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1

I like "Brownian Motion Calculus" by Ubbo F. Wiersema (2008). I found it more approachable than other books I've seen. But it depends on your level of mathematical sophistication (I'm a mathematical hick).


0

partial answer here 1)Measure, Integral and Probability by Capinski 2)Probability Essentials by Jacods 3)Knowing the odds by Walsh are all good but more elemetary than gut


0

No. Let $\Omega = [0,1]$. Let $C_1 = [1/4,1/2]$ and let $C_2 = [1/3,2/3]$. Then $$ \sigma(C_1) = \{\emptyset, [1/4,1/2], [0,1/4) \cup (1/2,1], [0,1]\}$$ and $$ \sigma(C_2) = \{\emptyset, [1/3,2/3], [0,1/3) \cup (2/3,1], [0,1]\}$$ so that $\sigma(C_1) \cap \sigma (C_2) = \{\emptyset, [0,1]\}$. On the other hand, $C_1 \cap C_2 = [1/3,1/2]$ so that $$ \sigma ...


1

I like the book Brownian Motion - An Introduction to Stochastic Processes by René Schilling & Lothar Partzsch pretty much. It does not only cover stochastic differential equations (in particular, several possibilites are presented how to solve SDEs, e.g. by transforming them into linear SDEs), but also contains a very interesting and detailed exposition ...


1

It suffices to prove that $\int_0^t \phi_s^2 \, ds$ is finite for all $t \in [0,\infty)$. This allows us to define the stochastic integral $$\int_0^t \phi(s) \, dB_s$$ for any $t \in [0,\infty)$ (as a local integral). This means that your proof is correct. In fact, the very argumentation shows that $\Lambda_{\text{loc}}^2$ contains all functions $f$ ...


1

Lemma: Let $(X,\|\cdot\|_X)$ and $(Y,\|\cdot\|_Y)$ be Banach spaces and $D \subseteq X$ dense. If $f:D \to Y$ is an isometry, i.e. $$\|f(x)-f(y)\|_Y = \|x-y\|_X$$ for all $x,y \in D$, then $f$ admits a unique continuous extension which is again an isometry. Proof: For any $x \in X$, we can choose $(x_n)_n \subseteq D$ such that $x_n \to x$. Then it ...


1

Your inequality is equivalent to $$ E(Z^2) \leq \left(\sum_{k=n}^{K_n} \lambda_k \sqrt{E(X_k^2)}\right)^2 $$ i.e. $$ \sum_{k=n}^{K_n} \sum_{j=n}^{K_n} \lambda_k \lambda_j E(X_k X_j) \leq \sum_{k=n}^{K_n} \sum_{j=n}^{K_n} \lambda_k \lambda_j \sqrt{E(X_k^2)}\sqrt{E(X_j^2)} $$ Then just remark by Cauchy inequality we have $$E(X_k X_j) \leq ...


2

We have $dX_t = dW_t - \frac{a}{2}dt$ and $d[X]_t = d[W]_t = dt$ since the deterministic piece $-at/2$ doesn't contribute to the quadratic variation. So, $$ f(X_t) = f(X_0)+\int_0^t f'(X_s)\, dW_s - \frac{a}{2} \int_0^t f'(X_s) \,ds + \frac{1}{2} \int_0^t f''(X_s) \, ds \\ = f(X_0)+\int_0^t f'(X_s)\, dW_s + \frac{1}{2} \int_0^t \big[f''(X_s) - af'(X_s) ...


1

Of course. In order to see this, just write that for $A\in F_s$, \begin{align} E\left[ 1_A \int_s^t fdW \right] &= \lim E\left[1_A \sum f(u_i)(W_{u_{i+1}} - W_{u_i}) \right] \\&= \lim E\left[1_A\right] E\left[ \sum f(u_i)(W_{u_{i+1}} - W_{u_i})\right] \\&= E\left[ 1_A\right] E\left[\int_s^t fdW\right] \end{align}


0

$$ \mathbb{E} F(Y)= \mathbb{E} F(g(X)) = \int F(g(x)) f_X(x) \, \text{d} x = [y=g(x), x=g^{-1}(y)]=\\ =\int F(y) f_X(g^{-1}(y)) | \det \mathcal{J}g^{-1}(y)| \, \text{d}y$$ $\Longrightarrow f_Y(y) =f_X(g^{-1}(y)) $


2

This is straightforward using the change of variable formula, and the characterization of the law of a variable $X$ by the application $$ f \ge 0 \to E[f(X)] $$


1

These pages will help. Here there is the theorem for $\mathbb R^2$.


0

$d Z_t = g(t) d W_t$. If you have problems with this, you should redo basics of stochastic calculus and Brownian motion.


2

The following result can be found as Lemma A.3 of these lecture notes: Let $X_n$ be a sequence of normally distributed random variables with mean zero and variances $\sigma_n^2 \in [0,\infty)$. Suppose $X_n \to X$ in distribution. Then $\sigma_n^2$ converges to some $\sigma^2 \in [0,\infty)$, and $X$ is normally distributed with mean zero and variance ...


2

For any $\omega \in \Omega$, we can construct a sequence $(\Pi_n)_n=(\Pi_n(\omega))_n$ of partitions of $[0,T]$ such that the mesh size $|\Pi_n|$ tends to $0$ as $n \to \infty$ and $$\lim_{n \to \infty} \sum_{t_j \in \Pi_n} (B_{t_j}(\omega)-B_{t_{j-1}}(\omega))^2 = \infty.$$ This means in particular that $$[B,B](T)(\omega) := \sup_{\Pi} \sum_{t_j \in \Pi} ...


1

If you have found a way to solve for $$dY_t = \sqrt{ 1 + Y_t } dt + \sqrt{ 1 + Y_t } dW_t,$$ and want to find a solution for your equation, I suggest you use Girsanov's theorem. This way you will have a weak solution and will be able any probability and expectation you need. You could find the details on how to do this by slightly modifying the arguments in ...


5

It seems that you missed $X_t$ next to $\frac{1}{2}$ term. If this is the case then we want to find a solution of $$\mathrm{d}X_t =\left(\sqrt{1+X^2_t}+ \frac{1}{2}X_t\right) \mathrm{d}t + \sqrt{1+X^2_t}\mathrm{d}B_t.$$ We can re-write it as $$\mathrm{d}X_t = \sqrt{1+X^2}\mathrm{d}t + \sqrt{1+X_t^2}\mathrm{d}B_t + \frac{1}{2}X_t(\mathrm{d}B_t)^2.$$ We may ...


2

Be careful: the chain rule is replaced with the Ito rule. This makes the integration step wrong in your solution.


1

The argument is that the random variables $Z((2j-1)/2^n)$ are not just any random variables and the sigma-algebras $\mathcal F_n$ are not just any sigma-algebras... Each $Z((2j-1)/2^n)$ is proportional to $$Y=B(u+2v)+B(u)-2B(u+v),\qquad u=(j-1)/2^n,\qquad v=1/2^{n+1},$$ and $\mathcal F_n\subseteq\mathcal G_n$ where $$\mathcal G_n=\sigma(B(k/2^n);1\leqslant ...



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