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0

If the paths are continuous then, for fixed $t$, you have $(W_s^{(N)} -W_t^{(N)})^2 \to 0$ and by dominated convergence, the expectation goes to zero (for fixed $N$). On the other hand, $ E(W_s^{(N)} -W_t^{(N)})^2)^2= Var(W_s^{(N)}) +Var(W_t^{(N)}) \ge Var(W_t^{(N)})$ is close to $Var(W_t)$.So all you need additionally is that $Var(X_t)>0$.


0

If you are familiar with $\sigma$-algebras, then you could start by demonstrating that $$ P(F\cap G|{\mathcal C}) =P(F|{\mathcal C})\cdot P(G|{\mathcal C}),\quad\forall F\in{\mathcal A},G\in{\mathcal B} $$ is equivalent to $$ P(G|{\mathcal C}\vee{\mathcal A}) =P(G|{\mathcal C}),\quad\forall G\in{\mathcal B}. $$ Here ${\mathcal A}, {\mathcal B},{\mathcal C}$ ...


0

Your intuition is good, but there are some technicalities you should be careful about. WLOG $M_0 = 0$. Start by assuming $M$ is a continuous martingale of bounded variation. Then if $B$ is a bound on the variation of $M$, and $(t_i)$ is a partition of $[0,t]$, $$ E[M_t^2] = E\sum (M_{t_{i+1}}-M_{t_i})^2 \leq B E[\sup_i|M_{t_{i+1}}-M_{t_i}|]. $$ Since $M$ ...


1

Conditioning on $N$ you get $P(X=x)=\sum_{n=x}^{\infty}\binom{n}{x}p^x(1-p)^{(n-x)}e^{-\lambda}\frac{\lambda^n}{n!} $ $=e^{-\lambda}\frac{(\lambda p)^x}{x!}\sum_{n=x}^{\infty}\frac{(\lambda (1-p))^{n-x}}{(n-x)!}$ $=e^{-\lambda}\frac{(\lambda p)^x}{x!}e^{\lambda (1-p)}=e^{-\lambda p} \frac{(\lambda p)^x}{x!}$ Similarly you can get the distribution of Y. ...


1

\begin{align} \binom{y+x}{x} p^{x}(1-p)^{y}\cdot\frac{\lambda^{y+x}e^{-\lambda}}{(y+x)!} \\ =\frac{\ (y+x)!}{y!x!} p^{x}(1-p)^{y}\cdot\frac{\lambda^{y+x}e^{-\lambda}}{(y+x)!} \\ =\frac{\lambda^{x} p^{x} e^{-\lambda p}}{x!} \cdot\frac{\lambda^{y}(1-p)^{y}e^{-\lambda (1-p)}}{y!} \end{align} Hence X and Y are independent Poisson with λ and λ(1-p) respectively. ...


0

The question in its present form is unclearly stated and some assertions in the question are incorrect. Based on comments under the question, here is what I think is meant: Two Poisson processes have respective rates $\lambda_0$ and $\lambda_1$; They are independent. (This was in comments, not in the original question.); Each time an arrival occurs, the ...


1

Hint: A Process $N$ is a poisson process with intensity $\lambda$ iff: $N(0) = 0$ almost surely. $N(t) - N(s) \sim \mathrm{Poi}(\lambda(t-s))$ for all $0 \le s < t$. $N(t) - N(s)$ is independent of $\sigma(\{N(k) \mid 0 \le k \le s\})$ for all $0 \le s < t$. $N$ has almost surely càdlàg paths.


1

Processes that are not right-continuous are a gigantic pain to deal with. Most likely they include right-continuous in their definition because they always intend to work under that assumption, and don't want to have to keep writing it. As with any mathematical term, there doesn't have to be a universal "correct" definition of martingale; authors are free to ...


2

These still aren't the most general definitions for martingales. The most general definition I know of is: Let $I$ be a poset. A Family $F$ of sigma-algebras is said to be a filtration if for all $s, t \in I$ with $s \le t$ the inclusion $F_s \subset F_t$ holds. A family $X_i$ of (quasi)integrable random variables is called a martingale with respect to $F$, ...


1

Regarding your second point I think this is clearer if you know that $\mathcal{F}_{\sigma\wedge\tau}=\mathcal{F}_{\sigma}\cap\mathcal{F}_{\tau}$ because then instead of writing : $$A \in \mathcal{F}_{\sigma}: \quad A =A \cap \{\sigma \leq \tau\} \in \mathcal{F}_{\tau \wedge \sigma} = \mathcal{F}_{\sigma}$$ You would write : $$A \in ...


1

Let $\epsilon > 0$ and consider the stopping time $$\tau_\epsilon = \inf\{t \in [\tau_0, T] : X_t \ge \epsilon\} \wedge T.$$ Since $\tau_0 \le \tau_\epsilon$ almost surely, optional stopping gives $E[X_{\tau_\epsilon}] \le E[X_{\tau_0}]$. Now let consider the events $\{\tau_0 < T\}$ and $\{\tau_0 = T\}$ and write $$E[X_{\tau_\epsilon} ; \tau_0 < T] ...


0

Here are two approaches, one more elementary than the other. Modify your argument to show that $X(r) = 0$ a.s on the event $\{\tau_0< r\}$ for each $r\in (0,T]$. Deduce from this that $X(r)=0$ for every rational in $(\tau_0,T)$, a.s. on the event $\{\tau_0<T\}$. Now invoke the continuity of $X$. Let $B=\{\omega: \tau_0(\omega)<T, X(t,\omega)>0$ ...


2

You just need to look at more stopping times. For any $t\geq 0$, you have $${E}[\mathbf{1}_{\{\tau_0<T\}}X(T \wedge (\tau_0+t))] ={E}[\mathbf{1}_{\{\tau_0<T\}}E[X(T\wedge (\tau_0+t))|F_{\tau_0}]] \le {E}[\mathbf{1}_{\{\tau_0<T\}}X(\tau_0)]=0.$$ This gives $$P\biggl(X(T\wedge(\tau_0+t))=0\mbox{ for all }t\in\mathbb{Q}\cap[0,T]; \tau_0<T\biggr) ...


2

You can generalize your derivation from: $$E[\mathbf{1}_{\{\tau_0<T\}}X(T)]=E[\mathbf{1}_{\{\tau_0<T\}}E[X(T)|F_{\tau_0}]]\le E[\mathbf{1}_{\{\tau_0<T\}}X(\tau_0)]=0$$ to: $$E[\mathbf{1}_{\{\tau_0=s\}}X(t)]=E[\mathbf{1}_{\{\tau_0=s\}}E[X(t)|F_{\tau_0}]]\le E[\mathbf{1}_{\{\tau_0=s\}}X(\tau_0)]=0$$ where $t > s$. As you mentioned, using the fact ...


2

$\newcommand{\var}{\operatorname{var}}\newcommand{\E}{\operatorname{E}}$The distribution of $S_N$ is called a "compound Poisson distribution". $$ \var(S_N \mid N=n) = \var(S_n) = n\sigma^2. $$ Therefore $$ \var(S_N\mid N) = N\sigma^2, $$ so $$ \E(\var(S_N\mid N)) = \sigma^2\E(N) = \sigma^2\lambda. $$ Next we have $$ \E(S_N\mid N=n) = \E(S_n) = n\mu. $$ ...


1

Your first solution looks reasonable. I'm not sure I understand what your second solution even really means, with the expected value of the conditional variance and so on. Maybe it works, I'm just not sure how to parse it. Here is a different idea: write $$\text{Var}(S_N)=E[S_N^2]-(E[S_N])^2=E[S_N^2]-\mu^2 \lambda^2$$ using part 1. Now $$E[S_N^2]=\sum_n ...


0

Thank you Augustin and Bryon. It seems that: There can in fact be many eigen vectors for eigenvalue 1 which are not probability vectors. There can also be many probability vectors (aka stationary probability vectors), for the eigenvalue 1. thank you for your answers.


1

$\tau$ here strikes me as a Stopping Time. In that case, it is not a process but a random variable a) defined on the entire space of paths, b) measurable with respect to the filtration $\mathcal{F}_t$ in the sense that: $$\{\tau \leq t\}$$ is $\mathcal{F}_t$ measurable for all $t$. It is strange for me to think of it as a standard adapted process. Say for ...


1

Let $A = \left(\begin{array}{rr} 0 & 1\\ -1 & 0 \end{array} \right)$, $X_t = \left(\begin{array}{r} X_1(t)\\ X_2(t) \end{array} \right)$, and $B_t = \left(\begin{array}{r} B_1(t)\\ B_2(t) \end{array} \right)$. Then \begin{align*} dX_t = AXdt + \alpha dB_t. \end{align*} Note that \begin{align*} d\left(e^{-At} X_t \right) &= -Ae^{-At} X_t dt + ...


0

It is easy to check that \begin{align*} d\left(e^{-\frac{\alpha^2}{2}t - \alpha W_t } S_t \right) = e^{-\frac{\alpha^2}{2}t - \alpha W_t } f(t) dW_t. \end{align*} Then, $S_t$ can be solved subsequently.


2

I think you misspoke somewhat in your first statement: first hitting times are stopping times, but there are many other kinds of stopping times. What you've written is a last exit time, which is not a stopping time, because as you've said we cannot know whether $\tau \leq t$ by observing the process up to time $t$. This creates some difficulties in ...


1

The player meets monsters at the rate $\lambda$, and with probability $p$ they are winning monsters, so the player meets winning monsters at the rate $\lambda p$. We can extend the game up to $t=1$, with monsters appearing after the player's death meeting a dead player. Then, since the Poisson monsters appear uniformly in time, the probability distribution ...


1

Let's redefine $x^*$ according to my comment above: $$ x^* = \inf\left\{ y : \int_0^T 1_{\{x(t)\leq y\}} dt \geq \tau\right\} $$ Assume $x(t)$ takes values in a state space of nonnegative integers, so $x(t) \in \{0, 1, 2, \ldots\}$ for all $t$. Assume that $x(0)=0$. Then the infimum value is achieved by a particular integer $x^* \in \{0, 1, 2, \ldots\}$ ...


2

Fix $\omega \in \Omega$. Since the mapping $s \mapsto X_s(\omega)$ is càdlàg, we have $$\sup_{s \in [0,t]} |X_s(\omega)| < \infty$$ for any $t \geq 0$. As $f$ is continuous, this implies in particular that $$M:=\sup_{s \in [0,t]} |f(X_s(\omega))|<\infty.$$ Hence, $$\left| Y_t(\omega) - Y_r(\omega) \right| = \left| \int_r^t f(X_s(\omega)) \, ds ...


0

The conjecture is wrong. Here is a counterexample. For each $n$ consider a continuous-time Markov chain $X^n_t$ over two states 1 and 2 with 1 as initial state and only one transition from 1 to 2 with rate $n$. Denote by $P^n(t)$ the semigroup of $X^n$. Let $X_t$ be a Markov chain with semi group $P(0) = I$ and $P(t) = \begin {pmatrix} 0 & 1 \\ 0 & ...


1

Yes. A Drunkards Walk follows a similar principle to Binomial Distribution; but rather than successes (and failures) in a certain number of trials, you are counting steps forward and backwards. $\mathsf P(\eta_t=m)$ is the probability that in $t$ steps of $\pm 1$ units each, you will have moved $m$ units in total; consisting of $x$ steps forward, and $y$ ...


1

You could try a Monte Carlo approach. Basically, you can simulate a large number of strong solutions and then evaluate the sample mean and variance of the specific instant of interest. Depending on the structure of the diffusion coefficient, it is possible to perform exact simulation. In this case, no approximation error will be propagated to your ...


2

Consider the process $$X_s = W^{(1)}_s-W^{(2)}_s$$ At any given time, this is the difference of two mean zero, independent, normals with variance $s$. That means $X_s$ is a mean zero normal with variance $2s$. So the question is, what is the expected value of the absolute value of a normal random variable. This is called a Folded Normal Distribution. One ...


1

I claim that $\Pi$ is Borel measurable. Define the family $$ \mathscr{A} := \left\{\text{Borel measurable $E\subseteq\mathcal{M}(\mathbb{R})$}: \text{$\Pi^{-1}(E)$ is Borel measurable}\right\} \;. $$ It is enough to verify that $\mathscr{A}$ is a $\sigma$-algebra, and $\mathscr{A}$ contains a countable sub-family $\mathscr{C}$ that generates the weak ...


0

Consider $$ f_n(x,y)=f(x,y)+ \frac{h(y)}{\sum_k h(k/n) g_n(k)}\left(\int_\mathbb{R} f(z,y)g(z)dz-\sum_{j\in\mathbb{N}} f\left(\frac{j}{n},y\right)g_n(j)\right)$$ with $h_n(y) = ne^{-1/y}$ if $y>0$ and $h_n(y) = 0$ if $y \leq 0$ then it verifies: $$\sum_k f_n(x,k/n) g_n(k)=\sum f_n(x,k/n) g_n(k)+ \sum_k\frac{h_n(y/n)g_n(k)}{\sum_k h_n(k/n) ...


1

I believe your argument provides a good interpretation for total (not quadratic) variation, because it is the one associated to the first derivative. For quadratic variation, one should note first that squaring an infinitesimal element will cause it to shrink even more. So if you are adding up smaller things than before and it still results in a great ...


2

If $U(t)=e^{-\mu t}W(\frac{\sigma^2e^{2\mu t}}{2\mu})$, then \begin{eqnarray*}\mbox{Cov}[U(t),U(t+s)]&=&\mbox{Cov}\left[e^{-\mu t}W\left(\frac{\sigma^2e^{2\mu t}}{2\mu}\right),e^{-\mu (t+s)}W\left(\frac{\sigma^2e^{2\mu (t+s)}}{2\mu}\right)\right]\\[10pt] &=& e^{-\mu t} e^{-\mu (t+s)} \mbox{Cov}\left[W\left(\frac{\sigma^2e^{2\mu ...


0

Let we say that a sequence of $2n$ throws is balanced if at any point we are not winning any dollar, but with the last throw we are losing zero. It is well-known that the number of balanced sequences of length $2n$ is given by the Catalan number: $$ C_n = \frac{1}{n+1}\binom{2n}{n}. $$ If we say that a sequence of $2n$ throws is returning if at any point we ...


1

It isn't true as stated. The problem is that separate weak convergence doesn't keep track of correlations between what's going on the two halves of the interval. Indeed, if we just consider the case of a constant sequence where $P_n = Q$ for every $n$, the question becomes: If $P \circ \Pi_1^{-1} = Q \circ \Pi_1^{-1}$ and $P \circ \Pi_2^{-1} = Q ...


0

A first observation is that the sequence $(P_n)_{n\geqslant 1}$ is tight in $C[0,1]$. Indeed, for a fixed $\varepsilon$, there exists a compact set $A_1$ of $C[0,u]$ such that for each $n$, $P_n\circ \pi_1^{-1}(A_1)\gt 1-\varepsilon$. Similarly, we can find a compact set $A_2$ such that for each $n$, $P_n\circ \pi_2^{-1}(A_2)\gt 1-\varepsilon$. The set ...


1

Consider $$\tau_n = \frac{[2^n\tau] + 1}{2^n}$$ Note that $\tau_n$ is discrete and $\tau_n \downarrow \tau$. Now use the result you have for the discrete cases $$P(X_{\tau_n}>c\vert \tau_n=j)=\dfrac{P(X_{\tau_n}>c, \tau_n=j)}{P(\tau_n=j)}=\dfrac{P(X_{j}>c, \tau_n=j)}{P(\tau_n=j)}=P(X_{j}>c). \quad j \in \Bbb{Z}_+/ 2^n$$ Now to the continous ...


0

Yes, but this follows from a general property of backshift operators provided (in your case) that $(1+B)Y_1=Y_1+Y_0=Z_1$.


1

Yes. That's it. A Markov chain is a sequence of random variables $\{X_k\}_{k\in\{0..n\}}$ representing $n+1$ subsequent states of a system, such that for all supported values $\{i_k\}_{k\in\{0..b\}}$, and $i_c$, where $0\leq a< b< c\leq n$ we have: $$\mathsf P(X_c=i_c\mid \bigcap_{k\in\{0..b\}} X_k=i_k)= \mathsf P(X_c=i_c\mid X_b=i_b)$$ So if we ...


0

I am not sure I understand your notation, and don't generally understand a lot of notation from stochastic analysts but would like to improve so I'll give it a shot. Let $(\Omega,\mathbb{F})$ be the probability space and filtration, and $g_t(\omega)=1_{X_t(\omega)>c}$. The question, interpreted into this abstract notation, is to show that ...


0

The reason is that the compounded return of an investment has a "drag" when the intermediate returns are more volatile. Check for instance the compounded return of two return series with the same arithmetic mean, but different variance. You will see that the series with a lower variance has a higher compounded return (and geometric mean). The reason is that ...


0

Suppose we have an $L^1$ discrete time stochastic process $\{X_n\}$, i.e. $E|X_n|<\infty$ for all $n\in\mathbb{N}$. Then since the process is $L^1$ the RV $V_n=\sum_{i=1}^{n}\ln E_{i-1}\left[\exp(X_n-X_{n-1})\right]$ is well-defined. Thus we are generally able to form the exponential martingale $N_n=\exp[X_n-V_n]$. As an aside, I've found the discrete ...


-1

Nate proved it for weakly stopping times. For stopping times: if $\sigma$ or $\tau$ is zero, the other one is automatically greater than $t$. If not, say $\epsilon=\sigma+\tau-t>0$, then there is a rational $r\in [0,t]$ such that $\tau>r>\tau-\epsilon=t-\sigma$ and such that $\sigma>t-r$; otherwise, $\sigma\leq t-r$ and thus, ...


0

I came up with a proof that uses the theorem of martingale convergence, and I think it is a quick proof if you have some knowledge of martingales. Let $T$ be the first time $n$ such that $M_n=z$. $T$ is a stopping time. WLOG $z\ge0$. $(M_n)_n$ is a martingale, and the stopped martingale $(M^T_n)_n$ is also a martingale. So ...


6

Let $(B_t)_{t \geq 0}$ be a Brownian motion on a probability space $(\Omega_1,F_1,P_1)$ and let $(\Omega_2,F_2,P_2)$ be an arbitrary probability space. Define a new probability space $(\Omega,F,P)$ by $$\Omega := \Omega_1 \times \Omega_2 \qquad F := F_1 \otimes F_2 \qquad P := P_1 \otimes P_2.$$ If we set $$\tilde{B}_t(\omega_1,\omega_2) := B_t(\omega_1) ...


0

any proof of this result is going to be technical to say the least. KMT is now a classical result in probability that was very surprising at its time. It provides the optimal rate of growth for the skorohod embedding theorem. For a "simple" proof that relies heavily on stein's method see http://arxiv.org/pdf/0711.0501v3.pdf Is the simplest proof I've seen, ...


0

I think the issue is that the indicator functions are not considered. Let $a_k = k/n$, and $A_k = [ a_k, a_{k+1} ], k=0,...,n-1.$ Then, \begin{align*} \mathbf{E} \int_0^1 (X_n(t) - W(t) )^2dt &= \mathbf{E} \int_0^1 \left(\sum_{k=0}^{n-1} W(a_k)^21_{A_k}(t) + W(t)^2 - 2 \sum_{k=0}^{n-1} W(a_k)W(t)1_{A_k}(t) \right) dt \\ &= \sum_{k=0}^{n-1} a_k /n + ...


1

Rogers and Williams (Diffusions, Markov processes and Martingales Vol. 2, p. 314) define the sigma-algebra generated by all progressive processes. In fact, they define a progressive set as a subset $F$ of $[0,\infty)\times\Omega$ of which the indicator function $1_F$ is a progressive process. The collection of progressive sets forms a sigma-algebra. They ...


1

$$ dS_t = \Big(\mu + \frac{1}{t}\Big) S_t\,dt + \sigma dW_t $$ for example. Of course this comes with a restriction to $t > 0$.


2

$P(Z)$ = $P$(you have received your first mail after a week) $=$ $P$(you have not received any mail during first week) $=$ $(1-p)^7$


0

I really dislike writing things like "dt dt = 0". One direct way is if you know the Ito rule for processes in $R^n$. In that case, apply it to the $R^2$ process $Z_t = (X_t, Y_t)$ and the function $f(x,y) = xy$



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