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6

From the stochastic integral analogue of integration by part, which is a corollary of Ito's Lemma, $$d(W_t\ln t) = \frac{W_t}{t}dt+(\ln t)dW_t,$$ so $$I:=\int_0^t \frac{W_s}{s}ds=W_t\ln t-\int_0^t (\ln s)dW_s=\int_0^t \ln\frac{t}{s}dW_s$$ as $W_t\ln t\to 0$ in probability, as $t\to 0^+$. Since the random variable $I$ is a linear combination of independent ...

3

I think both of you and your book are right! The point here is $d(S_t)d(e^{-rt})=0$. The reason is $dt\cdot dW=0$ and $dt\cdot dt=0$.

3

Let $$X_1(t) = A(t) \sin (t) + B(t) \cos (t)\\ X_2(t) = A(t) \cos (t) - B(t) \sin (t)$$ then thanks to the Ito fornula: \begin{align} dX_1(t) &= dA(t) \sin (t) + A(t) \cos (t) dt + dB(t) \cos (t) - B(t)\sin(t) dt \\ &= X_2(t)dt + dA(t) \sin (t) + dB(t) \cos (t) \\ dX_2(t) &= dA(t) \cos (t) - A(t)\sin (t) dt - dB(t) \sin (t) - B(t) \cos (t) ...

3

Edit: This answer shows the identity $$\mathbb{E} \left( \left| \sum_{i=0}^{n-1} \int_{t_i^n}^{t_{i+1}^n} (W(t_i^n)-W(t)) \, dt \right|^2 \right) = \mathbb{E} \left( \sum_{i=0}^{n-1} \left[\int_{t_i^n}^{t_{i+1}^n} (W(t_i^n)-W(t)) \, dt\right]^2 \right). \tag{1}$$ Fix $i<j$. Then, by the tower property, \begin{align*} ... 2 Sure assume dy = \mu dt + \sigma dZ is an Ito process. f(y) = e^y Normally by the chain rule, df = \frac {\partial f}{\partial t}dt + \frac {\partial f}{\partial y}dy However, because (dy)^2 ~ O(dt) we cannot neglect (dy)^2 terms. Taylor expand df to order 2 in dy \Rightarrow df = \frac {\partial f}{\partial t}dt + \frac {\partial ... 2 Recall Ito's formula, written in differential form, df(X) = f'(X)dX + \frac12f''(X)d\langle X\rangle. $$Assuming w(t) is a semimartingale, since y(x):=e^x is a C^2 function, we can compute$$ dy(w(t)) = e^{w(t)}dw(t) + \frac12e^{w(t)}d\langle w \rangle(t). $$If by w(t) you meant a brownian motion, then the quadratic variation term is just ... 2 We should focus on simulating \int_0^t e^{\theta (s)}\, \mathrm{d}W_s. If f(s)=e^{\theta (s)} is continuously differentiable, you could use the fact that$$\sum_{i=1}^{[tn]}f(s_i^*)\Big(W(s_i)-W(s_{i-1})\Big)\to\int_0^tf(s)dW(s)$$in quadratic mean, for s_i^* \in [s_{i-1},s_i]. Note that you should use$$W(s_i)-W(s_{i-1}) \sim N(0,s_i-s_{i-1})$$and ... 2 for your equation i have found$$2\,{\frac {1}{\Gamma \left( n+1 \right) }{{\rm e}^{{\frac {n}{100}}}} \Gamma \left( n+1,{\frac {n}{100}} \right) }-2-{{\rm e}^{{\frac {n}{ 100}}}}=0 $$and a solution is \approx 69,31471806 2 We have that$$ V = W_t^2 - t$$hence (written with t and x, where we plug in S_t = W_t for x), we have$$ V(x,t) = x^2 - tNow \begin{align*} \frac{\partial V}{\partial t} &= -1\\ \frac{\partial V}{\partial x} &= 2x\\ \frac{\partial^2 V}{\partial x^2} &= 2 \end{align*} Hence (note that S_t = W_t) \begin{align*} ... 2dV = -\beta Vdt +\sigma dB, \space V(0)=0$$Treat this like an integrating factor ODE with initial condition V(0) =0. Define Y=e^{\beta t}V, using Ito: 1)$$dY = \beta e^{\beta t}Vdt + e^{\beta t}dV = \beta e^{\beta t}Vdt - \beta e^{\beta t}Vdt + e^{\beta t}cdB = e^{\beta t}cdB$$2)$$\Rightarrow Y(t) = Y(0) + \sigma \int_{0}^te^{\beta s}dB ...

2

Since $\mathbb{E}W_{\tau}=0$ and $W_{\tau} \in \{1,-1\}$, we have $$0 = \mathbb{E}(W_{\tau}) = \mathbb{E}(1 \cdot 1_{\{W_{\tau}=1\}} + (-1) \cdot 1_{\{W_{\tau}=-1\}}) = \mathbb{P}(W_{\tau}=1) - \mathbb{P}(W_{\tau}=-1).$$ Hence, $$\mathbb{P}(W_{\tau}=1) = \mathbb{P}(W_{\tau}=-1) = \frac{1}{2}.$$ This implies $$\mathbb{E}f(W_{\tau}) = \frac{1}{2} ... 2 Hint: Using Itô's formula, show that X_t := \sin(W_t) e^{t/2} satisfies$$dX_t = \cos(W_t) e^{t/2} \, dW_t.$$Conclude that$$\sin(W_T) = \int_0^T e^{-(T-t)/2} \cos(W_t) \, dW_t.$$Remark: The idea is to find a determinstic function f such that X_t := f(t) \sin(W_t) is a martingale. By Itô's formula, a sufficient condition is$$\frac{1}{2} ...

2

Let $E^I$ denote the set of all functions from $I$ to $E$. Note that $X$ can be viewed as a mapping from $\Omega$ into $E^I$, with $[X(\omega)](t) = X_t(\omega)$. Also, $X$ is measurable with respect to the $\sigma$-algebra $\mathcal{F}$ on $E^I$ which is generated by all of the projections $\pi_t: E^I\to E$, given by $\pi_t(f) = f(t)$. Hence there is a ...

2

Since $\cosh(x) \ge 1$ for all real $x$ you have $\frac{e^{t/2}}{\cosh X_t} \le e^{t/2}$. So on any finite time interval you have a bounded local martingale, which is a martingale. Alternatively, note that $\frac{\sinh(x)}{(\cosh(x))^2} \le \frac{1}{2}$ for all $x$. So for any $T$, $$\int_0^T E\left[\left|-\frac{e^{t/2} \sinh ... 2 Here is one argument that would work, assuming that one already knows the following facts: For any s>0, the process (B_{s+t}-B_s)_{t \geq 0} is a Brownian motion. The time inversion (tB_{\frac{1}{t}}) of a Brownian motion is a Brownian motion (defined to start at 0 at t=0). \limsup_{t \to \infty} B_t >0 (in fact, the lim sup is ... 2 By definition, the Riemann-integral of a function f equals$$\int_a^b f(t) \, dt = \lim_{n \to \infty} \sum_{j=1}^n f(t_j^n) (t_j^n-t_{j-1}^n)$$where \Pi^n = \{a=t_0^n < \ldots < t_n^n = b\} denotes a partition of the interval [a,b] and the mesh size |\Pi^n| converges to 0 as n \to \infty. Consequently,$$\int_a^b X(t)(\omega) \, dt = ...

2

For an Itô process $(X_t)_{t \geq 0}$ of the form $$X_t-X_0 = \int_0^t \sigma(s) \, dB_s + \int_0^t b(s) \, ds \tag{1}$$ Itô's formula reads $$f(X_t)-f(X_0) = \int_0^t f'(X_s) \sigma(s) \, dB_s + \int_0^t \left( \frac{1}{2} f''(X_s) \sigma^2(s) + f'(X_s) b(s) \right) \, ds.$$ Now: $X_t := \int_0^t s \, dB_s - \frac{t^3}{6}$ is an Itô process. Choose ...

1

Stochastic calculus is to do with mathematics that operates on stochastic processes. The best known stochastic process is the Wiener process used for modelling Brownian motion. Other key components are Ito calculus & Malliavin calculus. Stochastic calculus is used in finance where prices can be modelled to follow SDEs. In the Black-Scholes model, ...

1

Let $Y_t=ln(X_t)$, by Ito's lemma we derive that the process $Y_t$ follows the SDE $dY_t=\sigma dW_t,\quad Y_0=0$ which has solution $Y_t= \sigma W_t$. Because $W_t$ has distribution $N(0,t)$, $Y_t$ has distribution $N(0,\sigma^2t)$. Since $X_t=e^{Y_t}$ and $X_t$ has normal distribution, we conclude that $X_t$ has log-normal distribution, so the ...

1

Take any path $P$ from $(0,0)$ to $(n,a-b)$. By adding a $+1$ step from $(-1,-1)$ to $(0,0)$ we now have a path $P^{'}$ of length $n+1$ from $(-1,-1)$ to $(n,a-b)$. We now shift path $P^{'}$ one unit up and one unit right, giving us a path $P^{''}$ from $(0,0)$ to $(n+1,a-b+1)$. Now, for $i\geq 1,\;$ $P$ has all its $X_i\geq 0\;$ iff $\;P^{''}$ has all its ...

1

Consider the application $$F(\xi) : F(\xi)(T)=\xi_0+\int_0^Tf(\xi(t))dt+W(T)$$ $$\left[ F(\xi) - F(\zeta)\right] (T) = \int_0^T \left[f(\xi(t)) - f(\zeta(t))\right] dt$$ under the same assumptions as in the Cauchy Lipschitz theorem, you can prove that (for the right space and right norm) $F$ is Lipschitz with constant $<1$ (the proof is exactly ...

1

Well, when it's known what is constant and what it's not, the key to solving this equation is observation that it's first order inhomogeneous linear equation: $$\frac{d\omega}{dt} + c \left (\frac{be^{-bt}}{ae^{-bt}-C_1} \right) \cdot \omega = f(t)$$ General solution of this ODE is a linear combination of any solution of it and general solution of ...

1

First a remark concerning the definition of $X^{(n)}$: Instead of the closed interval $A_k^{(n)} := [t_{k-1}^{(n)},t_k^{(n)}]$ you should use the half-open interval $A_k^{(n)} := (t_{k-1}^{(n)},t_k^{(n)}]$ (otherwise you might run into trouble at the boundary points $\{t_k^{(n)}\}$). Then $$X^{(n)}(s,\omega) := X_0(\omega) 1_{\{0\}}(s) + \sum_{k=1}^{2^n} ... 1 Ignoring for the moment the fact that \sqrt{2t+B_t} will sometimes be negative, we have F(t,x) = \frac23(2t+x)^\frac32,\qquad \frac{\partial F}{\partial t}=2\sqrt{2t+x},\qquad \frac{\partial f}{\partial x} = \frac{1}{2\sqrt{2t+x}}, so \int_0^T\sqrt{2t+B_t}\;dB_t = \frac23(2T+B_T)^\frac32 -\int_0^T2\sqrt{2t+B_t}+\frac{1}{2\sqrt{2t+B_t}}\;dt. We ... 1 Let 0 \leq t_1<\ldots<t_n. Since (B_t)_{t \geq 0} is a Gaussian process, we know that (B_{t_1},\ldots,B_{t_n}) is Gaussian. This implies in particular that \sum_{j=1}^n B_{t_j} is Gaussian. Since a Gaussian random variable is uniquely characterized by its mean and variance, it remains to calculate those two. As \mathbb{E}B_t=0 for any t ... 1 Heuritically, d(S_t)d(e^{-rt})=-re^{-rt}dt(dS_t)\sim O(dt)^{3/2}. Thus, it is not considered in the SDE for S_te^{-rt}. 1 Part (i): Use$$M_t := \frac{W_{t \wedge \tau(-b)} + b}{b}.$$How to come up with this choice? Well, we are interested in$$\sup_{0 \leq t \leq \tau(-b)} W_t = \sup_{t \geq 0} W_{t \wedge \tau(-b)}.$$We know that (W_{t \wedge \tau(-b)})_{t \geq 0} is a martingale and that W_{t \wedge \tau(-b)} \to -b as t \to \infty. This means that, in order to get ... 1 Let (X_t)_{t \geq 0} be a weak solution of the given SDE, then there exists a Brownian motion (B_t)_{t \geq 0} such that$$X_t = X_0 + \int_0^t b(s,X_s) \, ds + \int_0^t \sigma(s,X_s) \, dB_s.$$Using the elementary estimate$$(a+b+c)^p \leq 3^p (a^p+b^p+c^p), \qquad a,b,c \geq 0,$$we find$$|X_t|^p \leq 3^p |X_0|^p + 3^p \left| \int_0^t b(s,X_s) ...

1

Define $m(t)$ and $f(t,s)$ as the mean and covariance functions: \begin{align} m(t) &= E[X_t]\\ f(t,s) &= E[X_tX_s] - E[X_t]E[X_s] \end{align} We are told that the functions $m(t)$ and $f(t,s)$ are continuous. In general, if a function $h(t)$ is continuous, we know that $\lim_{s\rightarrow t} h(s) = h(t)$. Thus: \begin{align} ...

1

If X and Y are independent, then the result follows from Cramer's Theorem. Here is a link. Note the requirement of independence here. In the non-independence case, if $Y=Z-X$, where $Z$ is normal, and $X$ has any distribution also works.

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