and every n, where the process (Z_n) is i.i.d. ... 1 Complementing the answer above to make explicit use of Ito's isometry as you requested. The appropriate version of ito's isometry to use in this case is the following:$$ \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}}.$$Edit: Numerically, the integral involving the ... 1 Let Y_t=\displaystyle\int_0^t\mathrm e^{au}\mathrm dW_u then, for every t, X_t=b\mathrm e^{-at}Y_t+z(t) where z(\ ) is deterministic hence$$\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 ...