# Tag Info

1

It's not necessary to prove that $\lim_{n\to\infty}\mathbb{P}(X_n\leq c)=0$. Instead, let $$A=\{\alpha\in\mathbb{R}:\lim_{n\to\infty}\mathbb{P}(X_n\leq \alpha)=0\}$$ and observe that if $\alpha<c$, then there exists $\gamma>0$ such that $\alpha<c-\gamma<c$, hence $$\mathbb{P}(X_n\leq \alpha)\leq \mathbb{P}(X_n<c-\gamma)$$ for all $n$, so $$... 1 Example: Let N(t) be a unit rate Poisson process. Then M(t):=N(t)-t, t\ge 0, is a martingale with paths of finite variation (on each finite time interval) and M(0)=0. 0 It's hard to tell what you need, since this depends on the level of complexity and sophistication you're targeting, and most importantly, what you aim to learn from these books. Here are a few references that might help. Convex analysis (in increasing order of "difficulty"): "Convex Optimization", S. Boyd et al. "Introductory Lectures on Convex ... 1 That's interesting. Seems to me the fact that you can replace (3) by something weaker follows from the Central Limit Theorem, or at least by a CLT-ish argument. I wouldn't know what the fanciest result in this direction would be - there are all sorts of versions of things like the CLT, but it seems clear that at the very least one could replace (3) by the ... 2 For any random variables Z and Y with Z integrable we have E(Z) = E[ E(Z\mid Y) ] . Apply this to$$Z:=g(X)h(Y)$$and note that$$E(g(X)h(Y)\mid Y ) = h(Y) E(g(X)\mid Y), $$since h(Y) is measurable with respect to \sigma(Y). 1 For any measurable set A \subseteq [0,T] \times \Omega it holds that$$\begin{align*} \{(t,\omega); (t,B_t(\omega)) \in A\} &= \int_0^T \! \int 1_A(t,B_t) \, d\mathbb{P} \, dt \\ &= \int_0^T \! \int 1_A(t,x) \frac{1}{\sqrt{2\pi t}} e^{-x^2/2t} \, dx \, dt. \end{align*}$$Since e^{-x^2/2t} is strictly positive, this implies$$(\lambda|_{[0,T]} ...

1

Let $D:=\{(t,\omega):f(t,B_t(\omega)\not=0\}$ and $C=\{(t,x):f(x,t)\not=0\}$. Observe that $D=\varphi^{-1}(C)$, where $\varphi(t,\omega) =(t,B_t(\omega))$ (mapping $[0,T]\times\Omega$ into $[0,T]\times\Bbb R$). Consequently, writing $\lambda$ for Lebesgue measure on $[0,T]$, $$0=\lambda\otimes\Bbb P(D) =\int\int 1_{C}(t,x){1\over\sqrt{2\pi ... 1 If f: (0,\infty) \times \Omega \to \mathbb{R} is progressively measurable and$$\mathbb{E} \left( \int_0^t |f(s)|^2 \, ds \right)<\infty \quad \text{for all $t \geq 0$}$$then$$M_t := \int_0^t f(s) \, dB_s, \qquad t \geq 0,$$is a martingale. This implies in particular$$\mathbb{E}(M_t) = \mathbb{E}(M_0) = 0.$$Since f(s,\omega) := ... 1 This isn't an answer per se, rather some thoughts to consider. If \{X_n:n=1,2,\ldots\} is a Markov chain and we define the holding times \{T_n:n=1,2,\ldots\} by T_n\sim\mathsf{Exp}(\lambda_{X_n}), then the sequence \{T_n\} only defines a Poisson process if \lambda_{X_n} is equal for all n. Otherwise, the intensity would be a stochastic process ... 1 Choose$$f(t, B_t) = \frac{B_t^3}{3}$$so that we have$$ \frac{B_t^3}{3} = \int_0^t B_s^2 dB_s + \int_0^t 0 ds + \frac{1}{2}\int_0^t 2B_s ds  \to E[\frac{B_t^3}{3}] = E[\int_0^t B_s^2 dB_s + \int_0^t 0 ds + \frac{1}{2}\int_0^t 2B_s ds]  \to E[\frac{B_t^3}{3}] = E[\int_0^t B_s^2 dB_s] + \frac{1}{2}\int_0^t 2E[B_s] ds  \to ...

0

I see it on page 22 in the 1985 edition. I do think the text is correct. If it is an error though, Liggett should be notified as I don't see it on the errata posted on his website. The reason I think it's correct: For any $\xi,\zeta\in W^S$ that are identical on $T$, $f_T(\xi)=f_T(\zeta)$ since $f_T$ only sees a fixed $\eta$ outside of $T$ (I would have ...

1

Let's show that $\mathbb{E}[\int_0^t (e^{Z_s}(1-X_su_s))^{2}ds] < \infty$ By Cauchy-Schwarz inequality: $\mathbb{E}[\int_0^t (e^{Z_s}(1-X_su_s))^{2}ds] \leq \mathbb{E}[\int_0^t e^{4 Z_s}ds]^{\frac{1}{2}} \mathbb{E}[\int_0^t (1-X_su_s)^{4}ds]^{\frac{1}{2}}$ Because $u$ is bounded, it is clear that $\mathbb{E}[\int_0^t (1-X_su_s)^{4}ds] < ... 1 It depends very much on what you know about stochastic integrals. If you know Novikov's condition, then this might be your first choice. If you know local martingales, then you can use the fact that$(e^{Z_t})_t$is a local martingale and the boundedness of$u$to conclude that$e^{Z_t} \in L^2$. In my answer, I'll present a solution which does not require ... 2 You have : $$\mathbb{P}(A_4=k|N_{10}= 1000)=\frac{\mathbb{P}(A_4=k,N_{10}=1000)}{\mathbb{P}(N_{10}=1000)}$$ $$\mathbb{P}(N_{10}=1000)=\frac{(10\times 100)^{1000}}{1000!}e^{-10\times 100}$$ You have : $$A_t = \sum_{i=1}^{N_t} \xi_i$$ where$\xi_i$are independent and identically distributed $$\mathbb{P}(\xi_i=1)=\mathbb{P}(\xi_i=0)=\frac{1}{2}$$ ... 1 Let's try to prove this rigorously. Lemma 1 Let$E$be a topological space$t\ge 0$and$t_0^{(n)},\ldots,t_n^{(n)}\ge 0$with $$0=t_0^{(n)}<\cdots<t_n^{(n)}=t$$ for some$n\in\mathbb NF:[0,t]\times H\to E$be continuous$(X_t)_{t\ge 0}$be a left-continuous$H$-valued stochastic process on$(\Omega,\mathcal A,\operatorname P)... 0 Well, \begin{align} S_n&:=\sum_{i=1}^n\int_{t_{i-1}}^{t_i}L_i\Phi_0\;{\rm d}W_s\\ &=\sum_{i=1}^n\int_{t_{i-1}}^{t_i}F_x(t_{i-1},X_{t_{i-1}})\Phi_0\;{\rm d}W_s\\ &=\int_0^t\bigg[\sum_{i=1}^nF_x(t_{i-1},X_{t_{i-1}})\Phi_0\mathbf{1}_{(t_{i-1},t_‌​i]}(s)\bigg]\;{\rm d}W_s. \end{align} Now ifF(s,x)$is continuous for$(s,x)\in[0,t]\times H$and ... 0 This assertion is true and we use the approximation theorem of Weirstrass for explaining why : We can approximate uniformly a continuous function defined on a segment by polynomial functions. Thus, it exists a polynomial function$(B^\epsilon_t)_{t\in[0,T]} \in \mathcal{P}$(where$\mathcal{P}$is the set of polynomial function with real coefficient). And ... 0 Let me split this answer into two parts: Part 1 Let$U$and$H$be arbitrary Hilbert spaces,$L\in\mathcal L(U,H)$and$x\in H$. As Q. Huang noted in his answer, the authors of Stochastic Differential Equations in Infinite Dimensions$^3$"define" $$(L^\ast x)u:=\langle Lu,x\rangle\;\;\;\text{for }u\in U\;.\tag 7$$ I hate that, it's awful. Why? Well, cause ... 1 There is exactly a definition of the term$\int_0^t\langle\Phi_s{\rm d}W_s,F_x(s,X_s)\rangle$. For$\Phi_s$taking values in$\mathfrak L_2(Q^\frac{1}{2}U,H)$and satisfying the condition that the integral of$\Phi_s$-'s square-norm in$\mathfrak L_2(Q^\frac{1}{2}U,H)$is a.s. finite (just called the "Energe Condition" privately), and for$\Psi_s$a ... 1 There are many questions here, and some of them should perhaps be split into their own posts. But to address the title question, consider the Ornstein–Uhlenbeck process, which can be defined by the SDE$dX_t = dB_t - X_t \,dt$, or in various other equivalent ways. (In Wikipedia's notation, I am using the parameters$\mu = 0$and$\theta = \sigma = ...

3

I think everything you're doing is correct, but perhaps it would benefit from exploiting some cancellations so things work out more smoothly. In your second line, with six terms, I claim if you take the first, fourth, fifth, and sixth terms and sum them, you get $$\boxed{ \lambda^2 s^2 -\lambda s -2\lambda t X_s}.$$ I used this grouping because lots of ...

1

The values of $N(0), R(0), a$ do not matter, $\frac {R(t)}{N(t)} \to 1$ with probability $1$. Intuitively, when few people know the rumor, the number who know it doubles every time step. This will overtake the linear growth rate of the population. When most of the population knows the rumor, everybody around hears it pretty quickly and the only ones who ...

0

See Liggett (1991), Stochastic Interacting Systems, part I, section 3, p.71. I'll provide some insight below in how to think about it. This should be helpful in understanding the difference in extinction time on the extreme ends of super- and sub-criticality. Of course, near criticality, the analysis becomes much harder. Subcritical: In the subcritical ...

2

Note that $$\left( \frac{N_t}{t}-\lambda \right)^2= \frac{1}{t^2} (N_t-t\lambda)^2 \leq \frac{1}{\sigma^2} (N_t-t\lambda)^2$$ for all $t \in [\sigma,\tau]$. Hence, $$\mathbb{E} \left[ \sup_{t \in [\sigma,\tau]} \left( \frac{N_t}{t}-\lambda \right)^2 \right] \leq \frac{1}{\sigma^2} \mathbb{E} \left[ \sup_{t \in [\sigma,\tau]} (N_t-t\lambda)^2 \right].$$ ...

1

The property you're looking for, namely that $X_t$ is $\mathcal{B}([0,t])\otimes\mathcal{F}_t$-measurable for all $t$, is called progressive measurability. Now your $X$ may not be progressively measurable, but as an adapted (and measurable) process, it has a progressively measurable modification $Y$, and it is easy to see that modification does not change ...

4

Your mistake is in the third step you take: $\sqrt{a+b} \neq \sqrt{a}+\sqrt{b}$. You should use the expansion of $\sqrt{1+u}$ when $u\to 0$, writing $$\sqrt{e^{\sigma^2h}-1} = \sqrt{\sigma^2h + \frac{\sigma^4h^2}{2!} + o(h^2)} = \sqrt{\sigma^2h}\sqrt{1 + \frac{\sigma^2h}{2!} + o(h)}$$ and expanding the right factor as mentioned above.

1

We just have to maximize $f(t)=at-\log \phi_x(t)$ where $\phi_x(t)=1+p(e^t-1)=q+pe^t$ where $q=1-p$. One possible way is to use calculus to maximize $f(t)=at-\log (q+pe^t)$. We know that the function is concave. Hence it suffices to differentiate it and equate it to zero, we have $$a-{pe^t}/({q+pe^t})=0$$ or equivalently ...

0

First write the bernouilli mgf $$\phi_x(t)=\ln(pe^t+(1-p))$$ Plugging this into your sup equation and finding first order maximum condition will lead you to the answer.

0

A visual way to see it is to do a qq-plot. Take a bunch of $N(0,K)$ random variables (like via MATLAB's randn) and plot the quantiles of the distributions against each other (qqplot command in MATLAB). If they line up on the $x=y$ line, that's good evidence they are the same distribution and thus the noise is normal. Another way are hypothesis tests. You ...

0

1) The conditional distribution of the number of fish caught in $[30,90]$ is $Bin(\frac{1}{2}, 10)$. Because $\frac{90-30}{120-0} = \frac{1}{2}$ and $n=10$. So the wanted probabilty is that all the $10$ fish were caught in $[0,30)\cup(90,120]$ is $\frac{1}{2^{10}}$. 2) Let $N^A(2)$ be the number of type $A$ fish in $[0,120]$. Let $X^A(1)$ be the number of ...

2

Without an additional regularisation (say with colored noise), the integral you are trying to compute is strongly divergent. The integrand is dominated by $\xi^2=\left(\frac{dW}{dt}\right)^2$ which is positive and of order $\frac{1}{dt}$ and there is thus no hope to give it a meaning in the stochastic integral sense. You are just integrating infinite ...

0

An intuitive way of thinking about the fluid limit is to think of it as how the process would look like if we "zoom out" and view the process from a large distance (dividing space by $n$) and at large timescale (multiplying time by $n$). This way, stochastic fluctuations of order $\sqrt n$ would be negligible and will not be seen from this "zoomed out" view. ...

1

It seems like a minor mistake to me, but it can be easily fixed: It suffices to prove the lemma for bounded $A$. To see this, define $A^m_t=A_{t\wedge\tau_m}$, where $\tau_m=\inf\{s:A_s\geq m\}$, and for any $k\in\mathbb{N}$ take $m(k)$ such that $$E\int_{\tau_{m(k)}}^kX_t^2dA_t\leq 1/k.$$ By hypothesis, the lemma is true for $A^m$, so one can find a ...

1

Let $(B_t)$ be a Brownian motion (BM) on a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ and let $(\mathcal{F}_t)$ denote the natural filtration of BM. As you noticed BM as a stochastic process is adapted to its natural filtration, and so point 2. is satisfied. Another property from the definition of BM is the continuity of its paths, that is, for ...

1

The process stays in state $1$ an exponential amount of time with mean $1/\lambda_1$. The process then transitions to state $1$ with probability $1/2$ or to state $2$ with probability $1/2$. Let $T$ be the time spent in state $1$ until the process reaches state $2$. As a hint, let us try to calculate $\mathbb{E}[T]$. By conditioning on the number of times ...

0

Check out "Analysis, Geometry, and Modeling in Finance: Advanced Methods in Option Pricing" by Pierre Henry-Labord$\grave{\text{e}}$re. While this book is very focused on applications, it has an interesting blend of both advanced ideas in manifold theory (connections on vector bundles) as well probability and stochastic processes. The only unfortunate part ...

2

The SDE can be solved similarly as in the Vasicek model. Define $F(t,r(t)) = e^{\alpha t}r(t)$, then \begin{cases} \displaystyle \frac{\partial F}{\partial t} &= \alpha e^{\alpha t} r(t) \\ \displaystyle \frac{\partial F}{\partial r(t)} &= e^{\alpha t} \\ \displaystyle \frac{\partial^2 F}{\partial r(t)^2} &= 0. \end{cases} Applying Itô's lemma ...

1

What you want to prove is simply not true as a general fact. Take as an example the special case in which $X(t,\omega)=Y(\omega)$ for all $t$, where $Y$ is a square-integrable random variable, and likewise $X_n(t,\omega)=Y_n(\omega)$. In this case your hypothesis amounts to the condition that $E[|Y_n-Y|^2]\to 0$ as $n\to\infty$. It is well-known that while ...

1

The function $u(t,x)=e^{t/2}\,\cos(x)$ satisfies the heat equation ${d\over dt}u+{1\over 2}\Delta_x u=0,$ so that, by Ito's formula, $u(t,W_t)=e^{t/2}\,\cos(W_t)$ is a martingale.

4

First of all, note that $\mathbb{E}(B_T^3)=0$, i.e. we have to find $F_t$ such that $$B_T^3 = \int_0^T F_s \, dB_s. \tag{1}$$ It follows from Itô's formula that $$B_t^3 = 3 \int_0^t B_s^2 \, dB_s + 3 \int_0^t B_s \, ds \tag{2}$$ (please compare this with what you wrote in your question; there are so many typos in there that I'm really not sure whether ...

0

Actually, I have managed to find a solution to the equation, I post it for people who followed the topic and if it could interest some students : $-v_t + sup_{u}(-\frac{1}{2}u^2v_{xx} -x^2 + \frac{1}{2}u^2) =0$ $v(1,x)= x^2$ This system can be rewritten : $-v_t + -x^2+ sup_{u}(\frac{1}{2}u^2(1-v_{xx})) =0$ $v(1,x)= x^2$ We make the assumption that we ...

0

To find $\mathbb{E}[\cos(W(t))]$ you could try integrating the density function. However, that might turn out to be messy. Instead, recall that $$\cos x = \frac{e^{ix}+e^{-ix}}{2},$$ and hence \begin{align} \mathbb{E}[\cos(W(t))] = \mathbb{E}\left[\frac{e^{iW(t)}+e^{-iW(t)}}{2}\right], \end{align} where $\mathbb{E}[e^{iW(t)}]$ is the characteristic function ...

0

If I understand the setup correctly, it is symmetric under inversion, i.e. the random walk is equally likely to start off in a given direction or the opposite direction, and the angle constraints for subsequent steps also respect this invariance. Then the expected displacement is zero by symmetry.

1

No, the singularity of their laws has nothing to do with the singularity of the Lebesgue and counting measures. There are many ways to show this singularity. The simplest is to say that the paths of $N$ live in $\mathbb{Z}^{[0,+\infty)}$, while those of $W$ belong here with probability $0$, whence the required singularity follows.

1

Obviously, we have $$0 \leq Y \leq \frac{1}{2\epsilon} \int_{t}^{t+2\epsilon} 1 \, ds = 1.$$ Hence, $$Y = \underbrace{Y}_{\leq 1} \cdot 1_{\{Y>1/2\}} + \underbrace{Y}_{\leq 1/2} \cdot 1_{\{Y \leq 1/2\}} \leq 1_{\{Y>1/2\}} + \frac{1}{2} 1_{\{Y \geq 1/2\}}.$$ Now the claim follows by taking the conditional expectation $\mathbb{E}( \cdot \mid ... 1 It means that $$lim_{k \to \infty}\Bbb E\left\{\left[\sum_{i = 0}^{k - 1} {X_{t_{i+1}} + X_{t_i}\over 2} \left( W_{t_{i+1}} - W_{t_i} \right) - \int_0^T X_t \circ dW_t\right]^2\right\}=0.$$ 1 Let$C(t)$be the cost accumulated up to time$t$,$M^{(i)}_n$be the$n^{\mathsf{th}}$lifetime of machine$i$and $$T_n=\sum_{k=1}^n \left(M^{(1)}_n+M^{(2)}_n\right).$$ Then$\{C(t):t\geqslant 0\}$is a regenerative process with epochs$\{T_n\}$, and so by the renewal-reward theorem, $$\lim_{t\to\infty} \int_0^t C(s)\ \mathsf ds = \frac{\mathbb ... 3 Since the Poisson process (N_t)_{t \geq 0} has non-decreasing sample paths, we have$$N_{\tau} = \sup_{n \in \mathbb{N}} N_{\tau \wedge n}.$$By the monotone convergence theorem, this implies that$$\mathbb{E}(N_{\tau}) = \sup_{n \in \mathbb{N}} \mathbb{E}(N_{n \wedge \tau}).$$1 "One Thousand Exercises in Probability" by Grimmett and Stirzaker is a possible suggestion, though INMHO not as promising as it sounds. Some pros and some cons: pros: covers many areas and 1000 exercises are a 1000 exercises! Graduate level: check \checkmark cons: does not have the, let's say standard, exercises in each subject, so INMHO it serves ... 1 The long run distribution of the age of the machine currently in use has density$$ g(x) = {1-F_T(x)\over E[T]},\qquad x>0.$$The long run probability that the age of the current machine is less than one (year) is therefore$\int_0^1 g(x)\,dx\$.

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