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2

You can see $\sigma$-algebras as gathering of information. Following your example, knowing $\mathcal{F}_{n}$ is entirely sufficient to know the value of $X_{1}, ..., X_{n}$ for every $\omega$ : if you give me some $\omega $ and if I know $\mathcal{F}_{n}$, I'll be able to find a (measurable) subset of the form $\{X_{n} = x\}$ and therefore I'll be sure that ...


1

Note that $\mathrm d\langle V,W\rangle_t=\sigma\rho\sqrt{V_t}\mathrm dt$ hence, for every $t\gt0$, $$\rho=\frac1{\sigma t}\int_0^t\frac{\mathrm d\langle V,W\rangle_s}{\sqrt{V_s}}.$$ Likewise, for every $t\gt0$, $$\sqrt{1-\rho^2}=\frac1{\sigma t}\int_0^t\frac{\mathrm d\langle V,Z\rangle_s}{\sqrt{V_s}}.$$ Edit: Numerically, the integral involving the ...


1

Your reasoning is in both cases correct. You actually already proved that $d_m$ is a stopping time: $$\{d_m = n\} = \bigcap_{j=m}^{n-1} \{S_j>0\} \cap \{S_n \leq 0\} \in \mathcal{F}_n$$ as $\{S_j>0\} \in \mathcal{F}_n$ for all $j=m,\ldots,n-1$ (since $S_j$ is $\mathcal{F}_j$-measurable, hence $\mathcal{F}_n$-measurable) and $\{S_n \leq 0\} \in ...


1

Let $Y_t=\displaystyle\int_0^t\mathrm e^{au}\mathrm dW_u$ then, for every $t$, $X_t=b\mathrm e^{-at}Y_t+z(t)$ where $z(\ )$ is deterministic hence $$\mathrm{cov}(X_t,X_s)=(b\mathrm e^{-at})(b\mathrm e^{-as})\mathrm{cov}(Y_t,Y_s),$$ and, for every $t$, $Y_t=\displaystyle\int_0^\infty g_t(u)dW_u$ where $g_t$ is deterministic since $g_t(u)=\mathrm e^{au}\mathbf ...


-1

Actually the answer was MUCH more complicated, for anyone interested it is possible to find a proof in Sato, Lèvy processes and infinitely divisible distributions, in chapter 2


2

Note that the equality $$X_t = 2 X_{t/2}$$ does not hold. It follows from stationarity of the increments that $$X_{t}-X_{t/2} \stackrel{d}{\sim} X_{t/2},$$ this means that they are equal in distribution and not almost surely. (Recall that $X \stackrel{d}{\sim}Y$ does not imply $X+Y = 2X$.) One of the most basic examples for processes with stationary and ...


3

Since the Markov chain is irreducible, it is possible to get from state $i$ to state $j$, so $p_{ij}^{(k)} > 0$ for some $k$. Then $p_{ij}^{(n+k)} \ge p_{ii}^{(n)} p_{ij}^{(k)}$, so $$\sum_{n\ge 0} p_{ij}^{(n)} \ge \sum_{n\ge k} p_{ij}^{(n)} = \sum_{n\ge 0} p_{ij}^{(n+k)}\ge p_{ij}^{(k)} \sum_{n \ge 0} p_{ii}^{(n)} = \infty$$


0

Re your first question, note that for every independent random variables $(X,Y)$, $X$ with density $f$, and every events $(A,B)$, $$P(X\in A,X+Y\in B)=\int_Af(x)P(Y+x\in B)\mathrm dx.$$ To show this, note that, in full generality, $$P(X\in A,X+Y\in B)=\iint \mathbf 1_{x\in A}\mathbf 1_{x+y\in B}\mathrm dP_{(X,Y)}(x,y),$$ hence, by the independence of ...


1

Here's a rough estimate which may give an idea of how to get a proper answer. Let $X_i(t)$ be the number of colors such that there are $i$ balls of that color. Let the level at time $t$, denoted by $L(t)$, be the largest $i$ such that $X_i(t)>0$. Suppose $L(t)=i$. We want the probability that $L(t+1)=i+1$. At time $t$, there are $i X_i(t)$ balls of ...


1

All these depend a lot on various circumstances that may or may not be the case in a particular application. For example, in (2) the Poisson model may be good if traffic is light and flowing freely, but not if it's a two-lane highway with few passing opportunities where traffic tends to bunch up behind the slower vehicles. Poisson would not be a good model ...


1

What about a look at the literature? Sato proves this result in the first chapter of his monograph Lévy processes and infinitely divisible distributions (Theorem 3.2) using elementary properties of conditional probability. A more elegant proof can be based on Campbell's formula, but then we need some results on stochastic integration with respect to jump ...


3

First part: For every $n$, the simple Markov property at time $nK$ yields $$P_j(T_i\gt(n+1)K\mid T_i\gt nK,X_{nK}=k)=u(k),\qquad u(k)=P_k(T_i\gt K),$$ and $u(\ )\leqslant1-\varepsilon$ uniformly by hypothesis hence $$P_j(T_i\gt(n+1)K,X_{nK}=k)\leqslant(1-\varepsilon)P_j(T_i\gt nK,X_{nK}=k).$$ Summing these over $k$ yields ...


3

Hints Brownian motion: Apply Itô's formula to $f(t,x) := \exp(\beta x - t \beta^2/2).$ Poisson process: Use that $(N_t)_{t \geq 0}$ has independent increments, i.e. $N_t-N_s$ is independent of $F_s^N$. Start with the easier one: $$\mathbb{E}(N_t-\lambda t \mid F_s^N) = \mathbb{E}((N_t-N_s)+N_s \mid F_s^N) - \lambda t = \ldots$$


0

http://www.renyi.hu/~major/articles/breuer.pdf . I think this paper will clear your doubts.


2

For each fixed $t$, the random variable $X_t=W_{2t}-W_t$ is centered normal with variance $t$ hence indeed distributed like $W_t$. But the process $(X_t)$ is not a Brownian motion. To see this, note that $X_2=W_4-W_2$ and $X_1=W_2-W_1$ are the increments of a Brownian motion on some disjoint time intervals hence they are independent while $W_2$ and $W_1$ ...


0

You could also check the fact that $X_t$ is not a martingale and hence cannot be a Brownian motion. In particular, $E[X_t|F_s] = 0 \neq X_s$ for $s\leq t$.


0

$X_\frac{1}{2} = W_1 - W_{\frac{1}{2}}$ and $X_1 - X_{\frac{1}{2}} = (W_2 - W_1)-(W_1 -W_\frac{1}{2})$. They are not independent, while a Brownian motion has independent increments. Remark that we can only say $W_{2t} = \sqrt{2}W_t$ in distribution, they are not equal almost surely.


0

Summary By default, $dB(t)^2 = dt$ is a stochastic differential equation, that is, an equation involving stochastic differentials of the type $dB^2$ and $d[B]$. In my opinion, mnemonic rules for calculations with stochastic differentials such as $(dt)^2=0$, $dB(t)dt = 0$, etc., more than anything else, adds noise and complexity to the proper use, if they ...


1

We use the fact that if there exists measurable function $f$ such that $Y = f(X_1, \cdots, X_n)$, then $\sigma(Y)\subset \sigma(X_1, \cdots, X_n)$. Thus obviously $\sigma(X_1 - X_0, \cdots, X_n - X_{n-1}) \subset \sigma(X_1, \cdots,X_n)$, since $X_0 = 0$ is deterministic. On the other hand, $X_k = \sum_{i=1}^k(X_{i} - X_{i-1})$, thus $\sigma(X_1, ...


1

Let $(t_i)_{i \in \mathbb{N}}$ be a dense subset of $[0,T]$ such that $t_k = T$ for some $k \geq 1$. If we set $\mathcal{H}_n := \sigma(B_{t_1},\ldots,B_{t_n})$ and $$\mathcal{H}_{\infty}:= \sigma \left( \bigcup_{n \in \mathbb{N}} \mathcal{H}_n \right) \qquad \quad \mathcal{F}_T := \sigma(B_s; s \leq T),$$ then $\mathcal{H}_{\infty} = \mathcal{F}_T$. ...


0

Let $W_t=E[M_t|H_t]$. For $s\leq t$, iterated conditioning gives $$ E[W_t|H_s]=E[E[M_t|H_t]|H_s]=E[M_t|H_s]. $$ Using iterated conditioning again, the RHS above equals $$ E[M_t|H_s]=E[E[M_t|F_s]|H_s]=E[M_s|H_s]=W_s $$ where the next to last equality above uses the Martingale property of $M$. Thus we indeed have $$ E[W_t|H_s]=W_s\text{ for }s\leq t. $$


0

For $s \leq t$, $E[E[M_t|F_s]|H_s] = E[M_t|H_s]$ due to iterated conditioning. Furthermore, $E[E[M_t|F_s]|H_s] = E[M_s|H_s] = M_s$ since $M_t$ is an $F$-martingale and $M_s$ is $H_s$ measurable. Therefore, $E[M_t|H_s] = M_s$.


0

Compare: $Z\sim U(-2;2)$ How do I calculate $P(|Z|\in [{3 \over 2} ; {5 \over 2}])$ ? Is this really a problem? Then we have a problem (Houston)... :-)


4

If $(X_n)_{n \in \mathbb{N}}$ is a martingale, then $$\mathbb{E}(X_{\tau}) = \mathbb{E}(X_{\nu}) \tag{1}$$ holds for any two bounded stopping times $\tau,\nu$. In contrast, if $(X_n)_{n \in \mathbb{N}}$ is a submartingale, then $$\mathbb{E}(X_{\tau}) \leq \mathbb{E}(X_{\nu})$$ for bounded stopping times $\tau \leq \nu$. These results are known as ...


2

Long form of angryavian's argument: Let $Y=\frac1aX=g(X)$, with $X$ having pdf $f_X$, and $a>0$. There is a standard way to find the pdf of Y: for any integrable function $h$, $$E(h(Y))=\int_{-\infty}^\infty h(y) f_Y(y) \mathrm{d}y$$ But you have also $$E(h(Y))=E(h(g(X)))=\int_{-\infty}^\infty h(g(x)) f_X(x) \mathrm{d}x=\int_{-\infty}^\infty h(x/a) ...


0

If you want your modified random number $Y$ to follow the distribution $\propto e^{-a^2 y^2}$, the calculation there shows that $aY$ follows the distribution $\propto e^{-x^2}$, so what you can do is generate a random number $X$ from $\propto e^{-x^2}$, and then $X/a$ will follow the distribution $\propto e^{-a^2 y^2}$.


2

Since $e^{-\lambda t}$ does not depend on $s$, we have for all $t$ $$ P(X_2 > t) = \int_0^\infty \Pr(X_2 > t \mid X_1 = s) \lambda e^{-\lambda s}\,ds = e^{-\lambda t}. $$ Thus $X_2$ has exponential distribution with parameter $\lambda$ and the identity $$ \Pr(X_2 > t | X_1 = s) = \Pr(X_1 > t) = \Pr(X_2 > t),\qquad t,s > 0 $$ shows that ...


1

Note that $$X_t := - \int_0^t H_s \, dB_s - \frac{1}{2} \int_0^t H_s^2 \, ds$$ defines an Itô process. Therefore we may apply Itô's formula to $f(x) := e^x$: $$\begin{align*} f(X_t)-f(X_0) &= \int_0^t f'(X_s) \, dX_s + \frac{1}{2} \int_0^t f''(X_s) \, d\langle X \rangle_s \\ \Leftrightarrow M_t - 1 &= - \int_0^t M_s H_s \, dB_s - \frac{1}{2} ...


0

Conditional independence is the relevant concept. Let $\cal F$ be the natural filtration of $(X_t)$ and $\cal G \supset \cal F$ a bigger filtration. Then $(X_t)$ is Markovian with respect to $\cal G$ means that $\cal F_\infty$ is conditionally independent of $\cal G_t$ given $\sigma(X_t)$. You can check that 1 and 2 (the case when $\cal G=\cal F$) are ...


1

I would look at Autoregressive or ARMA models. I'm not sure if they'll work like you want them to, but when you include memory in your model things get a lot more complicated. Instead of being independent of most values in the series, suddenly they're dependent on all of them and things get gradually more and more complex. If you discuss a little more about ...


1

As explained in the comments, differentiating exponentials of matrices such as $\bar P(s,t)$ is slightly more complicated than in the scalar case since, quite generally, $$\left.\frac{\mathrm d}{\mathrm dt}\mathrm e^{A+tB}\right|_{t=0}=\sum_{i\geqslant0}\sum_{k\geqslant0}\frac{A^iBA^k}{(i+k+1)!},$$ which is neither $B\mathrm e^A$ not $\mathrm e^AB$ in ...


3

Let $(X_t,\mathcal{F}_t)_{t \geq 0}$ be an (càdlàg) $L^2$-martingale, i.e. a martingale which satisfies $$\sup_{t < \infty} \mathbb{E}(X_t^2)<\infty.$$ Then it follows from the Doob-Meyer decomposition that there exists a unique increasing previsible process $(A_t)_{t \geq 0}$ such that $A_0=0$ and $$(X_t^2-A_t,\mathcal{F}_t)_{t \geq 0} \,\, ...


1

The simplest approach: $X_0=Y_0$ and, for every $n$, $$X_{n+1}=A(X_n,Y_{n+1}),\qquad A(x,y)=\max\{x,y\}.$$ And now, watch the results fall in line like dominoes: Initial distribution: the distribution of $Y_0$ Markov property: obvious since $X_{n+1}$ is a deterministic function of the present state $X_n$ and of a new input $Y_{n+1}$ which is independent ...


1

$\{W_0>0\} = \{\tau_+=0\}$ is not correct. Note that $\tau_+(\omega)=0$ if, and only if, there exists a sequence of numbers $(t_n)_n \subseteq [0,\infty)$ such that $t_n \to 0$ and $W_{t_n}(\omega)>0$. Show the following equalities: $$\begin{align*} \{\tau_+=0\} &= \{\omega; \exists (t_n)_n \subseteq [0,\infty), t_n \downarrow 0: ...


1

Hint: In full generality (if $U$ and $V$ are integrable), $$E(\color{red}{5}U+\color{blue}{3}V)=\color{red}{5}\cdot E(U)+\color{blue}{3}\cdot E(V).$$ If $U$ and $V$ are independent (and square integrable), $$\mathrm{var}(\color{red}{5}U+\color{blue}{3}V)=\color{red}{5}^2\cdot \mathrm{var}(U)+\color{blue}{3}^2\cdot \mathrm{var}(V).$$


0

Handbook of Stochastic Methods by C W Gardiner http://tberg.dk/books/Handbook_of_Stochastic_Methods_for_Physics,_Chemistry_and_the_Natural_Sciences_(Gardiner).pdf Stochastic Processes in Physics and Chemistry N G Van Kampen http://bookzz.org/s/?q=stochastic+processes+in+physics+and+chemistry+van+kampen&t=0


1

When branching processes are concerned, conditioning on the first generation often uncovers the recursive structure of the whole object, hence its properties. In the present case, call $N$ the total size of the random tree $T$ and $L$ the progeny of the ancestor, assuming the branching process starts from an initial population of $1$. If $L=0$, then ...


1

Using the definitions, one sees that $$P[X(t) - X(s) = 1 \mid X(t) = 4]=\frac{P[X(t) - X(s) = 1, X(t) = 4]}{P[ X(t) = 4]}=\frac{P[4 - X(s) = 1, X(t) = 4]}{P[ X(t) = 4]},$$ and this would equal your suggestion $$ P[4 - X(s) = 1]$$ only if the numerator could be rewritten as $$P[4 - X(s) = 1]\,P[ X(t) = 4],$$ that is, if $X(s)$ and $X(t)$ were independent. But ...


1

It may be helpful to look at the more familiar deterministic case first. In order to obtain an analogy to the stochastic case you shouldn't use the relation $x(t)\Longleftrightarrow X(f)$, but you have to consider the energy spectral density $|X(f)|^2$. Its inverse Fourier transform is the deterministic autocorrelation function $\rho_x(\tau)$: ...


0

To derive the Black-Scholes PDE you demonstrate that the value of the option can be replicated by a dynamic trading strategy that holds positions in the underlying asset and the risk-free asset. As the underlying asset price changes over time, the positions are rebalanced to ensure that the portfolio tracks the option value -- leading ultimately to the same ...


3

The mapping $\omega \mapsto X_{\xi(\omega)}(\omega)$ is measurable if the corresponding process is jointly measurable, i.e. if $$(t,\omega) \mapsto X(t,\omega) \tag{1}$$ is $\mathcal{B}([0,T]) \otimes \mathcal{A} /\mathcal{B}(\mathbb{R})$-measurable for $T>0$. This can be proved in the following way: Let $f$ be a continuously differentiable function, ...


1

By definition, $$\mathbb{E}(\chi_{\{X_t = i\}} \mid X_{s_n}) = \mathbb{P}(X_t = i \mid X_{s_n}). \tag{1}$$ Now recall that for any random variable $Y$ and $B \in \mathcal{B}(\mathbb{R})$ there exists by the factorization lemma a function $h$ such that $$\mathbb{P}(B \mid Y) = h(Y);$$ this function $h$ is denoted by $$h(y) = \mathbb{P}(B \mid Y=y).$$ ...


0

Let $T$ denote the time between two successive $X_2$ arrivals, then $T$ is exponential with parameter $\lambda_2$ and, conditionally on $T=t$, the distribution of $N$ is Poisson with parameter $\lambda_1t$. Thus, $E(s^N\mid T)=\mathrm e^{-\lambda_1T(1-s)}$ and $E(s^N)=E(\mathrm e^{-rT})$ with $r=\lambda_1(1-s)$. Recall that, for every $u\gt-\lambda_2$, ...


2

If $t\leqslant s$, then $N(t)\leqslant N(s)$ hence $$X(t)X(s)=\sum_{n,k}Y_nY_k\mathbf 1_{n\leqslant N(t),k\leqslant N(s)}=\sum_{n}Y_n^2\mathbf 1_{n\leqslant N(t)}+\sum_{n\ne k}Y_nY_k\mathbf 1_{n\leqslant N(t),k\leqslant N(s)},$$ which implies that $E(X(t)X(s))=E(Y^2)U+E(Y)^2(V-U)$ with $$U=\sum_{n}P(n\leqslant N(t))=E(N(t)),$$ and $$ V=\sum_{n,k}P(n\leqslant ...


1

Here is a complete solution to (d), which is provided in the hope that the OP will emulate it for the other items. Consider the results $(Z_n)_{n\geqslant1}$ of the die, these are some i.i.d. sequence and, for every $n$ and $k$, $$[B_n=k]=[Z_{n+k}=6]\cap\bigcap_{i=1}^{k-1}[Z_{n+i}\ne6].$$ The dynamics of $(B_n)$ is as follows: If $B_n=k$ with ...


2

For a fixed $t$, the series defining $X(t)$ has only one term which can be different from $0$ (for the potential index $n$ for which $nT\lt t\leqslant nT+T/2$). Therefore, we can switch the sum and the expectation and we find that $\mathbb E[X(t)]=0$ for each $t$ because $p(t-nT)$ is not random and $\mathbb E[A_n]=0$. For the covariance, by the argument ...


1

In http://en.wikipedia.org/wiki/Feynman%E2%80%93Kac_formula#cite_note-1 they reference as having a proof Pham, Huyên (2009). Continuous-time stochastic control and optimisation with financial applications. Springer-Verlag


1

Assume that $X$ solves $$ dX_t = \mu(t,X_t) dt + \sigma(t, X_t) dW_t. $$ In integral form, this means $$ X_t = X_0 + \int_0^t \mu(s,X_s) ds + \int_0^t \sigma(s,X_s) dW_s. $$ Now, the quadratic variation of a stochastic integral process $H\cdot Y$, where $(H\cdot Y)_t = \int_0^t H_s dY_s$, is $$ [H\cdot Y]_t = \int_0^t H_s^2 d[Y]_s. $$ You can find this ...


1

When each point of a homogeneous Poisson point process (HPPP) of intensity (rate) $\lambda$ is marked (selected) with a probability $p$, independently of the other points, the result is two independent HPPPs (consisting of the marked and non-marked points) with intensities $\lambda p$ and $\lambda (1-p)$, respectively. In this case, we have a HPPP ...


1

Actually if $\theta=b/(a+b)$ the drift is zero on $[b,+\infty)$ hence the chain is null recurrent. Positive recurrence holds for every $\theta\lt b/(a+b)$, which we assume from now on. As you explain, the stationary distribution solves $\pi=\pi P$. For every $n\geqslant a$, this reads $$\pi(n)=\theta\pi(n-a)+(1-\theta)\pi(n+b).$$ Such linear systems are ...



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