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5

In discrete time at least, the definitions I'm familiar with are fairly straightforward. Given an increasing filtration $\{\mathcal{F}_n\}_{n=0}^{\infty}$, a process $\{X_n\}_{n=0}^{\infty}$ is adapted if each $X_n$ is $\mathcal{F}_n$-measurable. For predictable processes, the random variables are measurable with respect to slightly smaller ...


5

Solution 1: Recall the following two statements. Lemma 1: Let $(Y_t)_{t \geq 0}$ be a supermartingale and $f$ an increasing concave function, then $(f(Y_t))_{t \geq 0}$ is a supermartingale. Lemma 2: Let $(Y_t)_{t \geq 0}$ be a locale martingale such that $Y_t \geq 0$ for all $t \geq 0$. Then $(Y_t)_{t \geq 0}$ is a supermartingale. Lemma 1 is a ...


4

You have an incorrect statement in your summation: you should have $$\displaystyle W_t - W_s = \sqrt{\Delta t} \cdot \sum_{n = 1+s/\Delta t}^{t/\Delta t} Z_n$$ and the variance of this is $$\Delta t \cdot \dfrac{t-s}{\Delta t} = t-s$$ as you would hope


3

Here is the complete solution to the problem including some special cases for an easy start. With analogy to the integrating factor method from ODEs it seems natural to rearrange \begin{align*} \mathrm{d}X_t = (a(t)X_t+ b(t)) \mathrm{d}t + (g(t)X_t+ h(t))\mathrm{d}B_t \end{align*} to the form \begin{align*}\mathrm{d}X_t - X_t \left( ...


3

On the right side of the case $k=0$ of formula (5.1.12) you have the Ito integral $\int_0^t\sigma(s,Z)\,dB_s$. For this to make sense we need $B$ to be a Brownian motion with respect to the filtration $(\mathcal F^Z_t)$, and the simplest way to ensure this is to assume that $Z$ is independent of $B$.


3

Being predictable (as in being measurable with respect to the predictable $\sigma$-algebra) always implies being progressive (as in being measurable with respect to the progressive $\sigma$-algebra), but the other implication only holds in special cases, such as when e.g. the underlying filtration is the usual completion of the filtration generated by a ...


2

I'll rewrite the proof in terms of the process $\tilde{S}_t$, rather than using stochastic integrals. Well, let me first say that the idea of Girsanov's theorem is to 'eliminate' the drift term of $\tilde{S}_t$ by changing the probability measure (to $\mathbb{Q}$). We have, $$d\tilde{S}_t = \tilde{S}_t(\mu-r)dt + \tilde{S}_t \sigma dB_t.$$ Since $\tilde{B}_t ...


2

First of all the product formulae which you wrote are incorrect, it should have been $$ d(X_tY_t) = X_t \mathrm{d}Y_t + Y_t\mathrm{d}X_t + \mathrm{d}X_t\mathrm{d}Y_t.$$ However, there is a lot of notation involved in the above. Let me start without the abuse of notation, for two Ito processes $X$ and $Y$ we have that $$( \star) \ \qquad X_tY_t = X_0Y_0 + ...


2

The error is that you have have assumed that the distribution of $\frac{V_1}{V_1+V_2}$ is distributed as $F(5,14)$. This is not the case, as $V_1$ and $V_1+V_2$ are not independent (have a look at the Characterisation section in here) - on the other hand, $\frac{V_1/5}{V_2/9}$ would be distributed as $F(5,9)$. Going back to the problem at hand, we can ...


2

You need three ingredients: 1. Any continuously differentiable function is a semimartingale, since it is of locally finite total variation, 2. for a Riemann-Stieltjes / Lebesgue-Stieltjes integral you have $d\phi(s) = \phi'(s) ds$ and 3. the covariation is defined in terms of the martingale parts, hence we have $[X,\phi]_t = [X,0]_t = 0$. Hence combining it ...


2

As already stated in the comment, we usually do not have the formula $\int x dx = x^2/2$, if we replace the ordinary Lebesgue/Riemannian integral by the Itô integral, which is exactly the statement of Itô's lemma, as we have $df(B_t) = f'(B_t) dB_t + \frac{1}{2} f''(B_t) dt$. So in your case, if you are interested in calculating $f(B_s) dB_s$ for a ...


2

Check out George Lowther's blog: https://almostsure.wordpress.com/2010/01/03/the-stochastic-integral/ Especially Lemma 3, answers your question "why cadlag?"


1

We have: $$d(X_tY_t)=X_tdY_t + Y_tdX_t+d\langle X,Y\rangle_t={X_t\cos(X_t+Y_t)dW_t}+{Y_t\sin(X_t+Y_t)dW_t}+{\sin(X_t+Y_t)\cos(X_t+Y_t)dt}$$ Can you take it from here or do you need me to continue? EDIT: the two Brownian motions are in fact independent so the last term is 0. As said in a comment, you have to proceed differently.


1

Express $Y$ as an Ito process: $$ dY_t=X_t\,dt = X_t\,dt +\ 0\,dB_t. $$


1

"Progressive" and "optional" are refinements of "adapted" involving some degree of joint measurability of $(\omega,t)\mapsto X_t(\omega)$. The three notions coalesce in discrete time. Consider, for example, the notion "optional" in a discrete-time setting. So let $(\Omega,(\mathcal F_n)_{n\ge 0},\mathcal F,\Bbb P)$ be a filtered probability space. The ...


1

Set $a(x) := x^3$ and $b(x) := x^2$, then $$Y_t := \int_{X_0}^{X_t} \frac{ds}{b(s)} = \left[ - \frac{1}{x} \right]_{X_0}^{X_t} = 1- \frac{1}{X_t}.$$ Now recall that Itô's formula states $$f(X_t) -f(X_0) = \int_0^t f'(X_s) \, dX_s + \frac{1}{2} \int_0^t f''(X_s) b^2(X_s) \, ds$$ where $$\int_0^t f'(X_s) \, dX_s = \int_0^t f'(X_s) \, a(X_s) \, ds + ...


1

Using the integration by parts formula we have $$ \phi(t)B_t=0+\int_0^t\phi(s)dB_s+\int_0^tB_s d\phi(s)+\langle B_s,\phi(s)\rangle_t=\int_0^t\phi(s)dB_s+\int_0^tB_s \phi'ds $$ using the fact that $B_0=0$ for a Brownian Motion and $\langle B_s,\phi(s)\rangle_t=0$ since $\phi$ is a deterministic function of time.


1

Your calculation is almost good. However, your application of Ito's lemma is not entirely correct. I will outline the details for you: Since $X(t) = \int_{0}^{t} \sigma(s) dW_s$, it follows that $dX(t) = \sigma(t)dW_t$. Define $Z(t) = \exp(iuX(t)) = f(t,X(t))$. Now, an easy calculation shows $$\begin{cases} \displaystyle\frac{\partial f}{\partial t} = 0, \\ ...


1

To answer your first question, stochastic integrals of this type are Ito integrals. They are defined as a specialized limit of a Riemann-like sum with respect to a partition. Under appropriate conditions on processes $X_s$ and $B_s,$ we can define the integral as a limit of left-hand (non-anticipatory) sums: $$\int_0^t X_s dB_s = \lim_{n \to \infty} ...


1

First, the Ito interpretation prevents you from using backwards Euler, you can only use forward Euler. Second, the difference you are considering is actually $W_{t_1}-W_{t_0}=W_h-W_0$, and the increments of the Wiener process follow the law $$ W_s-W_t\sim N(0,|s-t|)\sim N(0,1)·\sqrt{|s-t|} $$ which should explain where the square root comes from.


1

Yes, Itô's formula for jump processes can be proved using Taylor's formula. For simplicity of notation, I'll just discuss that $f$ does not depend on the time $t$, i.e. $Y_t := f(X_t)$. By Taylor's formula, we have $$f(X_{t+h})-f(X_t) = f'(X_t) (X_{t+h}-X_t) + \frac{1}{2} f''(X_t) (X_{t+h}-X_t)^2 + R(X_t,X_{t+h})$$ where $$R(x,y) := \left( \int_0^1 ...


1

Theorem Let $X$ be a nonnegative random variable. If $\mathbb E(X)<\infty$, then $\mathbb P(X<\infty)=1$. Proof Assume on the contrary that $\mathbb P(X<\infty)<1$, or equivalently, $\mathbb P(X=\infty)>0$. Then $$ \mathbb E(X)\ge\infty\cdot P(X=\infty)=\infty, $$ a contradiction. QED.


1

For Brownian motion it is true that $B=C$. This means that if $f\in B$ there exists $g\in C$ such that $f(\omega,t)=g(\omega,t)$ for $\Bbb R\times\mu$-a.e. $(\omega,t)\in\Omega\times[0,T]$. This does not mean that $t\mapsto f(\cdot,t)$ is a predictable process. The matter is discussed in some detail in Chapter 3 of Introduction to Stochastic Integration by ...



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