# Tag Info

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 ...

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 $$g(x)=\lambda e^{−\lambda(x-T)} ,~x\gt T$$

3

Using the martingale property, \begin{align} \text{E}(W_n) &= \text{E}[\text{E}(W_n | W_{n - 1})] \\ &= \text{E}(W_{n - 1}) \\ &= ... \\ &= \text{E}(W_1) , \end{align} and so you have a uniform bound on both $\text{Var}(W_n)$ and $\text{E}(W_n)^2$.

3

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.

2

You can solve it using Ito's lemma but I'll rewrite it in a slightly different way to avoid the issues you came upon. We want to show $dW(t) = t^2dZ(t)$. This is really shorthand for: $$W(t) = W(0) + \int_0^t s^2 dZ(s)$$ You are also given $$W(t) = t^2Z(t) - 2\int_0^t s dZ(s)$$ We can imply from there that $W(0) = 0$ and so our ultimate goal is to show ...

2

$X_n$ is an $(\mathcal{F}_n)$-martingale: $$\mathbb{E}[X_k(X_m-X_l)]=\mathbb{E}[X_k\mathbb{E}[(X_m-X_l)|\mathcal{F}_k]]$$ $$=\mathbb{E}[X_k(X_k-X_k)]=0$$

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 $$P_1\{\max_{0\le n\le ... 1 Assuming that Var(\epsilon_1)=\sigma_{\epsilon}^2 we have$$\mathbb{E}\beta^{2t}b_t^2= \sigma_{\epsilon}^2\frac{\beta^2}{(\beta^2-1)^2}+o(1)$$Hence, \beta^tb_t is L^2-bounded and uniformly integrable so that you can pass the limit inside the expectation... Edit: In the first step, you need to show that \beta^tb_t converges a.s. (to 0). This ... 1 First of all, note that we have to ensure that$$\exp \left( \int_0^t X_s \, dB_s \right) \in L^1. \tag{1}$$If your claim is true, then$$\mathbb{E} \exp \left( \int_0^t X_s \, dB_s \right) = \mathbb{E}\exp \left( \frac{1}{2} \int_0^t X_s^2 \, ds \right),$$i.e. (1) holds if$$\mathbb{E}\exp \left( \frac{1}{2} \int_0^t X_s^2 \, ds \right)< \infty. ...

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

Doob says: $B_{\tau \wedge t} = E[B_{t \wedge t}\mid \mathcal{F}_\tau]$. Jensen says: $$E[\lvert B_{\tau \wedge t} \rvert] = E[\lvert E[B_{t \wedge t}\mid \mathcal{F}_\tau]\rvert] \leq E[ E[\lvert B_{t \wedge t}\rvert\mid \mathcal{F}_\tau]] = E[\lvert B_t \rvert] < \infty$$

1

well if exponentiation preserves a property then it will continue to hold. For example, $f$ holder cts implies that $e^f$ is so yes that property goes over.

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 ...

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 ...

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