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3

Let $(X_t,\mathcal{F}_t)_{t \geq 0}$ be an (càdlàg) $L^2$-martingale, i.e. a martingale which satisfies $$\sup_{t < \infty} \mathbb{E}(X_t^2)<\infty.$$ Then it follows from the Doob-Meyer decomposition that there exists a unique increasing previsible process $(A_t)_{t \geq 0}$ such that $A_0=0$ and $$(X_t^2-A_t,\mathcal{F}_t)_{t \geq 0} \,\, ...


3

Yes. If $X$ and $Y$ are independent random variables, and $f, g$ are (measureable) functions, then we have that $f(X)$ and $g(Y)$ also are inependent random variables. This generalizes to functions of multiple variables, which gives your case.


3

The mapping $\omega \mapsto X_{\xi(\omega)}(\omega)$ is measurable if the corresponding process is jointly measurable, i.e. if $$(t,\omega) \mapsto X(t,\omega) \tag{1}$$ is $\mathcal{B}([0,T]) \otimes \mathcal{A} /\mathcal{B}(\mathbb{R})$-measurable for $T>0$. This can be proved in the following way: Let $f$ be a continuously differentiable function, ...


3

In answer to my own question, the equality can be shown as follows. First, we realize that $\int_0^{2\pi}\int_0^b\int_0^b r_1r_2\frac{J_1\left (\alpha\sqrt{r_1^2+r_2^2-2r_1r_2\cos(\theta)}\right )}{\alpha\sqrt{r_1^2+r_2^2-2r_1r_2\cos(\theta)}} dr_1dr_2d\theta\\=\frac{1}{2\pi}\int_0^{2\pi}\int_0^{2\pi}\int_0^b\int_0^b r_1r_2\frac{J_1\left ...


2

...we can't use Ito [l]emma because $\displaystyle\int_0^t W_sds$ is not in the form $X_t = \displaystyle\int_0 ^t \sigma_s dW_s + \int_0 ^t \mu_s ds$. Actually $X_t=\displaystyle\int_0^t W_sds$ is in the form $X_t = \displaystyle\int_0 ^t \sigma_s dW_s + \int_0 ^t \mu_s ds$ with, for every $t\geqslant0$, $$\sigma_t=0,\qquad \mu_t=W_t.$$


2

Well, in the first case, when you consider a Brownian motion $B_t$, the following is usually understood: You are considering a background probability triple $(\Omega,\mathcal{F},P)$, with $\Omega$ being some set, $\mathcal{F}$ being some $\sigma$-algebra and $P$ being some probability space, and you consider a measurable mapping $B:\Omega\to C[0,\infty)$ ...


2

I'm sorry, I was very stupid. We can just apply Ito's formula and get: \begin{align*} & \int_a^b f(t)dW_t=f(b)W_b-f(a)W_a-\int_a^b W_t f'(t)dt. \end{align*} This yields that indeed we can find the desired pathwise upper bound of $\int_a^b f(t)dW_t$ in terms of $||f||$, $||f'||$, $\sup_{t \in [a,b]}|W(t)|$.


1

The condition $$\int_{\mathbb{R} \backslash \{0\}} \min\{1,z^2\} \nu(dz)<\infty$$ is equivalent to $$\int_{|z| \leq 1} z^2 \, \nu(dz) < \infty \quad \text{and} \quad \int_{|z| \geq 1} \nu(dz) < \infty.$$ Let's discuss them separately; for simplicity of notation we consider the $1$-dimensional case. Any Lévy process $(X_t)_{t \geq 0}$ has ...


1

For simplicity, I'll assume that $(M_t)_{t \geq 0}$ is a continuous martingale. The argumentation is similar for a local martingale, but more technical. Set $$I_n(t) := \int_0^t I_{n-1}(s) \, dM_s, \qquad n \in \mathbb{N}, t \geq 0, \tag{1}$$ and suppose that $$(n-1) I_{n-2}(t) = I_{n-2}(t) M_t - I_{n-3}(t) \langle M \rangle_t, \qquad t \geq 0. \tag{2}$$ ...


1

In the second case it is not necessarily the case that $dX_t=\mu_t\,dt+\sigma_t\,dB_t$. But the second case should imply the first. What is $dX_t\cdot dX_t$ for the first case ?


1

$$\delta_1 + 2 \delta_2 +3 \delta_3+\cdots=\sum_{j\geqslant1}j\,\delta_j=\sum_{j\geqslant1}j\,E\left[\frac{(\lambda S)^j}{j!} e^{-\lambda S} \right]=\sum_{i\geqslant0}E\left[\frac{(\lambda S)^{i+1}}{i!} e^{-\lambda S} \right]=E\left[\lambda S\right]$$


1

You just have to be careful with the notation because $t$ can appear in two arguments. The forward rate is $$f(t,T)=f(0,T) + \int_{0}^t \alpha(s,T)ds+ \sigma W_t.$$ The short rate is $$r_t=f(t,t)=f(0,t) + \int_{0}^t \alpha(s,t)ds+ \sigma W_t = F(t,X_t),$$ where $F_X = 1$, $F_{XX} = 0$ and $$X_t = \int_{0}^t \alpha(s,t)ds+ \sigma W_t,\\dX_t = ...


1

In http://en.wikipedia.org/wiki/Feynman%E2%80%93Kac_formula#cite_note-1 they reference as having a proof Pham, Huyên (2009). Continuous-time stochastic control and optimisation with financial applications. Springer-Verlag


1

Actually if $\theta=b/(a+b)$ the drift is zero on $[b,+\infty)$ hence the chain is null recurrent. Positive recurrence holds for every $\theta\lt b/(a+b)$, which we assume from now on. As you explain, the stationary distribution solves $\pi=\pi P$. For every $n\geqslant a$, this reads $$\pi(n)=\theta\pi(n-a)+(1-\theta)\pi(n+b).$$ Such linear systems are ...


1

Hint Apply Itô's formula to $f(t,x) := 2+t+e^x$. This gives you an expression of the form $$X_t = X_0 + \int_0^t \sigma(s,e^{W_s}) \, dW_s + \int_0^t b(s,e^{W_s} ) \, ds.$$ Now use that $$e^{W_s} = X_s-2-s.$$


1

Such a process does not exist. Two possible argumentations: Since a stochastic integral with respect to Brownian motion is a martingale, the expectation is constant, i.e. $$\mathbb{E} \left( \int_0^t X_s \, dB_s \right) = \mathbb{E} \left( \int_0^t X_s \, dW_s \right) = 0.$$ (To be more precise: Here, we need some integrability assumptions on $(X_t)_{t ...


1

Let $(Z_t)_{t \geq 0}$ be a Markov process with generator $B$. Then we know from Dynkin's formula that $$u(Z_t)-u(Z_0)- \int_0^t Bu(Z_s) \, ds$$ is a martingale if $u$ is "nice" (e.g. bounded and in the domain of the generator). In particular, if $u$ satisfies $Bu=0$, then we see that $$\mathbb{E}^xu(Z_t) = \mathbb{E}^x u(Z_0) = u(x).$$ This means the ...


1

You get $$ \log\frac{X\vee Y}{X\wedge Y} \leq z \iff X\vee Y \leq \mathrm e^z(X\wedge Y) $$ $$ \iff X\leq\mathrm e^z Y,Y\leq X \text{ or }Y\leq\mathrm e^z X,Y\geq X $$ where $\wedge$ if for $\min$, $\vee$ is for $\max$, and the latter two events have intersection of zero probability.



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