# Tag Info

0

Beware! In your explanation of why your $2 \times 2$ matrix can be parameterized by $t_{22}$, you divide through by $x_1$, which is to say that you assume that $x_1 \neq 0$. Instead, you can explicitly write out your conditions on the elements of $\mathbf{t}$ in a matrix equation: $$\left[ \begin{matrix} x_1 & x_2 & 0 & 0 \\ 0 & 0 & ... 0 Consider a state space \mathcal S=\{0\}\cup \{\delta\}\cup A\cup B, where A=\{a_1,a_2,\ldots\} and B=\{b_1,b_2,\ldots\}, and transition probabilities$$ P_{ij} = \begin{cases} \frac13,& i=0, j\in\{\delta, a_1,b_1\}\\ 1,& i = j = \delta\\ \frac13,& i=j=a_1\\ \frac23,& i=a_n, j=a_{n+1}\\ \frac13,& i=a_{n+1}, j=a_n\\ \frac23,& ...

1

Here is a somewhat silly example. Consider the trivial probability measure on the trivial $\sigma$-algebra $\mathcal{X} = \{ \emptyset, \mathbb{R} \}$ on the reals. Then any (non-random) function $f: [0,\infty) \to \mathbb{R}$ gives rise to a stochastic process $A_t = f(t)$ with trivial finite-dimensional distributions. So there are continuous and ...

3

Without loss of generality, we may assume $x>0$. By the very definition of the stopping time, we have \begin{align*} \mathbb{P}_x(\tau_0>t) &= \mathbb{P}_x \left( \left\{w \in C[0,1]; \inf_{s \in [0,t]} (w(s)-x) +x >0 \right\} \right) \end{align*} Since $\mathbb{P}_x$ is defined by translational invariance (it is the image measure of the ...

3

$$E\left[H_{t_{i-1}}^2\left(B_{t_i}-B_{t_{i-1}}\right)^2\right] = E\left[E\left[H_{t_{i-1}}^2\left(B_{t_i}-B_{t_{i-1}}\right)^2 \mid \mathbb{F}_{t_{i-1}}\right] \right] = E\left[H_{t_{i-1}}^2E\left[\left(B_{t_i}-B_{t_{i-1}}\right)^2 \mid \mathbb{F}_{t_{i-1}}\right] \right] = E\left[H_{t_{i-1}}^2E\left[\left(B_{t_i}-B_{t_{i-1}}\right)^2 \right] \right] = ... 0 You should also mention that h_i is \mathbb{F}_{t_i}-measurable. Since i is not a time index, it is not clear what it means for (h_i) to be adapted. That being said, note that linear combinations of martingales are again martingales. So then we are done if we just show that whenever s < t$$E[ h_{i-1} (B_{t_i\wedge t} - B_{t_{i-1}\wedge t})\mid ...

0

Thanks to the hint with the Coupon Collector's problem I was able to figure out how to calculate that. Python code: #!/usr/bin/python3 import math bincnt = 24 throws = 99 def nPr(n, r): return math.factorial(n) // math.factorial(n - r) def nCr(n, r): return nPr(n, r) // math.factorial(r) psum = 0 for j in range(bincnt): term = ((-1) ...

1

Suppose $X, Z \sim \text{Pos}(\lambda)$. The event $\min (X, Z) = 1$ is the same as the event $X = 1, Z \neq 0$ or $Z = 1, X \neq 0$. As you stated, the two events have the same probability, but only when considered independently of one another. It is possible that $X = Z = 1$ and both events be satisfied. Using $$\mathbb{P}(A \cup B) = \mathbb{P}(A) + ... 0 The main point in (A Brownian motion B that is discontinuous at an independent, uniformly distributed random variable U(0,1)) is that the finite- dimensional distributions do not determine the distribution of the process. In the continuous case it is true, you can recover the process via the finite-dimensional distributions, (we say that finite ... 1 If two states are communicating, then both are either transient or recurrent. In the markov chain above, all states are communicating (see theorem 3.3 in http://www.maths.qmul.ac.uk/~ig/MAS338/Recurrence.pdf). Meaning you can get from state i to state j in finite number of steps. Note that there is also no absorbing state, hence all four states will be ... 1 So let's see that X_{\sigma^n} \to X_{\sigma} in L^1. Since \sigma^n \downarrow \sigma right continuity gives us that X_{\sigma^n} \to X_{\sigma} almost surely. Now call X(n) = X_{\sigma^n} and \mathcal{F}_n = \mathcal{F}_\sigma. Note that 1) X(n) is \mathcal{F}_n adapted 2) X(n) \in L^1 for every n 3)\Bbb{E}[X(n+1) - X(n)\vert ... 0 If f: \mathbb{R} \times \mathbb{R} \longrightarrow \mathbb{R} then Z_t = f(X_t,Y_t) is a stochastic process given by$$dZ_t = f_x dX_t + f_ydY_t + \frac{1}{2}f_{xx}d[X]_t+ \frac{1}{2}f_{yy}d[Y]_t + \frac{1}{2}f_{xy}d[X,Y]_t$$where d[X]_t is the quadratic variation of X_t and d[X,Y]_t is the quadratic covariation of X_t and Y_t. People often ... 1 Fubini's theorem on [0,t]\times\Omega gives$$\mathbb{E}\left[\int^{t}_{0} W^2_s\, ds\right] =\int_\Omega \int^t_0 W_s^2(\omega)\,ds\,\mathbb{P}(d\omega) =\int^t_0\int_\Omega W_s^2(\omega)\,\mathbb{P}(d\omega)\,ds = \int^{t}_{0} \mathbb{E}[W^2_s]\, ds.1 You do have a martingale! \begin{align*} \mathbb{E} [Z_t | \mathcal{F}_{t-1}] &= \mathbb{E} [X_t | \mathcal{F}_{t-1}] + \mathbb{E}[Z_{t-1}|\mathcal{F}_{t-1}] \\ &= 0 + Z_{t-1} \end{align*} 1 Let X_n be the corresponding Markov chain. Define N(j)=\sum_{n=0}^\infty 1\{X_n=j\} (number of visits to j), \tau_j=\inf\{n\ge 0:X_n=j\}, and (\theta_n(\omega))(m)=\omega(m+n) (shift operator). Then\mathbb{E}_iN(j)=\mathbb{E}_i\left[\sum_{n=0}^\infty 1\{X_n=j\}\right]=\sum_{n=0}^\infty \mathbb{E}_i[1\{X_n=j\}]=\sum_{n=0}^\infty p^n(i,j)$$... 0 First note that \int\kappa\left(X_s,dy\right)f(y) is a X_s- measurable function. If it holds$$\operatorname{E}_x\left[f\left(\left(X_{s+t}\right)_{t\in I}\right)\mid\mathcal{F}_s\right]=\int_{E^I}\kappa\left(X_s,dy\right)f(y)$$for every \hat{f}(y) = f_1(y(t_1))\ldots f_k(y(t_k)) then consider f_n(y) \uparrow 1_{A_1}(y(t_1)) \ldots ... 2 Recall that for any function f \geq 0 and any measure \mu we have$$\int f \, d\mu = 0 \implies f=0 \quad \mu\text{-a.s.} $$Therefore, as (\langle X \rangle_t)_{t \geq 0} is an increasing process,$$\mathbb{E} \left( \int_0^t e^{X_s} \, d\langle X \rangle_s \right)=0$$implies$$\int_0^t e^{X_s} \, d \langle X \rangle_s =0$$\mathbb{P}-almost ... 1 Denoting the current day as index i (so that yesterday is i-1 and the day before yesterday is i-2), the previous state comprises two elements, State_{n-1}=\{S_{i-1},S_{i-2}\}, where S_k is the state of the weather for day k. Thus the current state is State_{n}=\{S_{i},S_{i-1}\} Lettting R be the occurrence of rain on a particular day, so ... 1 As \mathcal{F}_{t_n} \supseteq \mathcal{F}_{t+}, we know that X_n is independent of \mathcal{F}_{t+}. Therefore,$$(B_{t+s_1}-B_t, \ldots,B_{t+s_m}-B_{t}) = \lim_{n \to \infty} (B_{t_n+s}-B_{t_n},\ldots,B_{t_n+s_m}-B_{t_n}) = \lim_{n \to \infty} X_n$$is also independent of \mathcal{F}_{t+}. Lemma: Let \mathcal{F} be a \sigma-algebra and ... 2 If X_n\to X almost surely and X_n is independent of \mathcal G, then X is independent of \mathcal G. We use actually this. Call X the left hand side of (2). We can show that for each continuous and bounded function f\colon\mathbf R^n\to\mathbf R and E\in\mathcal F_t, we have \mathbb E\left[f(X)\mathbf 1(E)\right]=\mathbb E[f(X)]\mu(E) ... 0 1) Monotone class theorem and \pi-\lambda-system of Dynkin are complementary ways to prove that a certain set of subsets contains a \sigma- algebra. One can show that M(G) the smallest monotone class of an algebra G is a \lambda- system. Similarly, one can show that \lambda(P) the smallest \lambda- system of a \pi- system G is a monotone ... 0 It suffices to apply (d) with \tilde{g}= \theta_{N+1} gto the expression in \tilde{X}_{\tilde{\theta}}.$$\tilde{X}_{\tilde{\theta}}(t) = \exp \bigg\{\langle \theta, \alpha(t)- \alpha(0) \rangle - \int_0^t \langle \theta, b(u) \rangle \, d\xi(u) -\frac{1}{2} \int_0^t \langle \theta, a(u) \theta \rangle \, du + \\ \theta_{N+1} g(t,\alpha(t)) - ...

1

Since $(X_t)$ is a Markov process, $\mathbb{E}(g(X_t)\mid {\cal F}_s)=P_{s,t}g(X_s)$ where $P_{s,t}$ is the transition kernel. Therefore $\mathbb{E}(g(X_t)\mid {\cal F}_s)$ is measurable with respect to $\sigma(X_s)=\sigma(Z_s).$

0

The Itô isometry is important because it allows you to extend the stochastic integrals that aren't in general path-by-path well defined to any function in $L^2$. Remember that the Brownian motion is of unbounded variation and the integral in Stieltjes sense is not readily available. If you would like to see a path-by-path construction of the integral you ...

1

Thanks a lot! So according to my case (for any $t_1<t_2 \leq t_3 <t_4$) it would be: \begin{eqnarray*} && P\left(X_{t_4}-X_{t_3}\leq s \;\bigcap\; X_{t_2}-X_{t_1}\leq t\right) \\ &&\qquad = P\left(\sum_{i=N_{t_3}+1}^{N_{t_4}} J_i\leq s \;\bigcap\; \sum_{i=N_{t_1}+1}^{N_{t_2}} J_i\leq t\right) \\ && \\ &&\qquad = ...

2

I follow the same path as you: \begin{eqnarray*} && P\left(X_{t_3}-X_{t_2}\leq s \;\bigcap\; X_{t_4}-X_{t_3}\leq t\right) \\ &&\qquad = P\left(\sum_{i=N_{t_2}+1}^{N_{t_3}} J_i\leq s \;\bigcap\; \sum_{i=N_{t_3}+1}^{N_{t_4}} J_i\leq t\right) \\ && \\ &&\qquad = P\left(\sum_{i=1}^{N_{t_3}-N_{t_2}} J_i\leq s \;\bigcap\; ...

3

It is true that the second property can be deduced from the first one. Indeed, the first property implies that $(B_t)$ has stationary, independent increments. Hence, the following statement would give you the implication in the forward direction: Proposition: Any real-valued stochastic process with stationary, independent increments has the elementary ...

4

Using Ito's Lemma we have $$d\log X_t=\left(v-\frac12\sigma^2\right)dt+\sigma dW_t \tag 1$$ Integrating $(1)$ between $t_1$ and $t_2$ yields \begin{align} \log(X_{t_2}/X_{t_1})&=\left(v-\frac12\sigma^2\right)\left(t_2-t_1\right)+\sigma\int_{t_1}^{t_2}dW_t\\\\ &=\left(v-\frac12\sigma^2\right)\left(t_2-t_1\right)+\sigma\left(W_{t_2}-W_{t_1}\right) ... 2 The increments are stationary. Since B_t - B_s is the increment over the interval [s, t], it is the same in distribution as the incremeent over the interval [s-s, t-s] = [0,t-s]. Hence,B_t-B_s \sim B_{t-s}-B_0.$$But B_0 = 0 almost surely, so that:$$B_t-B_s \sim B_{t-s}.$$Finally, B_{t-s} \sim \mathcal{N} (0,t-s). We didn't use the ... 1 Hints: Note that f does not depend on the time t, therefore the term \frac{\partial}{\partial t} f is superfluous. Take expectation on both sides, then the stochastic integral \dots dW_t vanishes, because it is a martingale. Use Fubini's theorem and the fundamental theorem of calculus,$$\frac{1}{t} \int_0^t \mathbb{E}^xg(X_s) \, ds \stackrel{t ...

1

If they are independent then $R_X(\tau)$ will be zero everywhere except at $\tau=0$, i.e. $R_X(\tau)=\sigma_X^2\delta(\tau)$, because $X_t$ is zero mean. Now, if you take the fourier transform of $R_X(\tau)=\sigma_X^2\delta(\tau)$ then you get simply $S_X(f)=\sigma_X^2$ for all $f$. There is no phase shift in this case because $\tau_0=0$, otherwise for ...

1

Here is one argument: The process $X_t$ has independent increments if $X_t-X_s$ is independent of $X_u$ for $u\leq s$. We have $$X_t -X_s = \sum^{N_t}_{i=1}J_i-\sum^{N_s}_{i=1}J_i = \sum^{N_t}_{i=N_s+1}J_i$$ The number of jumps in $(s, t]$ is $N_t - N_s$, which is a Poisson increment and hence is independent of of $N_u$ for $u\leq s$. The jumps are also ...

0

In general, if $T$ is a $n\times n$ stochastic matrix we can parametrize each row with $n-1$ elements and hence $T$ has $n(n-1)$ parameters.

1

I can recommend the following book "Probability and Statistics by Example: Volume 2, Markov Chains: A Primer in Random Processes and their Applications" by Yuri Suhov and Mark Kelbert This book covers most of your topics (depends of course how deep you want to dive into), is well written and explains everything with examples and solved exercises. So at ...

2

A good starting point would be Introduction to Probability Models by Sheldon Ross. I think it covers all the topics that you described. I like the book because it's easy to read and has plenty of problems to try out, which makes it ideal for self learning.

1

To keep notation simple, we write $S_t$ instead of $S_t^i$, $\sigma_t^j$ instead of $\sigma_{t}^{ij}$ and so on. That's okay, because $i$ is a fixed number throughout this calculation. Suppose $(X_t)_{t \geq 0}$ is an Itô process of the form $$dX_t = b(t) \, dt + \eta(t) \, dW_t$$ where $\eta = (\eta_1,\ldots,\eta_m)$ and $(W_t)_{t \geq 0}$ is an ...

0

This answer takes a while. We will need several steps which I list here to guide our lines: 1)($Z$ simple) $I^M_t(Z)^2 - \int_0^t Z^2(s) \, d\langle M\rangle_s$ is a continuous local martingale 2) ($X$ $\Bbb{F}$- progressively measurable) $I^M_t(X)^2 - \int_0^t X^2(s) \, d\langle M\rangle_s$ is a continuous local martingale 3) $\langle I^M(X) + ... 2 Set $$f_n(s) := \sum_{j=1}^n s_j 1_{[t_{j-1},t_j]}(s).$$ Then your calculation shows $$\Phi_{X^m}(s) = \exp \left(- \frac{1}{2} \sum_{k=1}^m \langle f_n, e_k \rangle^2 \right).$$ Letting$m \to \infty$, we obtain $$\Phi_X(s) = \exp \left(- \frac{1}{2} \sum_{k \geq 1} \langle f_n, e_k \rangle^2 \right). \tag{1}$$ Since$(e_k)_{k \geq 1}$is an ONB, we ... 1 Doob says:$B_{\tau \wedge t} = E[B_{t \wedge t}\mid \mathcal{F}_\tau]$. Jensen says: $$E[\lvert B_{\tau \wedge t} \rvert] = E[\lvert E[B_{t \wedge t}\mid \mathcal{F}_\tau]\rvert] \leq E[ E[\lvert B_{t \wedge t}\rvert\mid \mathcal{F}_\tau]] = E[\lvert B_t \rvert] < \infty$$ 4 We have to consider three cases separately:$S_{k-1}(\omega)<0$: Then$S_k(\omega) \leq 0$and therefore $$|S_k(\omega)|-|S_{k-1}(\omega)| = -S_k(\omega)+ S_{k-1}(\omega) = - X_k(\omega).$$$S_{k-1}(\omega)=0$: Then$|S_k(\omega)|=1$and therefore $$|S_k(\omega)|-|S_{k-1}(\omega)|=1.$$$S_{k-1}(\omega)>0$: Then$S_k(\omega) \geq 0$and therefore ... 1 The result is true, and has nothing to do with topology or countability. Let${\cal C}$be the collection of subsets of$E^I$of the form$\pi_i^{-1}(E)$for some$i\in I$and$E\in{\cal E}$. Then use the fact that$X^{-1}(\sigma({\cal C}))=\sigma(X^{-1}({\cal C})). 1 Write $$\sum_{k=0}^{\infty} \frac{e^{-\lambda t} (\lambda t)^k}{k!} (f(N_t+k)-f(N_t)) = I_1+I_2+I_3$$ where \begin{align*} I_1 &:= e^{-\lambda t} \cdot 1 \cdot (f(N_t+0)-f(N_t))=0, \\ I_2 &:= e^{-\lambda t} \lambda t (f(N_t+1)-f(N_t)) \end{align*} and $$I_3 := \sum_{k \geq 2} e^{-\lambda t} \frac{(\lambda t)^k}{k!} (f(N_t+k)-f(N_t)).$$ It ... 0 By Strong Law of Large Numbers, without any assumptions on the variance of\epsilon_i$, $$\frac{1}{n}\sum^n_{i=1}\epsilon_i \rightarrow \mathbb E[\epsilon_i]=0 \quad \text{almost surely}$$ So since$\lim_{n \rightarrow \infty} n\beta^n = 0$and$n^{-1}b_n \rightarrow 0$almost surely, we have that $$\beta^n b_n\rightarrow 0 \quad \text{almost surely}$$ ... 4$X_n$is an$(\mathcal{F}_n)$-martingale: $$\mathbb{E}[X_k(X_m-X_l)]=\mathbb{E}[X_k\mathbb{E}[(X_m-X_l)|\mathcal{F}_k]]$$ $$=\mathbb{E}[X_k(X_k-X_k)]=0$$ 0 ok thanks a lot for that corindo, do you believe we can do a similar transformation for the three-dimensional case: $$dX_t^1=(X_t^2-2X_t^1)\,dt+\sigma_1 \,dW_t^1,\quad X_0^1=x_1$$ $$dX_t^2=(X_t^1+X_t^3-2X_t^2)\,dt+\sigma_2 \,dW_t^2,\quad X_0^2=x_2$$ $$dX_t^3=(X_t^2-2X_t^3)\,dt+\sigma_3 \,dW_t^3,\quad X_0^3=x_3$$ My final aim is to simulate the ... 1 Assuming that$Var(\epsilon_1)=\sigma_{\epsilon}^2$we have $$\mathbb{E}\beta^{2t}b_t^2= \sigma_{\epsilon}^2\frac{\beta^2}{(\beta^2-1)^2}+o(1)$$ Hence,$\beta^tb_t$is$L^2$-bounded and uniformly integrable so that you can pass the limit inside the expectation... Edit: In the first step, you need to show that$\beta^tb_t$converges a.s. (to$0$). This ... 0 The transformation you suggested gives:$dA_t=\frac{1}{\sqrt{2}}(\sigma_1 dW_t^1+\sigma_2 dW_t^2)dB_t=\frac{1}{\sqrt{2}}(2X_t^2-2X_t^1+\sigma_1 dW_t^1-\sigma_2 dW_t^2)$How is this supposed to help me? 3 Memorylessness might suggest that if the inter-arrival time in the second process is$X$then it is simply shifted up by$T$so has density $$g(x)=\lambda e^{−\lambda(x-T)} ,~x\gt T$$ 0 You can simply try the transformation $$A_t = \frac{X_t^1 + X_t^2}{\sqrt{2}}, \quad B_t = \frac{X_t^1 - X_t^2}{\sqrt{2}}$$ The equations will get much simpler. 2 Let$T=\inf\{n:X_n>L\}$. Define$\mathbb{E}_x[\cdot]=\mathbb{E}[\cdot|X_0=x]$,$\mathcal{F}_n=\sigma(X_0,\dots,X_n)$, and$\mathcal{F}_T=\{A:A\bigcap\{T=n\}\in\mathcal{F}_n\text{ for all $n$}\}$. Then since $$\{X_n>L\text{ for some 0\le n\le m }\}=\{\max_{0\le n\le m}X_n>L\}=\{T\le m\}$$ we have$\$P_1\{\max_{0\le n\le ...

Top 50 recent answers are included