# Tag Info

9

Think of the Wiener process as a curve into which you could zoom in ever deeper and deeper and it will still be completely wiggly (= a fractal). That means that even if you tried to put a tangent line onto it, it would find no stable support (= no differentiability) - yet the whole curve is completely closed, i.e. could be drawn without raising your pen (= ...

8

We know that $(\tilde{W}_t) := (-W_t)$ is also a Wiener process so $$E[W_pW_qW_r] = E[\tilde{W}_p\tilde{W}_q\tilde{W}_r] = (-1)^3E[W_pW_qW_r]$$ and that implies that $E[W_pW_qW_r] = 0$.

7

The trick is to start with the highest power, rewrite it as something you know (a third order moment) and then work backwards on the remaining terms. By that I mean you can complete the cube as follows: $$E[W_t^3 - 3tW_t|\mathcal{F}_s] = E[(W_t-W_s)^3 - C -3tW_t|\mathcal{F}_s]$$ where you'll need to find $C$ such that the equality holds (i.e. $C=W_s^3 + ... 7 Firstly,$m_T=\min\limits_{t\in[0,T]} B_t = -\max\limits_{t\in[0,T]} -B_t \overset{Law}{=} -\max\limits_{t\in[0,T]} B_t = -M_T$. So, you can either consider the running maximum or minimum. Let$\tau$be a stopping time and$(B_t)a Brownian motion. Then, \begin{align*} W_t =\begin{cases} B_t & t\leq \tau, \\ 2B_\tau - B_t & t\geq \tau, \end{cases} ... 5 I think what is meant is that the increment of the brownian between discrete points(i-1)$and$i$is normally distributed with mean 0 and variance equal to$\delta t$which represents the length of the interval between the two discrete points. You can then write the increment as$\sqrt{\delta t}$times a standard normal random. So the equation is to be ... 3 The most basisc understanding of continuity of curve is: You can draw it with a pen/pencil without lifting your hand. Thus the curve has no jumps that will force you to raise/shift your palm in order to continue drawing. The function$f(x)=|x|is continuous but not differentiable at the origin. If you look at the relevant Wikipedia entries on ... 3 \begin{align*} E\Big(W_t^3-3tW_t \mid \mathcal{F}_s\Big) &= E\Big((W_t-W_s+W_s)^3-3t(W_t-W_s+W_s) \mid \mathcal{F}_s\Big) \\ &=E\Big((W_t-W_s)^3+W_s^3+3(W_t-W_s)^2W_s + 3 (W_t-W_s)W_s^2\\ &\qquad \qquad -3t(W_t-W_s)-3tW_s \mid \mathcal{F}_s\Big) \\ &=E\Big((W_t-W_s)^3\Big) + W_s^3+3W_sE\Big((W_t-W_s)^2\Big)\\ &\qquad \qquad + 3W_s^2 E(... 3 There are many examples of "non-random" curves that are continuous everywhere, and yet differentiable nowhere. For example, the one defined by the formula $$f(x) = \sum_{k=0}^\infty 2^{-k}\cos(2^kx).$$ You may think of it as a limit of partial sumsf_n(x) =\sum_{k=0}^n 2^{-k}\cos(2^kx)$. Each$f_n$is differentiable, and consists of a combination of ... 3 In order to apply Ito's lemma, your function needs to be a twice-differentiable function. There is no issue with the non-differentiability of the Wiener process.$\frac{dF}{dX}$involves differentiating F, not the Wiener process X. Using a simple analogy: instantaneous velocity ($\frac{dD}{dt}$) is the derivative of position (D) over time; what is ... 2 First note that paths are a.s continuous. Then by strong Markov property and reflection principle,$(W_\tau - W_t)$is a Brownian motion independant of the before tau part. Then you can verify that increments are independent and gaussian by decomposing them in before and after tau part. Or you can décompose the quadratic variation and use Lévy 's ... 2 We write the differential form of Ito formula for simplification. Actually, the differential form for Ito formula $$dF(W(t)) = 2W(t)dW(t) + dt$$ means the integral form for Ito formula, $$\int{dF} = \int{2W(t)dW(t)} + \int{dt}$$ which make sense in mathemaitcs. 1 You are right that a Wiener process can not be differenciated in the conventional way since the derivative in respect to time does not exist. For this reason Ito lemma should be used to integrate and differenciate Brownian or Wiener processes as these are considered ito processes. 1 You can use that$f(t,W_t)\in C^2$is Martingale iff:$$\partial_t f+\frac{1}{2}\partial_{WW}f= 0$$ We get:$$\partial_t f=-3W_t$$$$\partial_{WW}f=6W_t$$ Finally: $$-3W_t+3W_t= 0$$ q.e.d. The proof of theorem follows by writing out$f(t,W_t)$via Ito formula. Proof of theorem: 1 Note that$\{W_t \mid t \geq 0\}$is a martingale. Then, for$0<p<q<r, \begin{align*} E(W_pW_qW_r) &= E\Big( E(W_pW_qW_r \mid \mathcal{F}_q)\Big)\\ &=E\Big(W_pW_q E(W_r \mid \mathcal{F}_q)\Big)\\ &=E\Big(W_pW_q^2\Big)\\ &=E\Big(W_p(W_q-W_p+W_p)^2\Big)\\ &=E\Big(W_p(W_q-W_p)^2+W_p^3+2W_p^2(W_q-W_p) \Big)\\ &=E(W_p)E\Big((W_q-... 1 I think you are on the right track here. You made a sign error in the first line, unfortunately: $$E[W_p W_q W_r] = E[W_r W_p^2 + W_pW_q^2 - W_qW_p^2]=\\ E[(W_r-W_q)W_p^2]+E[W_pW_q^2]= E[W_pW_q^2]$$ The first term is0$by independence (as$p<\text{min}(r,q)\$ and the square does not affect independence). To take care of the second term we do the ...

Only top voted, non community-wiki answers of a minimum length are eligible