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0

In general, you can't expect an analytical solution for general $a$ and $b$ in all but the simplest of cases. An Euler scheme applied SDEs is known as the Euler-Maruyama method, and it will converge to give sample paths consistent with the solution if you take small enough time steps. There are higher order methods that converge faster that you might care ...


2

Well, in the first case, when you consider a Brownian motion $B_t$, the following is usually understood: You are considering a background probability triple $(\Omega,\mathcal{F},P)$, with $\Omega$ being some set, $\mathcal{F}$ being some $\sigma$-algebra and $P$ being some probability space, and you consider a measurable mapping $B:\Omega\to C[0,\infty)$ ...


6

Then can you help me to show that $\alpha$ is in fact stopping time? Well, not very easily I am afraid, since $\alpha$ is not always a stopping time for the filtration $(\mathcal F_t)_{t\geqslant0}$ defined by $\mathcal F_t=\sigma(X_s;0\leqslant s\leqslant t)$. For a counterexample, consider some Bernoulli random variable $U$ such that $P(U=1)=\frac12$ ...


1

If $2aT\geqslant1$, then $\exp(aW_T^2)$ is not integrable hence $E(\exp(aW_T^2)\mid F_t)$ does not exist. From now on, assume that $2aT\lt1$. Define $Z$ by $$W_T-W_t=\sqrt{T-t}\cdot Z,$$ then $Z$ is standard normal and independent of $F_t$ and $$W_T^2=W_t^2+2\sqrt{T-t}W_tZ+(T-t)Z^2,$$ hence $$ E(\exp(aW_T^2)\mid F_t)=\exp(aW_t^2)\cdot ...


-1

The answer of Mr.Spot seems to bring me closer to the right answer, I believe. I again assume that $\lambda_1$ and $\lambda_2$ are the means and $n$ the number of repetitions. So the sum of the $n$ observed values would be $\lambda_1n$ and $\lambda_2n$. Starting with what we had at the end: $E[N_1(t)+N_2(t) \mid N_2(t)=k] = \lambda_1n + \lambda_2n $ and ...


1

For every independent integrable random variables $X$ and $Y$, $$E(X+Y\mid Y)=E(X)+Y,$$ by linearity of conditional expectation, hence $$E(X+Y\mid Y=y)=E(X)+y.$$


1

The notation (or lack thereof) is confusing and what (I think) you are asking for makes no sense but... Let $\{N_1(t):t\ge 0\}, \{N_2(t);t \ge 0\} $ be independent Poisson processes with rates $\lambda_1,\lambda_2.$ For a given time $t_0$ we observe values of $N_1(t_0), N_2(t_0)$ and then repeat $n$ times and sum the $n $ observed values for each process. ...


0

Answering my own question, after thinking about it a while... I think the reason I'm having trouble figuring this out is that, in order to get z, I would need to know the stdev of x. Alternatively, I'd need the stdev of the stdev of S. I'd welcome any disagreement with that conclusion. ;-)


0

Each $x(t)$ depends on the random sequence $(\tau_n)$ hence $x(t)$ is random. For example, the event $[x(t)=0]$ corresponds to every rectangular function in the series evaluating to $0$, that is, to the event that $|t-\tau_n|\gt\frac12T$ for every $n$. Considering the random set $\mathfrak T=\{\tau_n\mid n\in\mathbb Z\}$, one gets $$[x(t)=0]=[\mathfrak ...


1

As explained in the comments, this is really the iteration of a transformation of distributions, not random variables, and the technique used to find the fixed points in the other question can be adapted to solve the asymptotics. To be brief (since this is a rehashing of arguments already explained), the sequence of generating functions ...


0

Choose $p_{2x}=a$ and $p_{2x-1}=b$ for every $x\geqslant1$, with $a\lt\frac12\lt b$. For every $x\geqslant1$, the chain starting from $2x$ hits $2x+2$ before $2x-2$ with probability $$p=\frac{ab}{1-a-b+2ab}.$$ Thus, the chain is transient if $p\gt\frac12$ and (positive) recurrent if $p\lt\frac12$. This solves the question because one can adjust the ...


0

I think you are very close to the solution but made too complicated. First, simply condition and use the chain rule for expectations: $E[\ln Y_n] = E[E[\ln Y_n|\mathcal{F_{n-1}]]}$ Then it is the calculation you made before, so assuming it is correct and in line with your notations: $E[\ln Y_n] =\frac{1}{2}(\ln(1+2A)+\ln(1-A))+E[\ln(Y_{n-1})]$ This is ...


0

This is a simple application of Law of large numbers. Note that $$E(Y_1) = \int^\infty_0 x\nu(\text{d}x)$$ Assume that $E(Y_1)<\infty$. Let $n$ be the largest integer smaller than $t$, then $$\frac{Y_t}{t}=\frac{Y_1+(Y_2-Y_1)+...+(Y_n-Y_{n-1})}{n}\frac{n}{t}+\frac{Y_t-Y_n}{t}$$ By the strong law of large numbers, the first part of the sum converges ...


3

If $(X_t)_{t \geq 0}$ is a càdlàg process, then it follows from the very definition of "càdlàg" that $$\lim_{t \downarrow 0} X_t(\omega) = \limsup_{t \downarrow 0} X_t(\omega) = X_0(\omega).$$ Consequently, the $\limsup$ is measurable since $\omega \mapsto X_0(\omega)$ is measurable (as $(X_t)_{t \geq 0}$ is a stochastic process). Edit (càglàd case) If ...


2

Let $p_{jk}=2k/j^2$ be the one step transition probability. Then from state $m+1$ we can only get to state $m$ if we jump there in the next step, so $W_{m+1,m}=p_{m+1,m}=2m/(m+1)^2.$ Note that this expression satisfies the general formula for $W$ so we can start the induction here and continue to $W_{m+2,m}, W_{m+3,m},...$ Starting from state $j$ we ...


1

Did told me this is a well known result, though Googling confirms that it is not very easy to find on the internet (as you probably realised) For a Brownian Bridge $X$ with $X_0=0=X_1$, the distribution is a Uniform distribution on $[0,1]$. This result is due to Levy. I found this in a book called Handbook of Beta distribution and its applications page 193. ...


2

Here is an alternative proof: By the Markov property, we have $$\mathbb{E}^{x}\big( \mathbb{E}^{X_t} g(X_s) \big) = \mathbb{E}^x g(X_{s+t})$$ for any $s,t \geq 0$. Applying Fubini's theorem therefore yields $$\mathbb{E}^{x}(R_{\alpha}g(X_t)) = \int_0^{\infty} e^{-\alpha s} \mathbb{E}^x g(X_{s+t}) \, ds.$$ Consequently, we find $$\begin{align*} ...


1

Start with the formula $$E^x\left[R_\alpha g(X_t)\right]=\alpha\int^\infty_0e^{-\alpha s}\int_t^{t+s}E^x[g(X_v)]dv\,ds.$$ Plugging in $t=0$ we get $$R_\alpha g(x)=\alpha\int^\infty_0e^{-\alpha s}\int_0^{s}E^x[g(X_v)]dv\,ds.$$ Subtracting and dividing by $t$ gives, $${E^x\left[R_\alpha g(X_t)\right]-R_\alpha g(x)\over t} = \alpha\int^\infty_0e^{-\alpha ...


1

Let me proceed the solution of JimmyK4542. He is absolutely correct and let me just prove that the solution indeed $$ q^\prime(x)=(I−P(x))^+P′(x)q(x) $$ Notice that the left eigenvector for $P(x)$ corresponding to 1 is the vector of ones. Let's denote such vector (normalized by 1) as $e$. Thus $$ e^T (I-P(x))q(x)^\prime= e^T P^\prime(x) q(x)=0 $$ Thus ...


2

Assume that $P(x)$ has only one eigenvalue equal to $1$. Let $q(x)$ be the unique unit-norm leading eigenvector of $P(x)$ for all $x$. For convenience, I'll drop the "$(x)$" in the computations. We require $Pq = q$, i.e. $(I-P)q = 0$ and $q^Tq = 0$ for all $x$. Differentiation yields: $(I-P)\dfrac{dq}{dx} - \dfrac{dP}{dx}q = 0$ and $q^T\dfrac{dq}{dx} + ...


0

Let $T_1$ be the first time it dips below $40$ and $T_2$ be the first time it rises to $60$, then $T_1<N$ and $T_2<N$ almost surely. By the Optional Sampling Theorem, $E S(0)=ES(T_1)=ES(T_2)$ but none of $40, 50, 60$ equals to each other. You have actually given more conditions than that is required, one of dipping below 40 or rising above 60 ...


1

If the rate is $r$ per unit of time then the parameter is $\lambda = rT$ so the likelihood function is $$(rT)^n \frac{e^{-rT}}{n!}$$ If you take the derivative of this with respect to $r$ and set this equal to $0$ to solve to find the maximum likelihood estimate of $r$, you do not get $T/n$. This is in fact obvious from dimensional analysis. But you do get ...


1

Looking in the obvious place, the non-homogeneous Poisson process uses the intensity function to construct a parameter for a Poisson process, which is a discrete probability distribution. \begin{align*} \lambda_{a,b} &= \int_a^b \lambda(t) ~\mathrm{d}t \\ &= \int_a^b \alpha \mathrm{e}^{-t} ~\mathrm{d}t \\ &= \alpha ( \cosh(a) - \sinh(a) - ...


2

The claim holds true if there exists a unique solution to the SDE $$dZ_t = b(Z_t) \, dt + dB_t, \qquad Z_0 = z \tag{1}$$ e.g. if $b$ is locally Lipschitz continuous. Proof: Let $(X_t)_{t \geq 0}$ and $(Y_t)_{t \geq 0}$ be solutions to the SDEs $$\begin{align*} dX_t &= b(X_t) \, dt + dB_t \qquad X_0 = x \tag{2} \\ dY_t &= b(Y_t) \, dt + dB_t ...


1

Your attempt breaks down at the second = sign when some mysterious $t$ appears (what is $t$ here?). Instead, one should write $$ (\ast)=P(T_n \leqslant s <T_{n+1})=P(T_n\leqslant s\lt T_n+\tau_{n+1}). $$ Now, $(T_n,\tau_{n+1})$ is independent and one knows the distributions of $T_n$ and $\tau_{n+1}$ hence one can evaluate the RHS. Say the distribution of ...


2

$\mu$ is the measure induced by the cdf $F$ of the corresponding rv so $\mu(-\infty,x]=F(x).$ With this integrand, the Lebesque-Stieljes integral is equivalent to the Riemann-Stieltjes: $$C(K)=\int_K^\infty (x-K)dF(x) $$ In the special case where $F$ has density $f$: $$ C(K)=\int_K^\infty xdF(x)-K(1-F(K)) $$ $$C^\prime (K)=-Kf(K)-(1-F(K))+Kf(K)=F(K)-1 $$ ...


0

First of all, note that you should always specify the filtration since the answer depends, in general, on the given filtration. The following statements are both correct: Let $(X_t,\mathcal{F}_t^X)_{t \geq 0}$, $(Y_t,\mathcal{F}_t^Y)_{t \geq 0}$ be continuous local martingales with respect to their natural filtration. Then $(X_t Y_t, \mathcal{F}_t^{X ...


1

The estimate by Fatou's lemma says that $$ Y^{(n)}_t(\omega) \rightarrow X_t(\omega) $$ in $L^2(\Omega \times [0, T])$. Since $\sigma(t,x)$ is Lipschitz in $x$, $$ \sigma(t, Y^{(n)}_t(\omega)) \rightarrow \sigma(t, X_t(\omega)) $$ in $L^2(\Omega \times [0, T])$ also. So Ito isometry tells you that $$ \int_0 ^t \sigma(s, Y^{(n)}_s(\omega)) dB_s ...


8

Once $k>2$ the answer to #1 depends on the graph; even in the limiting determinstic case of $p=1$, and taking $n_0=1$, the expected time till full zombification can range from about $C \log n$ (for an expander graph) to at least $n/k$ (when $k$ is even, the node set is ${\bf Z}/n{\bf Z}$, and each node is joined to its $k$ nearest neighbors). As to #2: ...


1

Because, if $f(t)$ is a continuous function, then the existence of any $t \in [0,1]$ such that $f(t) > \epsilon$ implies the existence of a rational $s \in [0,1]$ such that $f(s) > \epsilon$. In other words, for any $n \in \mathbb{N}$, the two sets $$\{\omega : \exists t \in [0,1] \text{ s.t. } |f(t,\omega)| > 1/n\} = \{\omega : \exists t \in [0,1] ...


0

Is it also correct if we replace $S_k$ by a right-continuous stochastic process $X_t$, a right-continuous filtration $\mathcal{F}_t$, an open set $B$ and a stopping time $S$ with values in $[0,\infty[$? Or to be more precise i formulate it as a Theorem: Theorem: Assume we have a right-continuous stochastic process $X_t$, a right-continuous filtration ...


1

Whenever we consider a process $(Y_t)_t$ of the form $Y_t = f(B_t)$ (where $f$ is a "nice" function and $(B_t)_t$ a Brownian motion), then by Itô's formula $$Y_t - Y_0 = \int_0^t f'(B_s) \, dB_s + \frac{1}{2} \int_0^t f''(B_s) \, ds. \tag{1}$$ Since the stochastic integral is a martingale, this yields in particular that $$\mathbb{E}Y_t = \mathbb{E} \left( ...


1

The key point is to write $B_T^2$ in a clever way: $$\begin{align*} B_T^2 &= \sum_{j=1}^n (B_{t_j}^2-B_{t_{j-1}}^2) \\ &= \sum_{j=1}^n \big( (B_{t_j}-B_{t_j^{\ast}})+B_{t_j^{\ast}} \big)^2 - \big( (B_{t_{j-1}}-B_{t_j^{\ast}})+B_{t_j^{\ast}} \big)^2 \\ &= 2 \sum_{j=1}^n B_{t_j^{\ast}} (B_{t_j}-B_{t_{j-1}}) + \sum_{j=1}^n ...


0

That $S_n$ is square integrable should be clear; for the martingale property write $S_n=S_{n-1}+\xi_n$ and use linearity of conditional expectation. Notice that $$M_n=(S_{n-1}+\xi_n)^2-\mathbb E[\xi_n^2\mid \mathcal F_{n-1}]-V_{n-1}=M_{n-1}+2\xi_nS_{n-1}+\xi_n^2-\mathbb E[\xi_n^2\mid \mathcal F_{n-1}]$$ and $M_k$ is $\mathcal F_k$-measurable.


1

The proof is based on this standard fact from functional analysis: Theorem. Let $H$ be a Hilbert space and let $E \subset H$ be a linear subspace. Let $E^\perp := \{ x \in H : \langle x,y \rangle = 0 \text{ for all } y \in E\}$. Then $E$ is dense in $H$ iff $E^\perp = \{0\}$. So for this proof, $H = L^2(\mathcal{F}_T, P)$ and $E$ is the linear span ...


1

First of all, $$X_t := \int_0^t W_s \, ds \tag{1}$$ is an Itô process, i.e. an process of the form $$Y_t - Y_0 = \int_0^t \sigma_s \, dW_s + \int_0^t b_s \, ds$$ (here $\sigma_t =0$, $b_t = W_t$). For an Itô process $(Y_t)_{t \geq 0}$ Itô's formula reads $$f(Y_t)-f(Y_0) = \int_0^t f'(Y_s) \, dY_s + \int_0^t f(Y_s) + \frac{1}{2} f''(Y_s) \, d \langle Y ...


0

The following horrible formula for the joint distribution of max, min and end value of a Brownian motion was copied without guarantees from the Handbook Of Brownian Motion (Borodin/Salminen), 1.15.8, p.271. First, for simplicity, this is only written for $\sigma=1,t=1$, and the more general case comes directly from scaling. If we shorten W as the Brownian ...


1

It can be written as a markov chain to obtain the probabilities. But there are 101 states and couldn't find a simple closed form like in the case of the original gambler's ruin. If the limit was 5 units instead, the markov chain is written as: \begin{align*} A &= \left(\begin{array}{rrrrrr} r & q & 0 & 0 & 0 & p \\ p & r ...


1

Those are two equivalent descriptions of the same semi-group. You mentioned how to deduce $q$ from $(p_t)$, note that, conversely, $p_t=\mathrm e^{tq}$ for every $t\geqslant0$.


2

The condition means that, for every $n$, there exists some measurable set $A_n$ such that $$\tau\leqslant n\iff(X_1,\ldots,X_n)\in A_n.$$ Thus, every positive stopping time $\tau$ can be rewritten as $$\tau=\inf\{n\geqslant1\mid(X_1,\ldots,X_n)\in A_n\},$$ for some well-chosen sequence $(A_n)$.


1

Let $u_n$ be the probability that $n$ Bernoulli trials result in an even number of successes. This occurs if an initial failure is followed by an even number of successes, or an initial success is followed by an odd number of successes. Therefore $u_0=1$ and for $n\geq 1$ $$u_n=q u_{n-1}+p(1-u_{n-1}).$$ Multiplying by $s^n$ and adding over $n$ we see that ...


0

Here's a way to do by using a Markov chain: \begin{align*} A &= \begin{pmatrix} q & p & 0 \\ 0 & q & p \\ q & p & 0 \end{pmatrix} \end{align*} Answer will be the third column of the first row in $A^n$, and the generating function for all entries can be obtained directly from the matrix by computing \begin{align*} ...


1

For Polish $X$: Optimal measure $M^*$ exists: Topics in Optimal Transportation (Theorem 1.3 on page 19) and Optimal Transport (Theorem 4.1 on page 32) Equality $K_d=W_d$ holds: Topics in Optimal Transportation (Theorem 1.14 on page 34) and Optimal Transport (Remark 6.5 on page 95). The first of two books is more analysis-oriented, and may be easier to ...


3

Your definition of $\tau$ is flawed, actually, $$\tau=\int_0^\infty\mathbf 1_{W_t\gt t}\,\mathrm dt,$$ hence $$E(\tau)=\int_0^\infty P(W_t\gt t)\,\mathrm dt.$$ For every $t$, $W_t$ is normal centered with variance $t$ hence $$P(W_t\gt t)=P(Z\gt\sqrt{t}),$$ where $Z$ is standard normal. By symmetry, $P(Z\gt\sqrt{t})=P(Z\lt-\sqrt{t})$ hence $$ ...


1

Let $R_j$ denote the first return time to $j$, then (strong) Markov property at time $R_j$ yields$$P(X_n=j\mid X_0=j)=\sum_{m=1}^nP(R_j=m\mid X_0=j)P(X_{n-m}=j\mid X_0=j).$$ Nota: It might be time for you to get a good textbook since the proof above is in all those on the subject. You might try Markov chains by James Norris, available on the web.


1

Consider a trajectory of $n$ transitions, ending in state $j$. Either transition $n$ is the first return to state $j$, which has by definition has probability $f_j^{(n)} = f_j^{(n)} p_{jj}^{(0)}$, or there was a return at some earlier transition(s), the first of which we can label transition $m$. If first return was at $m \neq n$ (which itself has ...


1

Hint: First, note that $\psi$ is defined to be continuous (see top of proof). Then, if you define $<f, g> := \int fg$, and derive a norm from it, it is not too hard to show that the integral of the product of a continuous and a bounded function (which is the case here) is continuous (use Cauchy-Schwartz). Note also that this is a "for each $\omega$" ...


1

There is a quote on this point from page 808 in J.L. Doob's Classical Potential Theory and Its Probabilistic Counterpart. Before martingales had been formally christened, Lévy [1, 1935; 2, 1937], Bernstein [1, 1937], and other mathematicians had analyzed some of their properties in special contexts; usually the martingales in question arose as ...


1

As far as I know, this naming has its origin in the connection between Brownian motion and harmonic functions. Let $B$ be a $d$-dimensional Brownian motion. Let $f:\mathbb{R}^d\to\mathbb{R}$ be twice continuously differentiable. By Ito's formula, we have $$ f(B_t) = f(0) + \sum_{i=1}^d\int_0^t \frac{\partial f}{\partial x_i}(B_s) dB^i_s + ...


1

It all depends on what you want the set $\mathcal{V}^{m\times n}$ to be used for. In Oksendal's monograph, what it denotes is the following. Each set $\mathcal{V}^{m\times n}(0,T)$ denotes the set of $m\times n$ matrix processes with entries integrable over $(0,T]$. With Oksensdal's definition of $\mathcal{V}^{m\times n}$, this set becomes the set of ...



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