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4

Perhaps the most commonly used example is the natural filtration of Brownian motion, i.e. $\mathcal{F}_t=\sigma(W_s:s \in [0,t])$. This sort of thing is usually how filtrations are created.


3

Any càdlàg function $f: [0,T] \to \mathbb{R}$ has at most finitely many jumps with jump size $>\epsilon$ for any (fixed) $\epsilon>0$, see e.g. this answer. The estimate "$\leq [X,X]_t$" is not used to conclude that $\log V_t$ is of bounded variation, but to show that the series $\sum_{s \leq t} (\log(1+U_s)-U_s)$ is absolutely convergent. Recall ...


3

What you've written is the result that you get when you perform the simplest approximation of Ito's formula. In this case Ito's formula reads $$e^{B_t} = 1 + \int_0^t e^{B_s} dB_s + \int_0^t \frac{1}{2} e^{B_s} ds.$$ For small $t$, $\int_0^t e^{B_s} dB_s \approx B_t$ (replacing the integrand by its value at $0$, which is $1$) and $\int_0^t \frac{1}{2} ...


3

$E^x$ is the expectation under the path measure of $W(\cdot)$ started at $W(0) = x$. Example: $E^x[W(t)] = E^x[W(t)-W(0)+W(0)] = E^x[W(t)-W(0)] + x = x$ since $W(t)-W(0) \sim \mathcal{N}(0,t)$. In plain English, $E^x$ is just ordinary expectation with the understanding that the starting point of BM is a variable, $x$.


3

$\nu \circ f^{-1}$ denotes the push-forward measure (or image measure) of $f$ with respect to $\nu$, i.e. $$\nu_1(B) = (\nu \circ f^{-1})(B) = \int 1_B(f(x)) \, \nu(dx)$$ for any Borel set $B$. This implies $$\int g(x) \, \nu_1(dx) = \int g(f(x)) \, \nu(dx)$$ for any $g \in L^1(\nu_1)$.


3

Let me give you first a non-rigorous "proof" of the statement; afterwards I'll explain how to justify it. If we define $$f_n(x) := n x 1_{[0,1/n]}(x) + 1_{[1/n,\infty)}(x),$$ then $f_n \uparrow 1_{(0,\infty)}$. Obviously, $f_n$ is not differentiable at $x=0$ at $x=1/n$, but for the moment we forget about this (that's the non-rigorous part) and apply ...


3

$$E\left[H_{t_{i-1}}^2\left(B_{t_i}-B_{t_{i-1}}\right)^2\right] = E\left[E\left[H_{t_{i-1}}^2\left(B_{t_i}-B_{t_{i-1}}\right)^2 \mid \mathbb{F}_{t_{i-1}}\right] \right] = E\left[H_{t_{i-1}}^2E\left[\left(B_{t_i}-B_{t_{i-1}}\right)^2 \mid \mathbb{F}_{t_{i-1}}\right] \right] = E\left[H_{t_{i-1}}^2E\left[\left(B_{t_i}-B_{t_{i-1}}\right)^2 \right] \right] = ...


2

It is typical, at least in my experience, that "random walks" unless otherwise stated are sums of iid random variables. You can form two random walks with iid copies of normal distributions with means $\mu, \tilde{\mu}$ and variances $\sigma^2$ respectively. Then by the law of large numbers, the averages of their partial sums will converge to the their ...


2

If $X_n\to X$ almost surely and $X_n$ is independent of $\mathcal G$, then $X$ is independent of $\mathcal G$. We use actually this. Call $X$ the left hand side of (2). We can show that for each continuous and bounded function $f\colon\mathbf R^n\to\mathbf R$ and $E\in\mathcal F_t$, we have $\mathbb E\left[f(X)\mathbf 1(E)\right]=\mathbb E[f(X)]\mu(E)$ ...


2

Although it is correct to do it this way, it's straight overkill; from my point of view. Here is a much easier solution: Since $W_t \sim N(0,t)$, we have $$\mathbb{E}e^{\lambda W_t} = \exp \left(\frac{1}{2} \lambda^2 t \right).$$ Differentiating both sides with respect to $\lambda$ yields $$\mathbb{E}(W_t e^{\lambda W_t}) = \lambda t \exp \left( ...


2

You can look at $$\mathcal B_r=\Big\{A\subseteq [0,1]\times[0,1]\ \bigg|\ A\cap [0,r]\times[0,1]\text{ is Borel, }\land (A\cap(r,1]\times[0,1]\in\{(r,1]\times[0,1],\varnothing\})\Big\}$$ For $r\in[0,1]$. Where Borel means the standard Borel sets of $[0,1]\times[0,1]$. Then for $r\leq s$ we have $\cal B_r\subseteq B_s$, since whenever $A\in\cal B_r$, we ...


2

An example of a filtered probability space is $([0,1), \mathbf{B}_{[0,1)}, \{\mathcal{F}_t\}_{t\geq 0} , \mathbb{P})$, where, for each $t\in \mathbb{R}, t\geq 0$, define $$ \mathcal{F}_t = \mathbf{B}_{\left[0\,,\,1-\frac{1}{t+1}\right]}=\textrm{ the Borel $\sigma$-algebra defined in } \left [0\,,\,1-\frac{1}{t+1} \right] $$ Note that, by definition of ...


1

No, we cannot apply the dominated convergence theorem in this way (see all the comments above). For fixed $R>0$ denote by $$\tau := \inf\{t>0; |B_t| \geq R\}$$ the exit time from $(-R,R)$. Moreover, we denote by $$H \bullet B(T) := (H \bullet B)(T) := \int_0^T H(s) \, dB_s$$ the stochastic integral of $H$. By Markov's inequality and Itô's ...


1

Is your question just about $\sigma$-algebras or about filtration?? If you are just looking for an example of an uncountable increasing family of $\sigma$-algebras, there are many examples. One of the simplest examples is: for each $t\in \mathbb{R}, t\geq 0$, define $ \mathcal{F}_t = \mathbf{B}_{[0,t+1]}\textrm{ the Borel $\sigma$-algebra defined in } ...


1

Assuming $T$ is just a number and not a stopping time, $$ E\int_0^T S_t e^{r(T-t)}dt $$ is just just a double integral of a positive function, so we can exchange the order of integration $$ = \int_0^T E [S_t] e^{r(T-t)} dt $$ $$ = \int_0^T S_0 e^{\mu t} e^{r(T-t)}dt $$ and I presume you can finish from here as nothing left is random.


1

Indeed, the composition of measurable functions remains measurable. In your case, both, $M(t)$ and $x\to x²$ are measurable.


1

Fix a ball $B(x,r)$ and consider $\tau= \inf{t>0, |B_t-x| = r}$. Now compute by Itô's formula $$f(B_\tau) = f(B_0) + \int_0^\tau f'(B_s)\, dB_s + \frac{1}{2}\int_0^\tau \Delta f(B_s)\, ds $$ Take expectations: $$\Bbb{E}[f(B_\tau)] = f(B_0) + \Bbb{E}\bigg[\int_0^\tau f'(B_s)\, dB_s\bigg] + \frac{1}{2}\int_0^\tau \Delta f(B_s)\, ds $$ Since ...


1

As $\mathcal{F}_{t_n} \supseteq \mathcal{F}_{t+}$, we know that $X_n$ is independent of $\mathcal{F}_{t+}$. Therefore, $$(B_{t+s_1}-B_t, \ldots,B_{t+s_m}-B_{t}) = \lim_{n \to \infty} (B_{t_n+s}-B_{t_n},\ldots,B_{t_n+s_m}-B_{t_n}) = \lim_{n \to \infty} X_n$$ is also independent of $\mathcal{F}_{t+}$. Lemma: Let $\mathcal{F}$ be a $\sigma$-algebra and ...


1

This is only an answer for a part of the question To answer your second question $2)$: No, in general it's not true that a function bounded by a function of finite variation is itself of finite variation, for example take $$ f(x)=\begin{cases} 0, x=0 \\ \sin(\frac{1}{x}), \text{otherwise}\end{cases} $$ which is not of bounded variation although $f(x)$ is ...


1

If you compare your expression with $$ u(x,t) = E^Q\left[ \int_t^T e^{- \int_t^r V(X_\tau,\tau)\, d\tau}f(X_r,r)dr + e^{-\int_t^T V(X_\tau,\tau)\, d\tau}\psi(X_T) \Bigg| X_t=x \right] $$ where $dX = \mu(X,t)\,dt + \sigma(X,t)\,dW^Q$ (this is pasted from the wikipedia article ), we see that in your case the process $X_t$ is simply $B_t$ so $\mu =0$ and ...


1

$$\frac{d P}{d x} = \frac{2f'(x)}{\sigma^2}P$$ $$\frac{d^2 P}{d x^2} = \frac{2f''(x)}{\sigma^2}P + \frac{2f'(x)}{\sigma^2}P' = \frac{2f''(x)}{\sigma^2}P + \frac{4(f'(x))^2}{\sigma^4}P$$ It appears that you included the second term in the above derivation but forgot the first term. Adding it in gives you zero.


1

I believe you are confusing the mean and mode of your distribution, which is asymetric. $S_t$ follows a lognormal distribution with parameters $\mu = -\frac{1}{2} t$ and $\sigma^2 = t$. So, $E [S_t] = e^{\mu + \frac{1}{2} \sigma^2} = e^{-\frac{1}{2}t + \frac{1}{2} t} = e^0 = 1$, but the mode of the distribution is $e^{\mu - \sigma^2} = e^{-\frac{3}{2}t}$. ...


1

Fubini's theorem on $[0,t]\times\Omega$ gives $$\mathbb{E}\left[\int^{t}_{0} W^2_s\, ds\right] =\int_\Omega \int^t_0 W_s^2(\omega)\,ds\,\mathbb{P}(d\omega) =\int^t_0\int_\Omega W_s^2(\omega)\,\mathbb{P}(d\omega)\,ds = \int^{t}_{0} \mathbb{E}[W^2_s]\, ds.$$


1

There is not one answer to this question. Your question is really one about modeling, rather than being strictly about mathematics, so the best answer depends on what you're trying to model. Two answers that come to mind for me are as follows. One would be to instead consider applying small iid normally distributed perturbations every $\Delta t$ and ...


1

so we want to compute the Stratonovich $S$ integral, which is defined $$ S:=\int_0^TW_t\circ dW_t:=\lim_{n\to\infty}\sum_{i=0}^{n-1}\frac{1}{2}(W_{t_{i+1}}+W_{t_{i}})(W_{t_{i+1}}-W_{t_{i}}) $$ while we have a partition $P$ of $[0,T]$ with stepsize $h=T/n$, so $$ t_0=0,t_1=h,\dots,t_k=kh,\dots,t_n=T $$ so as $n\to\infty$ we have $|P|\to 0$. Now let's ...


1

So let's see that $X_{\sigma^n} \to X_{\sigma}$ in $L^1$. Since $\sigma^n \downarrow \sigma$ right continuity gives us that $X_{\sigma^n} \to X_{\sigma}$ almost surely. Now call $X(n) = X_{\sigma^n}$ and $\mathcal{F}_n = \mathcal{F}_\sigma$. Note that 1) $X(n)$ is $\mathcal{F}_n$ adapted 2) $X(n) \in L^1$ for every $n$ 3)$\Bbb{E}[X(n+1) - X(n)\vert ...



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