New answers tagged

0

I have not been able to find bibliography on this, but I think the result is correct. I also think your proof is fine, you use properties of the Brownian motion very nicely. Here is just an alternative proof which I personally find clearer. Using Fubini (and assuming $f$ is nice enough): \begin{align} g_T - g_0 = & \int_0^T \big(f(s, T) - f(s, s) \big) ...


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 $\...


0

Hint Using Itô formula, if $f(x,t)=e^{\sigma x+\mu t}$ $$X_T=1+\int_0^T \left(\mu+\frac{1}{2}\sigma ^2\right)e^{\sigma B_t+\mu t}\,\mathrm d t+\int_0^T\sigma e^{\sigma B_t+\mu t}\,\mathrm d W_t.$$ So, $X_t$ is a martingale if $$ \int_0^T \left(\mu+\frac{1}{2}\sigma ^2\right)e^{\sigma B_t+\mu t}\,\mathrm d t=0\quad \text{and}\quad \sigma e^{\sigma B_{\cdot }...


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.


0

The censoring function is the same as the absolute value plus the identity.


0

I am probably very late in answering this, but maybe the proof is useful for someone like me coming just now to this question. Using the integral form of $f$: \begin{align*} Y_t = & - \int_t^T \left[ f(0, u) + \int_0^t \alpha(s, u)ds + \int_0^t \sigma(s, u) dw_s \right] du \,, \end{align*} Using Fubini's theorem (twice): \begin{align*} Y_t = & - \...


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 ...


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^{-...


0

First, your sentence "$W_t=S_t-x$ is a Brownian motion" is wrong, or Brownian motion is bounded below by |x| almost surely. The aim of the integral is precisely to compensate for the absolute value making this not to be a Brownian motion. Have you checked out Revuz-Yor or Chaumont-Yor on this ? (I think your exercise in the latter book)


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 \...


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(...


0

Hint: There is a typo in these notes. Instead of $t$, it should say $\tau_G$ in the limit of integration. The stochastic integral (with $t \wedge \tau_G$ as the limit of integration, that is) is automatically a local martingale starting in 0. It is bounded by the definition of $\tau_G$ and the facts that $G$ is bounded and $u \in C^2(\overline{G})$. Hence, ...


-1

A stochastic integral (from $0$ to $t$) of a local martingale with respect to a Brownian Motion is, by construction, a local martingale starting in $0$. Now, a bounded local martingale $X_t$is an actual martingale by dominated convergence (pick a localising sequence $(T_n)_{n\in \mathbb{N}},$ then $X_{t\wedge T_n}\to X_t$ in $L^1$ as $n\to\infty$, and the ...


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 ...


0

First, it is important to notice that (not unusually) Roger and Williams have, at the point you are looking at, restricted their integrator $M$ to be a continuous $L^2$-bounded martingale. This means that at this stage, Brownian motion is not an allowed integrator, since it is not $L^2$-bounded. The punchline will be that by localisation, once we have the ...


0

Recall that a measurable function $u$ can be decomposed into $u=u^+-u^-$ with $u^\pm$ measurable positive functions and that $|u|=u^+ +u^-$. $u$ is Lebesgue inegrable, if $u^\pm$ are both measurable and $\int u^\pm d\mu<\infty$. So $$\int |u|d\mu=\int u^+ d\mu+\int u^-d\mu <\infty.$$ Also $$\mathbb{E}X=\int_{\Omega} X~d\mathbb{P}.$$


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 $E|X|=EX^{+}+EX^{-} <\infty$.


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 ...


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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