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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. ...


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 ...


2

Let $X_t=\displaystyle\int_0^ts\mathrm dW_s$ and $Y_t=\displaystyle\int_0^tW_s\mathrm ds$, then $X_t+Y_t=tW_t$ hence $E(X_t\mid W_t)=tW_t-E(Y_t\mid W_t)$. Furthermore, $E(W_s\mid W_t)=sW_t/t$ for every $s$ in $(0,t)$ hence $$E(Y_t\mid W_t)=\displaystyle\int_0^tE(W_s\mid W_t)\mathrm ds=(W_t/t)\displaystyle\int_0^ts\mathrm ds=tW_t/2.$$ Thus, $E(X_t\mid ...


1

Hi as mentonned in my comment the proof given does not allows us to conlcude by localization. I have finally found a rigourous and elementary proof of the fact that a CIR process possesses moments of all orders which is necessary to get the result (order 1 is enough for our need) . First two obersvations : by Yamada-Watanabe's theorem that the CIR's ...


1

Lets assume that we are in the regime in which $V_t > 0$ for every $t>0$. Consider the sequence of stopping times $$\tau_n = \inf \left\{ t: V_t > n \right\}, \quad n \in \mathbf{N},$$ then, as you mentioned before, if we can show that $\tau_n \to \infty$ as $n \to \infty$ in probability the result follows by standard localization: the expectation ...


1

It simply depends whether one wants to consider integrals with infinite time-horizon or not. Usually, stochastic integrals of the form $$\int_0^T K_s \, dW_s \tag{1}$$ are considered where $T<\infty$. In this case, $$\mathbb{E} \left[ \left( \int_0^T K_s \, dW_s \right)^2 \right] = \mathbb{E}\left( \int_0^T K_s^2 \, ds \right) \tag{2}$$ is called Itô's ...


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

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) ...



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