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1

Your solution to (b) looks fine. You can solve (a) in a similar fashion, but it may be easier to first note that $X_n$ is a function of $S_0, \dots, S_n$ (write $Y_k = S_k - S_{k-1}$). Then to show it is a martingale, it is equivalent to show that $\mathbb{E}[X_{n+1} - X_n \mid S_0, \dots, S_n] = 0$, which saves you from having to work with the sum. Each ...


3

Hint: By flipping the sign of $X_j$, all $j=1,2,\dots,n$, we have $\mathbb{P}(S_n=s\mid S_n^2=s^2)=\mathbb{P}(S_n=-s\mid S_n=s^2)$ for all $s$, so ...


2

Let $X_t$ denote the number of parrots at time $t$ and let $Y_t$ denote the number of crows at time $t$. $$\begin{aligned}P\left(X_{t}=k\mid X_{t}+Y_{t}=n\right) & =\frac{P\left(X_{t}=k\wedge Y_{t}=n-k\right)}{P\left(X_{t}+Y_{t}=n\right)}\\ & =\frac{P\left(X_{t}=k\right)P\left(Y_{t}=n-k\right)}{P\left(X_{t}+Y_{t}=n\right)}\\ & =\frac{e^{-\...


2

Let $\{t_1, t_2, t_3, ...\}$ be arrival times of a Poisson process of rate $\lambda>0$. Let $X(t)$ be a real-valued random process that possibly depends (causally) on the Poisson process. Assume that $\overline{X}$ is a real number such that: $$ \lim_{T\rightarrow\infty} \frac{1}{T}\int_0^T X(t)dt = \overline{X} \quad (\mbox{ with prob 1}) $$ The PASTA ...


1

According to "A first course in Stochastic Processes" by Samuel Karlin and Howard M. Taylor, this process is called "Multi-Type Branching Process". The extinction criterion for it is: $\lim\limits_{n \to \infty} P(v_n = 0) = 1$ iff the absolute value of all eigenvalues of $EA$ does not exceed $1$.


1

No you can't. Let $Y,Z$ be independent copies of the two state symmetric Markov chain starting with the invariant distribution. Let $X=1_{Y=Z}-1_{Y\neq Z}$. It is easy to check $$ \mathbb{E}[X\mid Y]=\mathbb{E}[X\mid Z]=0, $$ but of course $X\neq 0$.


1

Independence is not necessary for the weak law of large numbers. For a sequence of correlated random variables $X_i$, if the covariance $r(j) = Cov(X_k,X_{k+j})$ goes to zero as $j \to \infty$, then the weak law of large numbers still holds. To see this, suppose $X_1, X_2, \dots$ are a sequence of correlated random variables such that $r(j)=Cov(X_k,X_{k+j}...


1

We use the strong Markov property to remove the condition on $\mathcal{F}^W_{\tau_a}$ and say that $X(t-\tau_a)$ is just another Brownian motion, hence $$ \mathbb{P}(X(t-\tau_a)<0\mid\mathcal{F}^W_{\tau_a})=\frac12 $$ and we conclude $$ \mathbb{E}\left[\chi_{\sup_{s\in[0,t]} W(s)\geq a}\mathbb{P}(X(t-\tau_a)<0\mid\mathcal{F}^W_{\tau_a})\right]= \mathbb{...


1

If you take $f(x) = x$, you see that $W_t$ is a continuous local martingale. Then, taking $f(x) = x^2$ gives us that $W_t^2 - W_0^2 - t$ is also a continuous local martingale so that $W_t$ has quadratic variation at time $t$ equal to $t$. Now apply Levy's characterisation of Brownian motion to conclude.


0

No. You'll have to make a decision about those missing entries in the covariance matrix, and by extension the sort of function you want out. Choose them all zero and you'll get a function that is continuous nowhere. Choose something like the 'Orstein-Uhlenbeck exponential kernel' (referenced in the comments in your link) and you'll get something much more ...


3

$W(1)$ is a random variable and you cannot just suppose that it is equal to $a$. In fact, you know its distribution. If $W$ is a standard Brownian motion then $W(t) \sim \mathcal{N}(0,t)$ for every $t$. In particular, $W(1) \sim \mathcal{N}(0,1)$ which is why $\mathbb{E}[W(1)] = 0$.


0

Allow me to rewrite the PDE in question in my own notation. We are seeking to solve: \begin{equation} \frac{\partial }{\partial t} P(x,t) = -\frac{\partial}{\partial x}\left[\left(\sin(k x) + F \right) P(x,t)\right] + \frac{\partial ^2}{\partial x^2} \left[D \cdot P(x,t)\right] \end{equation} where $D$, $F$ and $k$ are constants. By using separation of ...


0

The hypothesis can be written as $Ee^{-tX}=e^{-\sqrt {2t} b}$ for all $t>0$. Differentiating w.r.t. $t$ (from the right) and setting $t=0$ we get $EX=\infty$.


0

If $I$ and $J$ are two disjoint intervals, then $1_I 1_J =0$ and therefore $$\big(x 1_{I}(t) + y 1_{J}(t) \big)^2 = x^2 1_I(t) + y^2 1_J(t)$$ for any real numbers $x,y \in \mathbb{R}$. This implies $$f(t)^2 = (W_2-W_1)^2 1_{[2,3)}(t) + (W_3-W_1)^2 1_{[3,5)}.$$ Hence, $$\int_0^{\infty} \mathbb{E}(f(t)^2) \, dt = \int_2^3 \mathbb{E}((W_2-W_1)^2) \, dt + \...


0

I think we need some extra conditions on the functions $g$ and $\omega$ (e.g. $\mathcal{C}^1_b$) in order to justify the derivation under the integral sign. In order to show this, let us simplify the problem by supposing $\omega\equiv 0$. We denote $X_t^x = xe^{\mu t+\sigma B_t}$ and assume $\sigma \geq 0$ and $x>0$. Hence, the problem is reduced to \...


0

First of all this is not really a Fokker-Planck equation since in this form the probability flux is not conserved but let us put aside this problem now. In order to solve the partial differential equation(PDE) we make an ansatz $f(x,t) = X(x) T(t)$. Inserting this into the PDE we get: \begin{eqnarray} (k_2 x+D) X^{''}(x) + (k_1 x-v) X^{'}(x) + E \cdot X(x)=...


2

Here are two methods that don't use any particularly advanced facts about B.M. (which you prefer will depend on what things you know about B.M.) (Via the reflection principle) Let $M(t) = \sup_{0 \leq s \leq t} B(t)$. The Reflection principle states that $$\mathbb{P}(M(t) \geq a) = 2 \mathbb{P}(B(t) > a) = 2 - 2\Phi(\frac{a}{\sqrt{t}})$$ where $\Phi$ is ...


1

This may be criticized as an overkill but a one line answer to your question comes from the Law of Iterated Logarithm for BM: $\lim \inf_{ t\to 0} \frac {B_t} {\sqrt {2t\log\, \log\, (\frac 1 t)}} =-1$ a.s.. This implies that every interval $(0,\delta)$ contains points $t$ with $B_t <0$.


2

$\newcommand{\Q}{\Bbb{Q}}\newcommand{\1}{\mathbf{1}}$Some hints: you will want to find the discounted $\Q$-expectation of the given payoff function. To do this, note that the $K_2\1_{K_1 < S_T < K_2}$ part is relatively easy to handle, assuming you know the (lognormal) distribution of $S_T$ (and remember that the $\Q$-expectation of an indicator is ...


2

The (simple) Markov property $$\mathbb{P}(X_t \in A \mid \mathcal{F}_s) = \mathbb{P}(X_t \in A \mid X_s) \tag{1}$$ makes perfect sense in any dimension $n \geq 1$. If, say, $(X_t)_{t \geq 0}$ is a continuous stochastic process which takes values in $\mathbb{R}^n$, then $(1)$ is well-defined for any Borel set $A \in \mathcal{B}(\mathbb{R}^n)$. The ...


0

Define the event $E_m$ as $$E_m =\{\text{ the $m$th ball is yellow }\}.$$ Then, you can compute the conditional expectation using $$\mathbf E[X_{m+1}|X_m] = \mathbf E[X_{m+1},E_m|X_m] + \mathbf E[X_{m+1},\neg E_m|X_m].$$ Now note that $$\mathbf E[X_{m+1},E_m|X_m]=\mathbf E[X_{m+1}|E_m,X_m]P(E_m|X_m),$$ with a similar equation for $\mathbf E[X_{m+1},\neg ...


1

Let $m=3$, with $T=\{1,2,3\}$ and suppose $(X_1,X_2)$ was uniform on $\{(0,1),(1,0)\}$ and so was $(X_1,X_3)$ and so was $(X_2,X_3)$. And that the $X_i$ are individually uniform on $\{0,1\}$. This specifies all $1$ and $2$-dimensional margins for $\{X_t:t\in S\}$ for all $S\subset T$ with $| S|<3$. There is no joint probability for $(X_1,X_2,X_3)$, ...


1

Yes, of course, as guaranteed by the Kolmogorov extension theorem. For instance, one can take $\lambda^{\mathbb R}$ where $\lambda$ is Lebesgue measure on $[0,1]$. But, as the Wikipedia article makes clear, it's not very useful Kolmogorov's extension theorem applies when $T$ is uncountable, but the price to pay for this level of generality is that the ...


2

Suppose $s\geq t$. $$E(W_s^2W_t^2) = E((W_s-W_t)^2W_t^2+2W_t^3W_s-W_t^4) \\ = E(W_{s-t}^2)E(W_t^2)+2E(W_t^3W_s)-E(W_t^4) =\\ (s-t)t+2E(W_t^3(W_s-W_t))+E(W_t^4) = \\ (s-t)t+2E(W_t^3)E(W_{s-t})+3t^2 = (s-t)t+3t^2 = 2t^2+st $$


0

I finally understood your problem... (but be rigorous in your questions, because you neither defined $\mathcal S_d$ nor $\mathcal R^d$, and we are not supposed to guess them). Instead of doing the proof in $\mathbb R^n$, do it in $\mathbb R$ (the principle is exactly the same and easier to see). We want to prove that $\sigma (\mathcal S)=\mathcal R$ ...


1

The first question is equivalent to a random walk on $\{-K,-(K-1),\dots,0,\dots,K-1,K\}$ starting at $0$ and taking a step to the right with probability $p$ or to the left with probability $1-p$, and asking the probability that the walk reaches $K$ before $-K$; presumably this is a standard problem that has been answered before. For the second question, the ...


0

Nothing can happen while $t<0$ since we don't consider the negative time. Now, $N(0) \ne 0$ means the event happens exactly at $t=0$. What is the Probability that the event happens at $t=0$ exactly? More generally, what is the probability, $Pr(X=c)$ when $X$ is a continuous random number and $c$ is a specific number? Even though the process counts ...


0

If I understand your question correctly, then I believe such a distribution is possible. Your original probability mass function was defined by $$P(n\text{ arrivals in }(0,t])=p(n)=\frac{(\lambda t)^n e^{-\lambda t}}{n!}$$ If you wish to limit the “number of arrivals” to $N$, then the new probability mass function will be given by the following, for $n\le N$...


3

I'm very new in this field, but I'm giving a try anyways. So please be critical and let me know if something is wrong (We are here to learn so...). The assertion is not true in general, indeed we have a couple of counterexamples. A simple counterexample. Let $\mu\equiv 1$ and $\sigma\equiv 0$. Define $$A:=(0,1)$$ Then we have for $X_0=1$, the explicit ...


3

Hints: Since $(B_t)_{t \geq 0}$ is a martingale it follows from the optional stopping theorem that $$\mathbb{E}(B_{T_R \wedge T_0}) = x.$$ Deduce from $B_{T_R \wedge T_0} \in \{0,R\}$ that $$\mathbb{P}(B_{T_R \wedge T_0}=R) = \frac{x}{R} \qquad \mathbb{P}(B_{T_0 \wedge T_R} = 0) = \frac{R-x}{R}. \tag{1}$$ Conclude that $$\mathbb{P}(T_R<T_0) = \frac{x}{R}....


0

No! You need much more than just $W_t-W_{k-1}\sim N(0,t-k+1)$. You want to show that $W_{t+k}-W_k$ is another Wiener process, i.e., you want to show it is almost surely $=0$ at $t=0$, independent Gaussian increments, and continuous paths with probability 1 These follows from $W_t$ being a Wiener process (why?).


1

Feller obtains in his book An Introduction to Probability Theory. Vol II (p.342) the following result (... unfortunately he does not give a detailed proof): Let $(B_t^x)_{t \geq 0}$ be a Brownian motion started at $x \in \mathbb{R}^d$ (i.e. $B_t^x = x+B_t$ where $(B_t)_{t \geq 0}$ is a standard Brownian motion). For fixed $a>0$ define a stopping time $\...


2

If you do not know the graph, then you need to think about something general. One piece of useful information that you do have is the degree of each vertex. Intuitively, it sound plausible that the invariant distribution is related to the degree of each vertex. If you draw some simple graphs, then you can see that this is indeed the case and that the ...


0

Finally, I think that I proved that $Y$ is adapted. We define $\tau_k$ as the time of the $k$-th jump ($\tau_0 := 0$). This is equivalent to the recursive definition $$ \tau_{k+1} := \inf \{ t \in (\tau_k;T] | X_t \ne X_{\tau_k} \}. $$ To prove that $\tau_{k+1}$ is a stopping time, note that $$ \{ \tau_{k+1} > t \} = \{ \tau_k \ge t \} \cup ( \{ \tau_k &...


1

They are not that complicatedly nested. You just need the following result: Let $T$ be a stopping time, $r$ be a $\mathcal F_T$-measurable positive random variable and $$ S=\inf\{t>T\mid d(B_t,B_T)=r\}. $$ Then $S$ is a stopping time. Proof: $\begin{align*} \{S\le t\}&=\{T<t\}\cap\left(\bigcup_{u\in]T,t]}\{d(B_u,B_T)=r\}\right)\\ &=\{T<t\...


0

This statement is true. For any $q \in \mathbb{Q}$, there exists a null set $N_q$ such that $$\lim_{n \to \infty}\sum_{s \in \pi_n}(B_{s'\wedge t}(\omega)-B_{s\wedge t}(\omega))=t$$ for every $\omega \in \Omega \setminus N_q$. Define $$\Omega'=\bigcap_{q \in \mathbb{Q}}\Omega\setminus N_q.$$ We have $$P(\Omega')=1.$$ Let $\omega \in \Omega'$. Then for ...


5

Write $\| \mu - \nu \|_{TV} := \frac{1}{2}\sum_{i \in S} |\mu(i) - \nu(i)|$ for the total-variation distance between two measures $\mu$ and $\nu$ on the countable state space $S$. We know the following facts: Fact. Let $X, Y$ be RVs with marginal distributions $\mu$ and $\nu$ under $\mathbf{P}$, respectively. Then $$ \| \mu - \nu \|_{TV} \leq \mathbf{P}(...


4

Approach I (via characteristic functions): The identity $$E(1_A g(B_{t_1},\ldots,B_{t_n})) = P(A) E(g(B_{t_1},\ldots,B_{t_n})) \tag{1}$$ can be easily extended to complex-valued continuous functions (just write $g= \text{Re} g + i \, \text{Im g}$ and apply $(1)$ separately to the real and imaginary part of $g$). Choosing $$g(x_1,\ldots,x_n) := \exp \left( i ...


1

Please feel free to amend the solution to improve it or correct some mistakes or you can even propose a new one of your own. Solution to Exercise 1.19 : 1 . Almost sure continuity results from the composition of the continuous mapping $<x,.>$ with the a.s. continuous trajectories of $X_t(\omega)$. So we only have to prove that increments over the ...


0

Take $m=2$. Let $\mathcal F_t$ be the natural filtration and consider a variable $X$ such that $$F_{X,B_{1,1},B_{2,1}}(x,y,z) = 2F(x)F(y)F(z) \text{ if } xyz>0, 0 \text{ otherwise}$$ where $F$ stands for the standard normal cdf, and this definition is extended by "independence" (i.e. you build the rest of your Brownian motions by drawing mutually ...


1

The fundamental matrix of an ergodic Markov chain is $$Z = \left(I-P+W\right)^{-1}$$ where $P$ is the transition matrix and $W$ is a matrix where each row is the fixed probability vector $w = (w_i)$ (the limiting distribution). You can calculate the mean first passage times $m_{ij}$ (expected number of steps to reach state $j$ when starting from the state ...


0

You can't draw that conclusion from the model. Suppose there are $4$ men and $4$ women each divided into two groups of $2$. Pair the groups and have each mixed sex group of $4$ shake hands all around their group. Then everyone shakes twice so the average is $2$ but hlaf the people are clearly disease free. Of course that distribution of handshakes is not ...


2

This can be expressed as: 1. go from state 1 to 2, and stay there for at least N-1 units of time. ie we have: $p_{12}p_{22}^{N-1}$


2

You formula is not correct. The answer should be $P_{12} P_{22}^{N-1}$, the first step is move from state 1 to state 2, then the chain has to stay in state 2 for the next $N-1$ steps.


0

After doing some digging around, I think I have found the answer. I am posting it here so that it might help somebody who may have the same question. As written in An Introduction to Computational Stochastic PDEs (page 314) and Oksendal's Stochastic Differential Equations: An Introduction with Applications, 6e (page 65), a White Noise process in continuous ...


0

Your notation $(x,y)=\{(2,0),(0,2),(1,1)\}$ is fine, but $P(E_3|E_1E_2) = \frac{1}{3}$ is not true. Let the two aces be $A\clubsuit$ and $A\spadesuit$. Case 1: $(x,y)=(2,0)$. There is only one way to have $2$ aces in $E_3$ (both $A\clubsuit$ and $A\spadesuit$) and ${24\choose 11}$ ways to have non-aces in $E_3$, while overall there are ${26\choose 13}$ ...


0

Another way to compute $P(E_3\mid E_1,E_2)$ (from your first attempt) is to consider the third and fourth piles as an array of $26$ locations where cards can be put, one card in each location. (Top of the third pile, card next to the top of the third pile, second card from the top of the third pile, etc.) There are two aces that can be put in two of these $...


1

The probability for selecting 1 from 2 aces and 12 from the remaining 24 cards when selecting 13 from 26 cards is: $$\begin{align}\mathsf P(E_3\mid E_1,E_2)&=\left.\dbinom{2}1\dbinom{24}{12}\middle/\dbinom{26}{13}\right.\\[1ex] &=\dfrac{13}{25}\end{align}$$ Let $(x,y)= (\text{number of ace in the third pile}, \text{number of ace in the fourth ...


2

[1] $E[Y_kY_i]=E[E[Y_kY_i|Y_i]]=E[Y_iE[Y_k|Y_i]]=E[Y^2_i]$ changing $k \rightarrow j$ and subtructing we have the first equality. [2] $E[(Y_k-Y_j)^2|F_i]=E[Y_k^2|F_i]+E[Y_j^2|F_i]-2E[Y_kY_j|F_i]$ but now: $E[Y_kY_j|F_i]=E[E[Y_kY_j|F_j]|F_i]$ $=E[Y_jE[Y_k|F_j]|F_i]=E[Y^2_j|F_i]$ This way we have the second equality. Use has been made as suggested ...


1

The answer above is circular to me (since the proof of martingale convergence that I know goes through the upcrossing inequality), so I will try to give a different answer (I will be paraphrasing Durrett PTE version 5 Theorem 4.2.10 and 4.2.11, so feel free to read those up; I think they are very accessible). The intuition is the following: since a ...


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