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0

My preferred way to answer this is to consider the related Markov chain, whose state at any time is not the number, but the set of marbles in the first urn. By symmetry, one shows that in equilibrium, all subsets of the whole set of marbles are equally probable. Thus the probability that the set has $k$ of the $n$ marbles is the number of subsets of size ...


-1

Suppose we sample uniformly from 1 cycle of the sine wave, then it is quite clear (I would suggest drawing a picture) that the cumulative distribution function is: $$ F(x)=\frac{\arcsin(x)}{\pi}+\frac{1}{2},\ \ \textrm{where}\ -1\le x\le1 $$ A longer explanation of this can be found here. Then density can then be found by differentiating this to get: $$ ...


3

If $D$ is a domain in $\mathbb{R}^n$, and $x_0 \in D$ is a point, then the distribution of Brownian motion started at $x_0$ stopped at the time when it leaves $D$ is the harmonic measure of $\partial D$ with base point $x_0$. In the plane you can often use conformal invariance to calculate harmonic measure. Since you know that the harmonic measure of the ...


1

$P(B_1 = \max_{t\in[0,1]}B_t) = P(\max_{t\in[0,1]}(B_t-B_1) = 0)= P(\max_{t\in[0,1]}(B_{1-t}-B_1) = 0)$ You can easily check that $(B_{1-t} - B_1)_{t\in [0,1]}$ is itself a Brownian motion, so $P(\max_{t\in[0,1]}(B_{1-t}-B_1) = 0) = P(\max_{t\in[0,1]}B_t = 0) = P(|B_1| = 0) = 0$


1

The joint distribution of the running maximum $M_1 = \max_{0 \leq t \leq 1}B_t$ and $B_1$ is $$f(m,b) = \frac{2(2m-b)}{\sqrt{2\pi}}\exp(-\frac{(2m-b)^2}{2}), \qquad m\geq 0, b \leq m $$ Hence, $P(M_1 = B_1) = 0$ Note : In your approach, you replaced $M_1$ with $\vert B_1 \vert $ which is not true! They have same distribution but the random variables are ...


1

The question is asking about the first and second moments of the random variable $$X=\int_0^Tdt_1\int_0^{t_1}dt_2 B(t_1)B(t_2).$$ By symmetry, $$2X=\int_0^Tdt_1\int_0^Tdt_2 B(t_1)B(t_2)=Y^2,$$ where $$Y=\int_0^Tdt_1B(t_1).$$ Since the family $(B(t))$ is centered gaussian, $Y$ is centered gaussian hence $$X=\langle X\rangle Z^2,$$ where $Z$ is standard ...


1

Hint: The random variables $S_A$ and $S_{A+B}-S_A$ are independent and $S_{A+B}-S_A$ is distributed like $S_B$. Now, you ought to know the distribution of each $S_A$ hence you ought to be able to finish this.


1

Let $N(t)=M(t+\tau)$ and $G(t)=F(t+\tau)$ then, for every nonnegative $t$, $$N(t)=G(t)+\int_0^tN(t-x)\,\mathrm dF(x)$$ hence $$\tilde{N}(s)=\frac{\tilde{G}(s)}{1-\tilde{f}(s)}.$$


1

Not sure I understand the bafflement here... Let $X=\limsup\limits_{t \to 0^{+} } B_t/\sqrt{t}$. For every positive $s$, $X$ is equal to the same limsup restricted to the values of $t$ in the interval $(0,s)$ hence $X$ is $\mathcal F_s$-measurable, that is, for every Borel subset $A$, $[X\in A]$ belongs to $\mathcal F_s$. Thus, $[X\in A]$ belongs to ...


1

Hint: Use $$\left\{ \limsup_{t \to 0+} \frac{B_t}{\sqrt{t}} > k \right\} = \bigcup_{\ell \in \mathbb{N}} \left\{ \limsup_{t \to 0+} \frac{B_t}{\sqrt{t}} > k+ \frac{1}{\ell} \right\}.$$


0

Hint: For the second SDE use the substitution $$Z_t := \exp(-\sqrt{Y_t}).$$ For the general approach see e.g. (the second part of) this answer. Remark: As @user144410 pointed out, the solution of the first SDE can be obtained from the second one using the transformation $X_t = \sqrt{Y_t}$.


1

First of all, your second and third paragraphs. Recall that if $\xi$ is $F$-measurable and $F\subseteq G$ then $\xi$ is $G$-measurable: that follows from the definition of measurability. Now, if $\Bbb F = (F_n)_{n\geq 0}$ is a filtration and $\xi = (\xi_n)_{n\geq 0}$ is $\Bbb F$-adapted, then for each $n$ we have $\xi_n$ is $F_n$-measurable. Since ...


0

Hint: $$E(X^2)=2\int_0^1\int_0^tE(W(t)^2W(s)^2)\mathrm ds\mathrm dt$$ Can you compute $E(W(t)^2W(s)^2)$ for $s\lt t$?


0

I seem to recall a theorem, in which if an absorbing state is accessible from a communicating class $C$, then all states in $C$ are transient.


0

Throwing MATLAB at the problem, I find that all but one eigenvalue has magnitude less than 1, and this one eigenvalue has left eigenvector corresponding to the distribution which is only at $E$, so my analysis agrees with yours. Note that there is a pair of complex eigenvalues, which represent cycles (for example the cycle between $A$ and $C$), but because ...


0

great to see you are interested in GPs. Someone asked a similar question about how to make predictions on a GP and I think I gave a fairly comprehensive work through with sample data that they gave. I think seeing it in action will answer your question. Here is the link.


3

We have that $$E(B(t)W(1))=E \left( (W(t)-tW(1))W(1) \right) =E(W(t)W(1)-tW(1)W(1)) $$ Using the linearity of the expectation we also have that this expression is equal to $$ E(W(t)W(1))-E(tW(1)W(1))=E(W(t)W(1))-tE(W(1)W(1))$$ As @PetiteEtincelle said, $E(W(t)W(s))=\min(t,s)$, so finally we conclude that $$ E(B(t)W(1))= \min (t,1)-t\min(1,1)=t-t=0$$


1

It’s a little confusing to have both a constant $\mu$ and a function $\mathbf{\mu}$, so I’m going to replace the former by $\nu$ for this answer. In addition, there appears to be a typo in the first line: the argument makes sense only if the expression on the righthand side of the first equals sign is $$\lambda\sum_{n\ge 1}np_{n-1}(t)-\lambda\sum_{n\ge ...


2

Basically, the definition of expected value gives $$\sum_{n=0}^\infty n p_n(t)=\vec{\mu}(t),$$ and the definition of a probability distribution gives $$\sum_{n=0}^\infty p_n(t)=1.$$ In particular, $$\sum_{n=0}^\infty (n+1)p_n(t) = \vec{\mu}(t)+1.$$ Being explicit: \begin{align*} &\phantom{{}={}} \lambda \sum_{n = 0}^\infty(n+1)p_n(t) - \lambda ...


1

If $G_n'(s) =\sum_{r=1}^\infty \frac{n^{r-1}}{(n+1)^{r+1}}\frac{d}{dx}\left[s^r\right] = \sum_{r=1}^\infty\frac{rn^{r-1}}{(n+1)^{r+1}}s^{r-1} $, then $G_n'(1) = \sum_{r=1}^\infty\frac{rn^{r-1}}{(n+1)^{r+1}} = \frac1{n(n+1)}\sum_{r=1}^\infty\frac{rn^{r}}{(n+1)^{r}} = \frac1{n(n+1)}\sum_{r=1}^\infty rz^r $, where $z = \frac{n}{n+1} $. Since $\sum_{r=1}^\infty ...


1

Note that your series for $G_n$ is a geometric series. You can find a simple closed form for the sum. Then take the derivative.


2

A simple approach (which avoids the task of checking the nonnegativity of the function in your question) is to compare the hitting times $\tau_1$ and $\tau_2$ with the hitting time $\theta$ of $\pm2\pi$ by the process $$dY=\mathrm dW-\mathrm{sgn}(Y)\,\mathrm dt,$$ starting from some $|y|\leqslant2\pi$. Since $|\sin|\leqslant1$, $\tau_1$ and $\tau_2$ are ...


0

Whenever you right down a definition of your stochastic process as such a graph, you force it to be a Markov Chain (MC), at least if the graph is given its usual interpretation. But in that case you consider every node to be a separate state, so that the "1" below would be a state different from the top-left "1". We call that Labeled MCs: whereas an MC is ...


0

The result is true, is given as e.g. Theorem I.50 of Protter's book "Stochastic integration and differential equations", and your argument is correct. In detail, to obtain the result, assume that $M$ is a local martingale. Let $(T_n)$ be a localising sequence, such that $M^{T_n}$ is a martingale for each $n$. Then $M^{T_n\land n}$ is a uniformly integrable ...


0

If $T$ is a general stopping time, then no, as you can select e.g. $T=\infty$ and the family you claim to be uniformly integrable is $(M_{\tau_n})$ for some increasing sequence of stopping times $(\tau_n)$. Choosing $\tau_n = n$ would yield that $(M_n)_{n\ge1}$ is uniformly integrable for all martingales $M$, which is absurd. If $T$ is a bounded stopping ...


2

You can study this using a Markov chain with two states (1=post-win, 2 = post-loss), and transition matrix $$ \pmatrix{ p & 1-p\cr q & 1-q\cr}$$ The equilibrium probabilities are $q/(q+1-p)$ and $(1-p)/(q+1-p)$. The long term proportion of wins is then $q/(q+1-p)$.


1

In addition to your comment: Let $P$ be a stochastic, irreducible matrix. Since $P$ is irreducible it will be either aperiodic or periodic. $P$ aperiodic$\iff\displaystyle\lim_{k\to\infty}P^k$ exists. $P$ periodic matrix with period $d>1$ $\iff \displaystyle\lim_{k\to\infty}P^k$ does not exist. We can prove the latter using cyclic subclasses.


1

$\dfrac{\log X_n}{n} = \dfrac{\sum_{m \le n} \log Y_m}{n} \to E\log Y_1$ almost surely by the strong law of large number. And by Jensen's inequality, $E\log Y_1 < \log EY_1 =0$ since $P(Y = 1) < 1$ So we get that $\dfrac{\log X_n}{n}$ converges to a strictly negative number alomost surely, thus $\log X_n \to -\infty$, i.e. $X_n \to 0$


0

The distributions of $(X(u),X(u+v),X(u+v+w))$ and $(-X(u),-X(u+v),-X(u+v+w))$ coincide hence $E[X(u)X(u+v)X(u+v+w)]=-E[X(u)X(u+v)X(u+v+w)]=0$.


-1

I'll replace $X_t = X(t)$. Note that $\{X_t \mid t \geq 0\}$ is a martingale. Then, for $0<p<q<r$, \begin{align*} E(X_pX_qX_r) &= E\Big( E(X_pX_qX_r \mid \mathcal{F}_q)\Big)\\ &=E\Big(X_pX_q E(X_r \mid \mathcal{F}_q)\Big)\\ &=E\Big(X_pX_q^2\Big)\\ &=E\Big(X_p(X_q-X_p+X_p)^2\Big)\\ &=E\Big(X_p(X_q-X_p)^2+X_p^3+2X_p^2(X_q-X_p) ...


0

First of all, look up the term "Superposition of a Poisson process". a) The probability you wrote does not make sense, I guess some parenthesis is missing. First you need to know that the sum of any two independent Poisson random variables $X$ and $Y$ with rates, $\lambda_1$ and $\lambda_2$ is Poisson with the sum of rates. You can show this by computing ...


0

Hint on c): $$T>t\iff X_tY_t=0$$


1

The answer is certainly: yes. By definition of a martingale, $\mathsf E M_t = \mathsf E M_0$ (just using the tower property) and since $M_0 = 0$ in your case, all follows. Note that Ito integral may not be a martingale unless some integrability conditions are satisfied, in general it may be only a local martingale - so you need to see which integrability ...


1

I'll tell you how to do one, hopefully that helps you try the others. $F_T = \{A: A\cap \{T\leq t\} \in F_t\}$. 1) $T$ is a stopping time $\Rightarrow \{T\leq t\}\in F_t$. Now, to show that $T$ is measurable wrt a sigma algebra $F_T$ we have to show that $T^{-1}(B)\in F_T\,\forall\,B\in\mathcal{B}_\mathbb{R}$. Showing this for all Borel sets $B$ is too ...


0

For any $m\in\mathbb{N}$, we have: $$\mathbb{P}[X_1-X_2=m]=\sum_{n\in\mathbb{Z}}\mathbb{P}[X_1=m+n]\mathbb{P}[X_2=n]=\lambda^m e^{-2\lambda}\sum_{n=0}^{+\infty}\frac{\lambda^{2n}}{(m+n)!n!}$$ or: $$\mathbb{P}[X_1-X_2=m]=e^{-2\lambda}I_m(2\lambda)$$ where $I_m$ is a hyperbolic Bessel function. By symmetry, the probability mass function of $X_1-X_2$ is given ...


1

1) $\mathbb E(X_n|\mathfrak{F_{n-1}})\leq X_{n-1}$ respectively $\mathbb E(X_n|\mathfrak{F_{n-1}})\geq X_{n-1}$ 2) $\mathbb E(\mathbb E(X_n|\mathfrak{F_{n-1}}))=\mathbb E(X_n)$ gives you what you want


0

This is the discrete SIR model. Transitions occur when an individual of type S becomes of type I or when an individual of type I becomes of type R. Transitions S$\to$I occur at rate $\beta S(t)I(t)$ and transitions I$\to$R occur at rate $\gamma I(t)$, thus, if $I(t)=0$ then no further transition occurs. Encoding each state $(S(t),I(t),R(t))$ by the three ...


1

Suppose that $(X_t,Y_t)_t$ is a solution to the given system. Then in particular, we can write $$X_t-X_0 = - \underbrace{\int_0^t Y_s \, dB_s}_{\text{local martingale}} - \frac{1}{2} \underbrace{\int_0^t X_s \, ds}_{\text{bounded variation, continuous}}.$$ Applying the optional stopping theorem yields that $$(t,\omega) \mapsto \left( \int_0^{t \wedge ...


0

First of all $e^{A(t)}$ doesn't mean anything for time varying systems. You need to use state transition matrix $\Phi(t, t_0)$ as in the following: $$A_t = \Phi(t + T, t)$$ $$B_t = \int_t^{t+T} \Phi(t+T, \tau) B(\tau) d \tau$$ assuming $u_t$ is constant over time $[t, t+T)$. But in this case I believe $G_t$ cannot be written separately. The system becomes ...


2

You are confusing the measure on path space with Lebesgue measure. The "almost everywhere" refers to the former: almost every individual path can be taken to be continuous everywhere. Indeed, the Wikipedia page you link to says that Brownian motion is "almost surely everywhere continuous". In other words, if $\mathbb{P}$ is Wiener measure on a suitable ...


1

Conditioning on whether $X$ or $Y$ is the maximum of the two random variables you can write the required probability as: $$\begin{align*}P(\max\{X,Y\}>a\min\{X,Y\})&=P(X>aY\mid X\ge Y)P(X\ge Y)\\[0.2cm]&+P(Y>aX\mid Y\ge X)P(Y\ge X)\end{align*}$$ which due to symmetry of $X,Y$ can be simplified to: $$P(\max\{X,Y\}>a\min\{X,Y\})=2\cdot ...


5

Since events $\{X \geqslant Y\}$ and $\{X < Y\}$ are disjoint: $$ \begin{eqnarray} \Pr(\max(X,Y) > a \min(X,Y)) &=& \Pr(\max(X,Y) > a \min(X,Y), X \geqslant Y) \\ && + \Pr(\max(X,Y) > a \min(X,Y), X < Y) \\ &=& \Pr(X > a Y, X \geqslant Y) + \Pr(Y > a X, X < Y) \end{eqnarray} $$ Now proceed graphically. Since ...


2

Well the maximum of a brownian motion would be a singular function (and it's very useful). EDIT: you can also consider the local time $L_t$ (second EDIT: which you mentioned in your question), which gives you the Tanaka formula: $$|W_t| = \int_0^t sign(W_s)\,dW_s + L_t$$ It can be used to extend Ito's formula to convex functions (I think a simplified ...


0

It sounds like you are describing a $M/M/1/3$ system. Three in the queue and one in service sounds correct. State space is the total number in the system: (0,1,2,3,4), where 0 means no customers, 1 means one in service and empty queue, 2 means one in service and one in the queue and so on. This is a finite state, (0,1,2,3,4), birth and death process with ...


1

When you are not sure, there is not harm in treating entries as separate variables to apply Ito's formula. The only thing you have to take care of in that case are quadratic covariations.


1

The covariance is only well defined for square integrable random variables. If $X$ and $Y$ are square integrable and $E(Y\mid X)=0$ then $X$ and $Y$ are orthogonal in the Hilbert space of square integrable random variables, or equivalently uncorrelated, as $E(YX)=E(E(YX\mid X))=E(XE(Y\mid X))=0$


0

The requirements to be a Poisson process do depend on the filtration. Roughly speaking, to be adapted is easier when the filtration is larger, and the condition involving predictable processes is more difficult to meet since there are more predictable processes when the filtration is larger. But all this is offtopic to determine whether the distribution of ...


1

Why not apply the general argument, valid for every subcritical or critical branching process? Conditionally on $Z_1=k$, the probability to go extinct is $\theta^k$, where $\theta$ is the probability that the whole process, starting from $Z_0=1$, goes extinct. Thus, $$\theta=\sum_kP(Z_1=k)\theta^k=G(\theta),\qquad G(s)=E(s^{Z_1}),$$ where $G$ denotes the ...


0

$N(s_1)|N(t)=k$ is a Binomial random variable with $n=k$ and $p = \frac{s_1}{t}$. Now since $Var(X) = E(X^2)-E(X)^2$, $E(X^2) = Var(X)+E(X)^2$ and since the variance of a Bernoulli random variable is $np(1-p)$, it follows: $E(N^2(s_1)|N(t)=k) = k(\frac{s_1}{t})(1-\frac{s_1}{t}) + k^2\frac{s^2_1}{t^2}$


1

I think so... Here's my thought. Let $M$ be an irreducible, symmetric and positive-definite $n\times n$ stochastic matrix, with spectrum $\sigma(M)=\{\lambda_1, \lambda_2,\ldots, \lambda_n\}$. Since $M$ is stochastic, we have that $\lambda_1=1$. Since $M$ is symmetric we have that $M$ is diagonalizable. Since $M$ is positive definite, we have that all ...



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