Tag Info

Hot answers tagged

4

Do you know about conditional expectation? For random variable $X$ and continuous random variable $Y$ jointly defined on the same probability space, "$\mathbb{P}(X \geqslant Y)= \int_{\mathbb{R}} \mathbb{P}( X\geqslant y)\, f_Y(y) \text{d}y$" is true if the random variables in question are independent; it does not hold in general. More generally, using the ...


3

You didn't say your geographical area. In the United States, departments are typically not only willing to consider applicants without an MS--that most admitted PhD candidate have not had master's training is sort of the norm. As a consequence, the US degree process usually involve some number of years of course work (1 or 2 typically) with a ...


3

To use Ito's formula to calculate a stochastic integral, you want to find an Ito process whose differential is the integrand. When calculating $\int_0^t B_s dB_s$, you initially guess (by intuition from regular calculus) that the process might be $\frac{1}{2} B_t^2$. So you use Ito's formula to calculate its differential and get $d \left ( \frac{1}{2} B_t^2 ...


3

To your specific query of what expressions like $\text dY_t$ mean, I offer the following. If you are referring to such entities as they arise in, say, the study of Ito processes, such as $$ \text dY_t = \mu_t\text dt + \sigma_t \text dW_t,\ \ Y_0=y $$ such entities don't mean anything except only as meaningful shorthand for equivalent statements in terms of ...


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

I'm not an expert on this, but I did some research and found this Topics in Gaussian rough paths theory. I suggest you read it from the begining, pay special attention on page 5 and 6. From all the books and papers I found on Google this paper gives the most comprehensible explanation why you need those iterated integrals. Thy naturaly occur when you ...


2

Consider an equation like \begin{equation}\tag{1}dY_t = f(Y_t) dW_t\end{equation} where $Y_t$ is an unknown function and $W_t$ is a continuous, but not differentiable, function. If $W_t$ is Brownian motion then there is a classical theory for how to understand (1). Brownian motion has finite $p$-variation only for certain values of $p$. As you may know, ...


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

Levy's characterisation and the use of It$ô$'s lemma give the same answer. I think the independence argument in your approach is not quite enough to conclude that $b$ must be zero, unfortunately (please see my comment below). Levy's Characterisation: The aim here is to show that there is a choice of $a$ and $b$ such that ...


2

Im surprised you didn't know this, Doob, but anyway: Apply Ito on $X_t = t+e^{W_t}$ to get $$ dX_t = e^{W_t} dW_t + (1+\frac{1}{2}e^{W_t})dt $$ In integral form this is $$ X_t = X_0 + \int^t_0 e^{W_u} dW_u + \int^t_0 (1+\frac{1}{2}e^{W_u})du $$ The $dW_u$-integral is local Martingale since it is an integral against Brownian Motion, and the $du$-integral ...


2

Note that the random variable $\int_{0}^{t}e^{-rs}\text dW_r$ is not a function of solely $W_r$ but, rather, depends on the entire process $\{W_r\}_{0\leqslant r\leqslant t}$. This is why your choice of function, $$f(t,x) = S_0e^{rt} + \sigma e^{rt} \int_0^t e^{-rs} \text dx ,$$ is not a well-defined $C^{1,2}$ function (and hence, your intended version of ...


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

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 ...


1

For notational convenience, define $f_{n,t}:=\cos(n-1/2)\pi t$ and choose any finite set of $k$-distinct natural numbers $n_1, \ldots, n_k$. Then, the Itô integrals $\{\int_0^1 f_{n_1,t}~\text dW_t,\ \ldots,\ \int_0^1 f_{n_k,t}$$\text dW_t\}$ are jointly normal. Consequently, these intergrals are independent if any pair of integrals has zero correlation. So, ...


1

As you pointed out, we know that $\langle B \rangle_t = 0$ in probability for $t \geq 0$. In fact, $\langle B \rangle_t = 0$ almost surely (mind that the exceptional set may depend on $t$). Indeed: By definition, $$\langle B \rangle_t = \mathbb{P}-\lim_{|\Pi| \to 0} V_t(B,\Pi)$$ where $$V_t(B,\Pi) := \sum_{t_k \in \Pi} (B_{t_k}-B_{t_{k-1}})^2.$$ Here ...


1

Let $\Pi$ be a partition of the interval $[0,t]$, i.e. $\Pi : 0 = t_0 <t_1 <\ldots<t_m = t$. Note that as $\text{mesh}(\Pi) \rightarrow 0$ $$S(\Pi) := \sum_{i=0}^{m-1}\lambda(t_i)(W_{t_{i+1}}-W_{t_i}) \rightarrow \int_0^t\lambda(s)dW_s$$ This convergence is in $L^2$. Next, you can see that $S(\Pi)$ is a linear combination of the increments of ...


1

$\newcommand{\E}{\operatorname{E}}$ The characteristic function is \begin{align} s \mapsto {} & \E\left( e^{isW_\tau} \right) = \E\left( \E\left( e^{isW_\tau} \mid\tau \right) \right) = \E\left( e^{-\tau s^2/2} \right) =\int_0^\infty e^{-t s^2/2} f_\tau(t)\, dt \\[6pt] = {} & \int_0^\infty e^{-t s^2/2} e^{-t}\,dt = \int_0^\infty e^{-(1+s^2/2)t} \, dt ...


1

For any ($L^2$-)martingale $(M_t)_{t \geq 0}$ the quadratic variation $(\langle M \rangle_t)_{t \geq 0}$ is defined such that $$M_t^2 - \langle M \rangle_t$$ is a martingale. Apply this to $M_t = W_t^2-t$. (There is no need for Itô's formula!)


1

As @Did points out, by definition of $Y_t$ in terms of stochastic integrals, $$ \text dY_t = \frac{1}{X_t}\text dX_t-\frac{1}{2}\frac{1}{X_t^2}\text d\langle X\rangle_t\tag{1} $$ And $$ Z_t^{-1}\text dZ_t = Z_t^{-1}\text de^{Y_t}=Z_t^{-1}e^{Y_t}(\text dY_t+\frac{1}{2}\text d\langle Y\rangle_t) = \text dY_t+\frac{1}{2}\text d\langle Y\rangle_t\tag{2} $$ ...


1

Use the Law of Iterated Expectation. $$\begin{align} \mathbb{P}(\lambda_1 X + \lambda_2 Y = a) & = \mathbb{E}[\mathbb{1}_{\lambda_1 X + \lambda_2 Y = a}] \\[2ex] & = \Bbb E[\Bbb E[\mathbf 1_{\lambda_1 X + \lambda_2 Y = a}\mid Y]] \\[2ex] & = \int_{\Bbb R}\Bbb E[\mathbb{1}_{\lambda_1 X + \lambda_2 Y = a}\mid Y](y) \, \text{d} \mathbb{P}_Y(y) ...


1

Hints: If $\ \Bbb E[\int_{t}^{T}|h_s|\,\text ds],\,\Bbb E[\int_{t}^{T}|M_sh_s|\,\text ds]<\infty\ $ then, in succession, use the Fubini-Tonelli theorem, the tower property of conditional expectation, the martingale property, and Fubini-Tonelli one more time, as follows: $$ \Bbb E \left[ M_T\int_t^T h_s \text ds \, \vert\, \mathcal F_t\right] = \int_t^T ...


1

For a) Simulating the SDE of an underlying stock price is not used to price the stock; that would yield nothing more than a self-fulfilling prophecy. Rather, practitioners use simulation methodologies to price financial derivatives of an underlying asset. For example, the "payoff" function for a simple European call option is $\max(0,P_T - K)$, where $K$ ...


1

I do not think you can get a solid technical understanding of SDE's without knowing some measure theory - it is worth learning the basics of measure theory (for example, you should be able to comfortably talk about dominated convergence ). Additionally a touch of functional analysis will not hurt - the Lebesgue spaces are quite crucial (even if not apparent ...


1

I think this is easier to understand in a somewhat broader context. You have a sequence of stochastic processes $f_n$. We think of this in three equivalent ways. We have sets $\Omega,T,A$, with $T$ the time set and $A$ the set of values of the process. We can think of each $f_n$ with any of the signatures: $\Omega \times T \to A$ (a function of the ...


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

Out of Borodin and Salminen Handbook of Brownian Motion - Facts and Formulae: $$ E_x\left[\exp\left(-\frac{\gamma^2}{2} \int_0^t W_s^2 ds \right) \right] = \frac{1}{\sqrt{\cosh(t\gamma)}}\exp\left( -\frac{x^2\gamma \sinh(t\gamma)}{2 \cosh(t\gamma)} \right) $$ You can obtain it by using Girsanov's theorem. Really nifty formula when $x=0$. I also recommend ...


1

Define a martingale by $$M_t := \begin{cases} \mathbb{E}(\phi(W_T) \mid \mathcal{F}_t), & t \leq T, \\ \phi(W_T), & t>T. \end{cases}$$ Then, by the martingale representation theorem, there exists a representation of the form $$M_t = c+ \int_0^t \beta_s \, dW_s.$$ For $t=T$, this proves the claim. Remark: One possibility to prove the mentioned ...


1

The differential is a linear mapping that approximates a differentiable function in a given nieghborhood of a point. In this question, you're being asked to construct this linear mapping. A differential equation,on the other hand, is a equation of a given function in terms of the derivative or derivatives of a given order of a given function of one or ...


1

The not-so-exact calculation goes as $$ d(2e^{B_t})=2e^{B_{t+dt}}-2e^{B_t}=2e^{B_t}(e^{dB_t}-1)\\ =2e^{B_t}\Bigl(1+dB_t+\tfrac12(dB_t)^2+O((dB_t)^3)-1\Bigr) $$ and using the rules $dB_t^2=dt$ and to ignore all higher powers results in $$ d(2e^{B_t})=2e^{B_t}(dB_t+\tfrac12 dt)=e^{B_t}dt+2e^{B_t}dB_t $$ as you also computed, probably using the Ito formula. ...


1

Since $$ e^{-\nu t}Y_t=X_{\tau}^{1-\frac{\delta}{2}}\,,\tag{1} $$ It$ô$'s lemma implies $$ -\nu e^{-\nu t}Y_t\text dt + e^{-\nu t}\text dY_t = \text d(\color{red}{e^{-\nu t}Y_t}) = \text d(\color{red}{X_{\tau}^{1-\frac{\delta}{2}}}) = (1-\frac{\delta}{2})X_{\tau}^{-\frac{\delta}{2}}\text ...



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