# Tag Info

9

The fundamental reason is that variances are additive, while standard deviations are not. If you didn't make the variance proportional to the time step, your discretizations would not behave consistently when you changed the time step. To be explicit, let's take $X_{i+1}=X_i+R_{i+1}\sqrt{\Delta t}$, where the $R_i$ are independent and standard normal. After ...

7

It seems to me that it is easy to show directly that $X_t$ is a martingale by verifying that $E[X_t \mid \mathcal{F}_s] = X_s$. (Here I assume that $B_t$ is a Brownian motion with respect to the filtration $\mathcal{F}_t$, and that you are trying to show $X_t$ is a martingale with respect to the same filtration.) One just writes $B_t = B_s + (B_t - B_s)$ ...

6

The Ito formula, for $\mathrm{d} X_t = - a X_t \mathrm{d} t + \mathrm{d} B_t$, you need is: $$\mathrm{d}\left( f(t, X_t) \right) = \left( \partial_t f(t,X_t) - a \partial_x f(t,X_t) + \frac{1}{2} \partial_{xx} f(t,X_t) \right) \mathrm{d} t + \left( \partial_x f(t,X_t) \right) \mathrm{d} B_t$$ HINT: use $f(t,X_t) = \mathrm{e}^{a t} ... 5 You can do this by different methods. One way I like is to use Ito's formula. For instance, consider$B_t^2$.$d(B_t^2) = 2B_t dB_t + dt$. Hence,$\int_{0}^{T} B_t dB_t = \frac{1}{2} \int_{0}^{T} (d(B_t^2) - dt) = \frac{1}{2} (B_T^2 - T)$. Similarly, consider$B_t^3$. Use Ito's formula,$d(B_t^3) = 3B_t^2 dB_t + 3 B_t dt$. Hence,$\int_{0}^{T} B_t^2 ...

5

The trick you shall learn when dealing with Ito integrals is that $$\int_a^bf_s\mathrm d W_s\quad\text{ and }\quad \int_b^cf_s\mathrm d W_s$$ are independent whenever $a<b<c$, it follows from independence of increments of $W_t$. As a result, if you assume that in your case $s\leq t$ then $$\mathsf E\left[\int_0^s f_u\mathrm dW_u \int_0^t ... 4 In order to determine the law of X_t it suffices to determine it for t=1 due to scaling property, X_t \stackrel{d}{=} t X_1. Using the vector Ito process V_t = (X_t, Y_t, Z_t) with$$ \mathbb{d} X_t = Y_t \mathrm{d} B^{(2)}_t \qquad \mathrm{d}Y_t = \mathrm{d} B^{(1)}_t \qquad \mathrm{d} Z_t = \mathrm{d} B^{(2)}_t \qquad V_0 = (0,0,0) $$... 4 Since (X(t))_t is a martingale in the filtration (\mathcal F_t)_t, one knows that \mathrm E(X(u)\mid\mathcal F_s) is X(u) is u\leqslant s and X(s) if u\geqslant s. Hence, for every s\leqslant t,$$ \mathrm E(g(t)\mid\mathcal F_s)=g(s)+\int_s^tf(u)\mathrm E\left(X(u)\mid F_s\right)\mathrm du=g(s)+X(s)\cdot \int_s^tf(u)\mathrm du. $$The only ... 4 Let f(x) := x^k, then by Itô's formula$$W_t^k = \int_0^t k \cdot W_s^{k-1} \, dW_s + \frac{1}{2} k \cdot (k-1) \cdot \int_0^t W_s^{k-2} \, ds$$Since (t,\omega) \mapsto \left(\int_0^t k \cdot W_s^{k-1} \, dW_s \right)(\omega) is a martingale, we have$$\mathbb{E}(W_t^k) = 0+ \frac{1}{2} k \cdot (k-1) \cdot \mathbb{E} \left( \int_0^t W_s^{k-2} \, ds ...

4

Let $Y_t = \mathrm{e}^{t- 2 W_t} X_t^2$. Then, applying Ito's lemma: $$\mathrm{d} Y_t = \mathrm{e}^{t- 2 W_t} \mathrm{d} (X_t^2) + X_t^2 \mathrm{d} \mathrm{e}^{t- 2 W_t} + \mathrm{d} \mathrm{e}^{t- 2 W_t} \cdot \mathrm{d} (X_t^2)$$ Since $\mathrm{d} \mathrm{e}^{t- 2 W_t} = \mathrm{e}^{t- 2 W_t} \left( 3 \mathrm{d} t - 2 \mathrm{d} W_t \right)$ and ...

4

An easy way to calculate an expected value of SDE's solutions is to find an equation on $m(t) = \mathsf{E}[X_t]$. More precisely, from $$dX_t = \mu X_tdt+\sigma X_t dB_t$$ taking expectations of both sides (while using that $\mathsf{E}[dB_t] = 0$) we obtain $$dm(t) = \mu m(t)dt,$$ which gives a desired answer. Indeed, from the latter equation follows ...

4

Let $\tau=\inf\{t\geqslant0\mid x_t=0\}$. Since $x_t=w_t-Mt+d$, $\tau=\inf\{t\geqslant0\mid w_t=Mt-d\}$. Thus $\tau$ is a stopping time for $(w_t)$. Assume that $\tau$ is integrable. Then the optional stopping theorem yields $\mathrm E(w_\tau)=\mathrm E(w_0)=0$, hence $\mathrm E(x_\tau)=d-M\mathrm E(\tau)$. Since $\tau$ is integrable, $\tau$ is almost surely ...

4

The obvious works: plugging in KL integral equation the value of $R$ and splitting the integral on $(0,T)$ into a sum of integrals on $(0,s)$ and on $(s,T)$, one gets $$\lambda_n\phi_n(s)=\lambda^2s\int_0^Tt\phi_n(t)\mathrm dt+\lambda\int_0^st\phi_n(t)\mathrm dt+\lambda s\int_s^T\phi_n(t)\mathrm dt.$$ Differentiating this twice yields $$... 4 "All kinds of stochastics" is quite wide formulation, but I guess it includes probability theory (PT). When studying stochastic processes/stochastic calculus/statistics you certainly need to know PT- so I would say this is the basic course here. Jonas has mentioned measure theory - and it is indeed important for the good understanding of probability ... 4 In the book you've cited the RDE refers to the equation of the form$$ \dot X = f(X,Y,t),\quad X(t_0) = X_0.\quad(\star) $$where X_0 is a r.v. and Y is a stochastic process. There see Section 5.4, p. 134 where it is written: Accordingly, we follow Sisky [Stochastic Differential Equation, 1967] and classify (\star) into three basic types: ... 3 The point is that Ito Lemma applies to the situation where you have a function of an Ito process, rather than a functional. E.g. you can apply it in case h(\phi_t), but you don't apply Ito Lemma to compute the differential of \int_0^t\phi_s\mathrm ds. In the latter case you rather have$$ \mathrm d\left(\int_0^t\phi_s\mathrm ds\right) = \phi_t\mathrm ...

3


3

Let's consider a simplest case: Let $f : \mathbb{R} \to \mathbb{R}$ satisfy global Lipschitz condition $$|f(x) - f(y)| \leq L |x - y|, \quad x, y \in \mathbb{R}$$ with Lipschitz constant $L$. Then by triangle inequality, we have $$|f(x)| \leq |f(x) - f(0)| + |f(0)| \leq L|x| + |f(0)| \leq C(|x| + 1)$$ for large $C \geq \max(L, |f(0)|)$. Same argument ...

3

The SDE $$\mathrm{d} X_t = - \lambda X_t \,\mathrm{d} t + \mathrm{d} W_t$$ has as solution $$X_t = \mathrm{e}^{- \lambda t} X_0 + \int_0^t \mathrm{e}^{- \lambda \left( t - s \right)} \,\mathrm{d} W_s$$ If you make the transformation $Y_t = \mathrm{e}^{\lambda t} X_t$, then by Ito's formula applied to $Y_t = f \left( t, X_t \right)$ and after ...

3

Let's assume that $\mathcal{F}_t$ is the natural filtration of the Brownian motion. Notice that, for $s<t$ $$\begin{eqnarray} \mathbb{E}\left(X_t|\mathcal{F}_s\right) &=& \mathbb{E}\left(\left( \mathrm{e}^{t/2} \cos(B_s + (B_t-B_s))\right)| \mathcal{F}_s \right) \\&=& \mathrm{e}^{t/2} \mathbb{E}\left( \cos(B_s) \cos(B_{t-s}) - ... 3 No, almost surely finite does not imply integrable. Consider a random variable X exponential with parameter 1 and Y=\mathrm e^X. Then X is almost surely finite hence Y is almost surely finite but, for every y\geqslant1, \mathrm P(Y\geqslant y)=\mathrm P(X\geqslant\log y)=\mathrm e^{-\log y}=1/y, hence Y is not integrable. 3 Every step is correct, including the identities Y_t=\displaystyle\int_0^tW_s\mathrm ds=\int_0^t(t-s)\mathrm dW_s and [Y]_t=0, except the underlying assertion that the process Y^2-[Y] should be a martingale and consequently that \mathbb E(Y_t^2) and \mathbb E([Y]_t) should coincide for every t. Recall that the quadratic variation process [Y] is ... 3 This addresses the question cited as a motivation. For every t\geqslant0, introduce X_t=\int\limits_{0}^{t}\cos(B_s)\,\textrm{d}s and m(t)=\mathrm E(\cos(B_t))=\mathrm E(\cos(\sqrt{t}Z)), where Z is standard normal. Then \mathrm E(X_t)=\int\limits_{0}^{t}m(s)\,\textrm{d}s and \mathrm E(X_t^2)=\int\limits_{0}^{t}\int\limits_{u}^{t}2\mathrm ... 3 The answer is that E(X_t)=x_0\exp(\mu t). The easiest way to see it is to start from the SDE and to note that \mathrm{d}E(X_t)=\mu E(X_t)\mathrm{d}t and E(X_0)=x_0. Hence a(t)=E(X_t) solves a'(t)=\mu a(t) and a(0)=x_0, that is, a(t)=x_0\exp(\mu t). Your solution goes astray when you solve the SDE, the factor of B_t is wrong and, in fact, ... 3 I think the given representation of the Brownian Bridge is not correct. It should be$$Y_t = a \cdot (1-t) + b \cdot t + (1-t) \cdot \underbrace{\int_0^t \frac{1}{1-s} \, dB_s}_{=:I_t} \tag{1}$$instead. Moreover, the covariance is defined as \mathbb{E}((Y_t-\mathbb{E}Y_t) \cdot (Y_s-\mathbb{E}Y_s)), so you forgot to subtract the expectation value of Y ... 3 You can see this book Introduction to Stochastic Integration p.109 and following; example 7.6.3 answers your question. Sorry for my english. 3 If f is any continuous function, then L'Hopital's rule gives $\lim_{t\to1}\; (1-t)\int_0^t\frac{f(s)}{(1-s)^2}\,ds = f(1).$ In terms of generalized functions, this says that if $\mu_t(s) = \frac{1-t}{(1-s)^2}1_{[0,t]}(s),$ then \mu_t\to\delta_1. The Brownian bridge can also be written in terms of generalized functions: \[ Y_t= a(1 ... 3 What to compute the integral means is unclear in this context but one can say this: for every t\geqslant0, the random variable$$ X_t=\int_0^tW_s\mathrm ds $$is centered normal vith variance \sigma_t^2 where$$ \sigma_t^2=\mathbb E(X_t^2)=2\int_0^t\int_0^s\mathbb E(W_sW_u)\mathrm du\mathrm ds=\int_0^t\int_0^s2u\mathrm du\mathrm ds=\frac{t^3}3. $$The ... 3 No, in general this is not correct. Applying Itô's formula, one can actually show that$$(t,w) \mapsto M(t,w) := \exp \left( \int_0^t X_s \, dB_s - \frac{1}{2} \int_0^t X_s^2 \, ds \right)$$is a local martingale. There are several sufficient conditions (on the process (X_t)_t) implying that (M_t)_{t \geq 0} is indeed a martingale. In this case, one ... 3 HINT: This is about applying Ito's lemma to f(X_t, Y_t) = X_t^2 + Y_t^2. Rewrite your vector valued process as$$ \mathrm{d} X_t = Y_t \mathrm{d} B_t \qquad \mathrm{d} Y_t = -X_t \mathrm{d} B_t \qquad X_0 = x \qquad Y_0 = y  \begin{eqnarray} f(X_t, Y_t) = f(X_0, Y_0) &+& \int_0^t \left( \frac{1}{2} Y_s^2 ...

3

As mike already wrote: $<X,B>_t = \sigma \cdot X_t \, dt$ - this fact follows straight from the definition of $<\cdot,\cdot>$. Another approach to solve the SDE: Solve the homogeneous SDE $dX_t = \sigma \cdot X_t \, dB_t$. Idea: Set $Z_t := \log X_t$ and apply Itô's formula to $f(x) := \log x$, then dZ_t = \frac{1}{X_t} \, dX_t - ...

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