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5

Yes, the right-hand side is a Riemann-Stieltjes integral. There is a general result which states that the Riemann-Stieltjes integral $\int_a^b g(t) \, df(t)$ is well-defined if $f$ is of bounded variation and $g$ is continuous. Since Brownian motion has continuous sample paths, this implies that the (Riemann-Stieltjes) integral $\int_a^b B_t \, df(t)$ is ...

3

There's actually two definitions of solution to SDE. Strong and weak. Strong solution. Given a probability space $(\Omega, \mathcal F, \mathcal F_t,P)$ and a Brownian motion $B(t)$ on that space adapted to $\mathcal F_t$, a solution to $dX=fdt+g~dB(t)$ with initial condition $x\in \mathbb R$ is a continuous process adapted to $\mathcal F_t$ such that $X(... 3 Recall that if$X$is a$\mathcal{N}(0, \sigma^2)$random variable then its moments are given by $$\mathbb{E}[X^n] = \begin{cases} 0 \qquad & n \text{ odd} \\ \sigma^n (n-1)!! \qquad & n \text{ even} \end{cases}$$ where$n \in \mathbb{N}$and$!!$is the double factorial. Since$W_s \sim \mathcal{N}(0,s)$we have, by an application of Fubini's ... 2 Your integration of both sides doesn't make any sense to me. You should get $$X_t = g(0) + \int_0^t g'(B_s)\,dB_s + \frac{1}{2} \int_0^t g''(B_s)\,ds$$ and then observe that a martingale has zero bounded variation part, so you will get a martingale only if the last term on the right side vanishes. You also need to ensure that the$dB_s$term is a ... 2 Why is$M^2E[[X^i]_T]\leq M^2 \text{sup}_{t\geq 0} E[(X_t^i)^2]$? ($[X^i]_T$is the variance process) By the definition of the quadratic variation (... you call it "variance process"...) we have $$\mathbb{E}([X]_T^i) = \mathbb{E}((X_T^i)^2)- \mathbb{E}((X_0^i)^2) \leq \mathbb{E}((X_T^i)^2).$$ Hence, in particular,$$\mathbb{E}([X]_T^i) \leq \sup_{t \... 1 You should write$Z_t=e^{Y_t}$and notice that you know the SDE of$Y_t$as it is provided in integral form. Then you can apply Ito's lemma on the exponential function. 1 In the vasicek model, the short rate follows the following dynamic:$dr_{t} = a(b-r_{t})dt + \sigma dW_{t}$With :$a$constant positive, which represents the return force$b$constant positive, which represents the long-term rate$\sigma$constant positive, which represents the volatility The parameters of the model then become$\vartheta$,$a$and$\...

1

I suppose $F(t) = \sigma(W_s,s\leq t)$. I recall that $M_t = \int_0^t\sigma_s dW_s$ is a Wienner integral. Therefore: $M_t \sim \mathcal{N}\left(0, \int_0^t\sigma_s^2ds\right)$ for $s\leq t$, $M_t - M_s$ is independant of $F_s$ We have then : \begin{align} E[e^{-r(t_2-t_1)}e^{\int_{t_1}^{t_2}(\mu_u-\frac{\sigma_u^2}{2})du+\sigma_udW_u}|F(t_1)] &= e^{-...

1

$EX <\infty$ does not imply $E|X| <\infty$ ($EX$ could be $-\infty$). But if $EX$ exists and is a real number then $E|X| <\infty$. This is a consequence of the basic definition of integration in measure theory. Indeed, if $EX$ exists and is finite then $EX^{+} <\infty$ and $EX^{-} <\infty$, so $E|X|=EX^{+}+EX^{-} <\infty$.

1

I would say first compute $f(t,u)=E[e^{-r(u-t)}(\mu(u)-rG_u)|F_t]$ and then compute the derivative. Otherwise you will need justify the derivation under the conditional expectation. Here we suppose that $\mu$ is in $L^1_{loc}(\mathbb{R})$ \begin{align} f(t,u) &= E[e^{-r(u-t)}(\mu(u)-rG_u)|F_t]\\ &= e^{-r(u-t)}\mu(u)- e^{-r(u-t)}r\left(G_t+\int_{t}^...

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