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3

Applying Itô's formula for the Itô process $$X_t := \int_0^t f(s) \, dW_s$$ and $f(x) := x^4$ gives $$X_t^4 = 4 \int_0^t X_s f(s) \, dW_s + \frac{4 \cdot 3}{2} \int_0^t X_s^2 f(s)^2 \, ds.$$ Taking expecation yields $$\begin{align*} \mathbb{E}(X_t^4) &= 6 \int_0^t \mathbb{E}(X_s^2 f(s)^2) \, ds \\ &\leq 6 \mathbb{E} \left(\sup_{s \leq t} |X_s|^2 ...


3

The definition of an Itō stochastic integral $ \int_0^t H_s \; dX_s$ (with respect to a semimartingale $X_t$) is the limit of $\sum_{j=1}^n H_{t_{j-1}} (X_{t_j} - X_{t_{j-1}})$ for partitions $0 = t_0 \le t_1 \le \ldots \le t_n= t$ of $[0,t]$ with mesh size going to $0$. For $H = 1$ the sum telescopes to $X_t - X_0$. So in this case the identity follows ...


3

It is just notation. Think of it as the measure of the 'infinitesimal' slice $dx$. I like the notation in that it is suggestive of a Darboux sum, but it is a little cumbersome. Alternative notations are $\int e^{i(u,x)} d \mu(x)$, or $\int f d\mu$ with $f(x) = e^{i(u,x)}$.


2

Here's a different argument. Define $$X_t = \int_0^t f(s,\omega) dW_s \\ Y_t = X_t^4 \\ Z_t = X_t^2.$$ Apply the Ito formula to $Y_t$ and $Z_t$: $$Y_t = 4 \int_0^t f(s,\omega) X_s^3 dW_s + 6 \int_0^t f(s,\omega)^2 X_s^2 ds \\ Z_t = 2 \int_0^t f(s,\omega) X_s dW_s + \int_0^t f(s,\omega)^2 ds$$ Now substitute and take expectations, thereby canceling the ...


2

Brownian motion is often defined as a probability measure on the space $C([0,\infty),\mathbb R)$ of real valued continuous functions defined on $[0,\infty)$. As such, a change of probability measure on this space can yield processes pretty different from the ones you have in mind: the means can change, the variances can change, the gaussianity can fail, etc. ...


2

Notice that $M_n$ is $\mathcal F_n$ measurable hence so is $S_n$, and $M_n$ is integrable since it is bounded. It remains to check that $\mathbb E(S_{n+1} \mid \mathcal F_n)=S_n$. To this aim, we have to compute $\mathbb E(e^{bM_{n+1}} \mid \mathcal F_n)$. As $$e^{bM_{n+1}}=\underbrace{e^{bM_{n+1}-bM_n}}_{\mbox{independent of }\mathcal F_n}\cdot ...


2

Suppose that $S_t$ and $S_t'$ are perfectly correlated for each $t>0$ with correlation $1$. Then it follows from the very definition that $S_t = S_t'$ for all $t>0$. Applying Itô's formula (with $f(x)=x^2$ yields $$\begin{align*} \underbrace{(S_T-S_T')^2}_{0} &= 2 \int_0^T \underbrace{(S_t-S_t')}_0 d(S_t-S_t') + \int_0^T \underbrace{\langle S-S' ...


2

The first important thing to notice is that we cannot expect convergence for fixed $\omega$, i.e. the sum $$\sum_{k=1}^K f(t_k) (X_{t_k}(\omega)-X_{t_{k-1}}(\omega))$$ does in general not converge as $K \to \infty$. Recall the following lemma: Lemma: Let $\alpha: [a,b] \to \mathbb{R}$. If $$A(f) := \int_a^b f(t) \, d\alpha(t) := \lim_{K \to \infty} ...


1

When you are not sure, there is not harm in treating entries as separate variables to apply Ito's formula. The only thing you have to take care of in that case are quadratic covariations.


1

$\dfrac{\log X_n}{n} = \dfrac{\sum_{m \le n} \log Y_m}{n} \to E\log Y_1$ almost surely by the strong law of large number. And by Jensen's inequality, $E\log Y_1 < \log EY_1 =0$ since $P(Y = 1) < 1$ So we get that $\dfrac{\log X_n}{n}$ converges to a strictly negative number alomost surely, thus $\log X_n \to -\infty$, i.e. $X_n \to 0$


1

To gain a working knowledge of stochastic calculus, you don't need all that functional analysis/ measure theory. What you need is a good foundation in probability, an understanding of stochastic processes (basic ones [markov chains, queues, renewals], what they are, what they look like, applications, markov properties), calculus 2-3 (Taylor expansions are ...


1

Hint: Solve the stochastic differential equation $$dM_t = M_t \, dW_t$$ in order to find an explicit formula for $M_t$ in terms of $t$ and $W_t$.


1

after thinking a while about this formula it's clear now. First of all we have to sum over $k$, which is the length of distinct factors $X_i$'s. Then the second sum is over all $r_j$ such that $r_1 + \dotsm + r_k = r$. This is necessary, because there are many possibilities of $r_i$'s such that $r_1 + \dotsm + r_k = r$. The factor is clear, by the ...


1

As @Did wrote, let's talk about the space $\Omega = C([0,\infty),\Bbb R)$. This is the space of continuous trajectories, so if we define a measure on $\Omega$ we are saying which of the trajectories are more likely than others. For example, we can define a Wiener measure $P$ on $\Omega$ - that is what you call a Brownian motion. If we define another measure ...


1

Suppose that $(X_t)_{t \geq 0}$ is a local martingale. Since $$X_0 + \int_0^t \alpha_s dW_s$$ is also a martingale, this means that $$M_t := X_t - \left( X_0 + \int_0^t \alpha_s \, dW_s \right) = \int_0^t \beta_s \, ds$$ is a local martingale. Moreover, $(M_t)_{t \geq 0}$ is of bounded variation and has continuous sample paths. It is widely known that ...


1

The laws of the processes do not agree. Time-reversal for stochastic processes is more complicated than just to swich signs in the drift compnent. Think of an Ornstein-Uhlenbeck process started from an equilibrium for example. Just google for time-reversed diffusions for more help.


1

The answer is certainly: yes. By definition of a martingale, $\mathsf E M_t = \mathsf E M_0$ (just using the tower property) and since $M_0 = 0$ in your case, all follows. Note that Ito integral may not be a martingale unless some integrability conditions are satisfied, in general it may be only a local martingale - so you need to see which integrability ...


1

It follows from Itô's formula that the solution to the SDE $$dM_t = M_t \sigma_t dW_t$$ equals $$M_t = M_0 \exp \left( \int_0^t \sigma_s \, dW_s - \frac{1}{2} \int_0^t \sigma_s^2 \, ds \right).$$ By assumption, $M_0 \geq 0$. Hence, $$\begin{align*} \sqrt{M_T} &= \sqrt{M_0} \exp \left( \frac{1}{2} \int_0^T \sigma_s \, dW_s - \frac{1}{4} \int_0^T ...


1

Hint: Fix $s \leq t$. Show that $$\langle M,N \rangle_t^s := \langle M,N \rangle_t - \langle M,N \rangle_s $$ defines a symmetric positiv definite bilinear form. Alternatively: Apply the first inequality (the one you already proved) to the shifted processes $$\tilde{M}_t := M_{t+s}-M_s \qquad \quad \tilde{N}_t := N_{t+s}-M_s, \qquad t \geq 0.$$


1

Hints: Continuous (adapted) processes are predictable. (Stochastic) integrals are continuous, i.e. both mappings $$t \mapsto \int_0^t b(s) \, ds \qquad \quad t \mapsto \int_0^t \sigma(s) \, dW_s$$ are continuous whenever the integrals make sense.


1

Suppose that $(X_t,Y_t)_t$ is a solution to the given system. Then in particular, we can write $$X_t-X_0 = - \underbrace{\int_0^t Y_s \, dB_s}_{\text{local martingale}} - \frac{1}{2} \underbrace{\int_0^t X_s \, ds}_{\text{bounded variation, continuous}}.$$ Applying the optional stopping theorem yields that $$(t,\omega) \mapsto \left( \int_0^{t \wedge ...


1

First of all, your second and third paragraphs. Recall that if $\xi$ is $F$-measurable and $F\subseteq G$ then $\xi$ is $G$-measurable: that follows from the definition of measurability. Now, if $\Bbb F = (F_n)_{n\geq 0}$ is a filtration and $\xi = (\xi_n)_{n\geq 0}$ is $\Bbb F$-adapted, then for each $n$ we have $\xi_n$ is $F_n$-measurable. Since ...


1

Instead of $$\forall \varepsilon>0 \exists \delta>0, A \in \mathcal{F}_T, \mathbb{P}(A) <\delta: \int_A \sup_{t \in [0,T]} e^{-rt} \Psi(S_t) \, d\mathbb{P}<\varepsilon$$ it should read $$\forall \varepsilon>0 \exists \delta>0: A \in \mathcal{F}_T, \mathbb{P}(A) <\delta \Rightarrow \int_A \sup_{t \in [0,T]} e^{-rt} \Psi(S_t) \, ...


1

You shouldn't start with a PDE. The Black-Scholes-Merton equation is a PDE that lists the relationship between the greeks. You will need to differentiate with the solution of the PDE . You have listed the solution to the PDE for a call option in your first part of the question. You will need to do partial differentiation for the solution. $\frac{\partial^2 ...



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