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5

I think it's pretty hard to find a book which covers martingale theory; usually, books either give just an introduction or they focus on one particular aspect of martingale theory. I'll list some books which might be of interest and sketch (roughly) which parts they cover: David Williams: Probability with Martingales (Basic properties, optional stopping, ...


4

I will treat the case where M is a continuous semimartingale. Unfortunately it is generally not the Riemann Stieltjes integral. You know that the Stieltjes measure of g, is only defined if g has finite variation. However, as you know, many stochastic processes does not have sample paths with finite variation, and therefore such an integral does not exist. ...


3

Let $X_t = \int_0^t e^s dB_s$. Then $$r_t = f(t, X_t) = .1 + .1e^{-t} + e^{-t}X_t$$ By Ito's Lemma: $$dr_t = \frac{\partial f}{\partial x}dX_t + (\frac{\partial f}{\partial t} + \frac{1}{2}\frac{\partial^2f}{dx^2}Var(X_t))dt$$ So: \begin{eqnarray*} \frac{\partial f}{\partial x}dX_t + (\frac{\partial f}{\partial t} + ...


3

It is true that the second property can be deduced from the first one. Indeed, the first property implies that $(B_t)$ has stationary, independent increments. Hence, the following statement would give you the implication in the forward direction: Proposition: Any real-valued stochastic process with stationary, independent increments has the elementary ...


3

You have correctly found $x_t$ as $$x_t=x_0e^{-\theta t}+\mu (1-e^{-\theta t})+\sigma e^{-\theta t}\int_0^te^{\theta s}dW_s$$ We can rewrite this as $$x_t=a_t-b_tc_t$$ where $$\begin{align} &a_t=x_0e^{-\theta t}+\mu (1-e^{-\theta t})\\\\ &b_t=-\sigma e^{-\theta t}\\\\ &c_t=\int_0^te^{\theta s}dW_s \end{align}$$ Note, ...


3

The RHS is direct to evaluate. The variance of $B_s$ is $s$ so: $$\int_0^t E[B_s^2] ds = \int_0^t s ds = \frac{t^2}{2}$$ You are right about use Ito's Lemma for the LHS. By Ito: $$B_t^2 = t + 2\int_0^t B_s dB_s$$ So \begin{eqnarray*} E[(\int_0^t B_s dB_s)^2] &=& E\left[\left(\frac{B_t^2 - t}{2}\right)^2 \right] \\ &=& ...


3

$X_n$ is an $(\mathcal{F}_n)$-martingale: $$\mathbb{E}[X_k(X_m-X_l)]=\mathbb{E}[X_k\mathbb{E}[(X_m-X_l)|\mathcal{F}_k]]$$ $$=\mathbb{E}[X_k(X_k-X_k)]=0$$


2

You have $Y_t = f(X_t, t) = \exp(X_t - t^3/6)$. Assume $dX_t = m(t) dt + s(t) dW_t$ so $(dX_t)^2 = s(t)^2 dt$. Now apply Ito's lemma to $Y_t$: $$ \begin{split} dY_t &= \exp(X_t - t^3/6) \frac{-t^2dt}{2} + \exp(X_t - t^3/6) (dX_t) + \exp(X_t - t^3/6) (dX_t)^2/2 \\ &= Y_t \left[ \frac{-t^2dt}{2} + ...


1

The calculation in the proof shows that $$\begin{align*} X_t := |M_t N_t - \langle M,N \rangle_t| &\leq \sup_{s \geq 0} |M_s| \cdot \sup_{s \geq 0} |N_s| + \sqrt{\langle M \rangle_{\infty}} \sqrt{\langle N \rangle_{\infty}} =: X \in L^1. \end{align*}$$ Consequently, $$\int_{X_t \geq R} X_t \, d\mathbb{P} \leq \int_{X \geq R} X \, d\mathbb{P}.$$ Since ...


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

The increments are stationary. Since $B_t - B_s$ is the increment over the interval $[s, t]$, it is the same in distribution as the incremeent over the interval $[s-s, t-s] = [0,t-s]$. Hence, $$B_t-B_s \sim B_{t-s}-B_0.$$ But $B_0 = 0$ almost surely, so that: $$B_t-B_s \sim B_{t-s}.$$ Finally, $B_{t-s} \sim \mathcal{N} (0,t-s)$. We didn't use the ...


1

As @muaddib pointed out, you have simply rewritten the definition of $\hat{W}_t$ - but this doesn't show that $(\hat{W}_t)_{t \geq 0}$ is a martingale. Hints: Show that $(X_t)_{t \geq 0}$ is a martingale with respect to its canonical filtration $$\mathcal{F}_t := \sigma(X_s; s \leq t) = \sigma(W_s; s \leq e^{\beta t}-1).$$ Conclude that $(\hat{W}_t)_{t ...


1

Regarding the first question: The $0$ and $t$ in the typical statement of Ito's Lemma aren't required. In general you can say: $$f(X_{v}) = f(X_{u}) + \int \limits_{u}^{v} f'(X_{s}) b(s, \omega) \,ds + a(s,\omega) \,dB_{s} + \frac{1}{2}\int \limits_{u}^{v} f''(X_{s}) a^{2}(s, \omega) \,ds. $$ Then set $u = 0, v = t$ for the typical statement, and $u = t, v ...


1

Recall two facts on progressive measurability: Let $(Y_t)_{t \geq 0}$ be a progressively measurable (real-valued) stochastic process and $f: \mathbb{R} \to \mathbb{R}$ Borel-measurable. Then $(f(Y_t))_{t \geq 0}$ is progressively measurable. Any adapted stochastic process with continuous sample paths is progressively measurable. Since $(X_t)_{t \geq 0}$ ...


1

Write $$\sum_{k=0}^{\infty} \frac{e^{-\lambda t} (\lambda t)^k}{k!} (f(N_t+k)-f(N_t)) = I_1+I_2+I_3$$ where \begin{align*} I_1 &:= e^{-\lambda t} \cdot 1 \cdot (f(N_t+0)-f(N_t))=0, \\ I_2 &:= e^{-\lambda t} \lambda t (f(N_t+1)-f(N_t)) \end{align*} and $$I_3 := \sum_{k \geq 2} e^{-\lambda t} \frac{(\lambda t)^k}{k!} (f(N_t+k)-f(N_t)).$$ It ...


1

As Brenton stated in the comments, the $u$ could be pretty much anything, because it goes away after you apply the bounds on the definite integral. Perhaps a bit clearer would have been to designate it $t'$ instead of $u$: $$\mathbb E I^2(t)=\mathbb E\int_0^t \Delta^2(t') dt'.$$ This makes it more semantically clear that you're integrating over time.


1

On your second question, you're correct that the first term is $Var(\hat{Y})$, and you are asking why $Var(\hat{Y})=\frac{1}{N}Var(\hat{f})$? $$Var(\hat{Y}) = Var\left(\frac{1}{N}\sum_{n=1}^N\hat{f^{(n)}}\right)$$ $$= \frac{1}{N^2}Var\left(\sum_{n=1}^N\hat{f^{(n)}}\right)$$ $$ = \frac{1}{N^2}(N)Var(\hat{f})$$ $$ = \frac{1}{N}Var(\hat{f}).$$


1

I think the notation of the Monte Carlo simulation is awkward for purposes of the proof. Let's simplify and unify the notation a little. This is a 'standard' result in the theory of estimation-- slightly disguised. MSE equals variance plus squared bias. You want to estimate the parameter $\tau$ of a distribution that produces observations $T_1, \dots, ...


1

Hints: Note that $f$ does not depend on the time $t$, therefore the term $\frac{\partial}{\partial t} f$ is superfluous. Take expectation on both sides, then the stochastic integral $\dots dW_t$ vanishes, because it is a martingale. Use Fubini's theorem and the fundamental theorem of calculus, $$\frac{1}{t} \int_0^t \mathbb{E}^xg(X_s) \, ds \stackrel{t ...


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

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 $\mathbb E\left[f(X)\mathbf 1(E)\right]=\mathbb E[f(X)]\mu(E)$ ...


1

By the definition of the Itô integral, we know that $$\sum_{j=1}^n V_j \cdot \Delta_j \to \int_0^t B_s \, dB_s. \tag{1}$$ Note that we can write $$I_1(n) = \sum_{j=1}^n V_{j+1} \Delta_j = \sum_{j=1}^n \underbrace{(V_{j+1}-V_j)}_{\Delta_j} \Delta_j + \sum_{j=1}^n V_j \Delta_j. \tag{2}$$ By $(1)$, the second term at the right-hand side converges to ...


1

Yes, it is a consequence of Fubini's theorem. By Fubini, we have $$\text{var}(I(T)) = \mathbb{E} \left( \int_0^T W(u) \, du \int_0^T W_v \, dv \right) = \int_0^T \int_0^T \underbrace{\mathbb{E}(W_u W_v)}_{\text{cov}(W_u,W_v)} \, du \, dv.$$ Using again Fubini's theorem, we find $$\begin{align*} \int_0^T \int_0^T \text{cov}(W_u,W_v) \, du \, dv &= ...


1

Hint: Apply Tonelli's (or Fubini's) theorem and use that $$\mathbb{E}\exp(W_u) = \exp \left( \frac{1}{2} u \right)$$ since $W_u \sim N(0,u)$.


1

The result is true, and has nothing to do with topology or countability. Let ${\cal C}$ be the collection of subsets of $E^I$ of the form $\pi_i^{-1}(E)$ for some $i\in I$ and $E\in{\cal E}$. Then use the fact that $X^{-1}(\sigma({\cal C}))=\sigma(X^{-1}({\cal C}))$.


1

Doob says: $B_{\tau \wedge t} = E[B_{t \wedge t}\mid \mathcal{F}_\tau]$. Jensen says: $$E[\lvert B_{\tau \wedge t} \rvert] = E[\lvert E[B_{t \wedge t}\mid \mathcal{F}_\tau]\rvert] \leq E[ E[\lvert B_{t \wedge t}\rvert\mid \mathcal{F}_\tau]] = E[\lvert B_t \rvert] < \infty$$


1

To keep notation simple, we write $S_t$ instead of $S_t^i$, $\sigma_t^j$ instead of $\sigma_{t}^{ij}$ and so on. That's okay, because $i$ is a fixed number throughout this calculation. Suppose $(X_t)_{t \geq 0}$ is an Itô process of the form $$dX_t = b(t) \, dt + \eta(t) \, dW_t$$ where $\eta = (\eta_1,\ldots,\eta_m)$ and $(W_t)_{t \geq 0}$ is an ...



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