Hot answers tagged

3

Remember that with measures, if two measurable sets $A$ and $B$ are disjoint, then $\lambda(A \cup B) = \lambda(A) + \lambda(B)$. With that in mind, also remember that a singleton set has Lebesgue measure $0$ (can you prove this?). So for each $x \in \Bbb R$, if $\lambda$ is Lebesgue measure, $\lambda( \{x \}) = 0$. Using the above two ideas, this means ...


2

The SDE can be solved similarly as in the Vasicek model. Define $F(t,r(t)) = e^{\alpha t}r(t)$, then \begin{cases} \displaystyle \frac{\partial F}{\partial t} &= \alpha e^{\alpha t} r(t) \\ \displaystyle \frac{\partial F}{\partial r(t)} &= e^{\alpha t} \\ \displaystyle \frac{\partial^2 F}{\partial r(t)^2} &= 0. \end{cases} Applying Itô's lemma ...


2

Without an additional regularisation (say with colored noise), the integral you are trying to compute is strongly divergent. The integrand is dominated by $\xi^2=\left(\frac{dW}{dt}\right)^2$ which is positive and of order $\frac{1}{dt}$ and there is thus no hope to give it a meaning in the stochastic integral sense. You are just integrating infinite ...


2

Here are some comments to your definition. You say 'the consistency of these distributions following from the continuity'; I am not sure it's true. The consistency follows from the fact that you use finite dimensional distributions of the Wiener process. You can use Kolmogorov's theorem without the assumption of continuity. You say $\mathscr{C}[0,\infty) ...


1

For a discrete version of Girsanov's theorem with adapted drift you need to consider a sequence $$ X_n = X_{n-1} + \mu_n +\sigma_n \epsilon_n, $$ where $\{\epsilon_n\}$ are iid standard Gaussian variables, and $\{\mu_n,\sigma_n\}$ are predictable (for simplicity, let $\mu_n$ and $\mu_n/sigma_n$ also be bounded). In order to construct the martingale density, ...


1

It means that $$ lim_{k \to \infty}\Bbb E\left\{\left[\sum_{i = 0}^{k - 1} {X_{t_{i+1}} + X_{t_i}\over 2} \left( W_{t_{i+1}} - W_{t_i} \right) - \int_0^T X_t \circ dW_t\right]^2\right\}=0. $$


1

To prove the desired identity, it is enough to show that for each fixed $x$, $$ H_n(x)=\frac{d^n}{d\alpha^n}e^{\alpha x-\frac{1}{2}\alpha^2}\Big|_{\alpha=0}$$ By completing the square, we may write $$ e^{\alpha x-\frac{1}{2}\alpha^2}=e^{\frac{x^2}{2}}e^{-\frac{(x-\alpha)^2}{2}}$$ and hence $$\frac{d^n}{d\alpha^n}e^{\alpha ...


1

Partial derivatives ignore implicit dependence. $\partial/\partial t f(x)=0$ since there is no explicit $t$ dependence, but $d/dt f$ might not be zero. In your last equation, you have done the actual derivative, not the partial derivatives with respect to $t$.


1

Hint: Apply the polarization identity in $\mathbb{R}$: $$xy = \frac{1}{4}((x+y)^2-(x-y)^2),$$ to $x = \displaystyle \int_{0}^{t} \cos(u) dB_u$ and $y = \displaystyle \int_{0}^{t} \sin(u) dB_u$, together with the Itô-isometry. This will lead to the result.


1

The property you're looking for, namely that $X_t$ is $\mathcal{B}([0,t])\otimes\mathcal{F}_t$-measurable for all $t$, is called progressive measurability. Now your $X$ may not be progressively measurable, but as an adapted (and measurable) process, it has a progressively measurable modification $Y$, and it is easy to see that modification does not change ...


1

The solution of the SDE is $$X_T = X_0 \exp\left\{\left(a-\frac{1}{2}b^2\right)T+bW_T\right\}.$$ Since $W_T \sim N(0,T)$, we have for any $x$ \begin{align} \mathbb{P}(W_T \leq x) &= \mathbb{P}\left(\phi \leq \frac{x}{\sqrt{T}}\right) \\ &= \mathbb{P}\left(\sqrt{T} \phi \leq x\right), \end{align} where $\phi \sim N(0,1)$. Hence, $W_T$ and $\sqrt{T} ...


1

If $f: (0,\infty) \times \Omega \to \mathbb{R}$ is progressively measurable and $$\mathbb{E} \left( \int_0^t |f(s)|^2 \, ds \right)<\infty \quad \text{for all $t \geq 0$}$$ then $$M_t := \int_0^t f(s) \, dB_s, \qquad t \geq 0,$$ is a martingale. This implies in particular $$\mathbb{E}(M_t) = \mathbb{E}(M_0) = 0.$$ Since $f(s,\omega) := ...


1

@webbster: I'm going to flip this around and deal instead with right-limited functions. So let $h:[0,1]\to\Bbb R$ have right limits ate each point of $[0,1)$. Extend $h$ to all of $[0,\infty)$ by setting $h(t)=h(1)$ for $t\ge 1$. It suffices to show that for a fixed $\epsilon>0$, the set $B:=\{t\in[0,1): |h(t)-h(t+)|\ge\epsilon\}$ is countable. For ...


1

If random variable $Y$ takes values in $\mathbb R$, you can find a measurable function $f: \mathbb R\to \mathbb R$, such that $f(N)$ has the same distribution as $Y$, where $N$ is a standard Gaussian random variable. Now, for a Brownian motion $B_t$, define $$ M_t = E(f(B_1)|\mathcal F^B_t)$$


1

Yes it is correct since $t\to e^{\beta t}$ has finite variation, you have no quadratic term in the Ito's lemma. More precisely, if $A$ is a finite variation process and $X$ a semi-martingale you have : $$d(AX)_t =A_t dX_t + X_t dA_t$$ If now $X$ and $Y$ are two semi-martingales you have : $$d(XY)_t =X_t dY_t + Y_t dX_t + d\left\langle X, Y\right\rangle_t$$ ...


1

It depends very much on what you know about stochastic integrals. If you know Novikov's condition, then this might be your first choice. If you know local martingales, then you can use the fact that $(e^{Z_t})_t$ is a local martingale and the boundedness of $u$ to conclude that $e^{Z_t} \in L^2$. In my answer, I'll present a solution which does not require ...


1

Let's show that $\mathbb{E}[\int_0^t (e^{Z_s}(1-X_su_s))^{2}ds] < \infty $ By Cauchy-Schwarz inequality: $\mathbb{E}[\int_0^t (e^{Z_s}(1-X_su_s))^{2}ds] \leq \mathbb{E}[\int_0^t e^{4 Z_s}ds]^{\frac{1}{2}} \mathbb{E}[\int_0^t (1-X_su_s)^{4}ds]^{\frac{1}{2}} $ Because $u$ is bounded, it is clear that $\mathbb{E}[\int_0^t (1-X_su_s)^{4}ds] < ...



Only top voted, non community-wiki answers of a minimum length are eligible