New answers tagged

0

Easy way: we know that $B_t$ is a Gaussian process, i.e. for any $t_1, \dots, t_n$ the random vector $(B_{t_1}, \dots, B_{t_n})$ has a jointly Gaussian distribution. Since any linear transformation of a jointly Gaussian vector is again jointly Gaussian, it follows that $W_t$ is also a Gaussian process, and likewise so is $Y_t$. Then it's easy to compute ...


0

For a simpler scenario without effective incoming rate, suppose there is a queue where customers arrive at rate $\lambda$ and they go into either of two lines with equal probability $\dfrac{1}{2}$. There is one server who serves the customer at rate $\lambda$. Regardless of the number of different lines, the one server can only serve one customer at a time ...


2

Below is an outline. The grey areas are spoilers — reveal them by placing your mouse over them if you want details, but I'd suggest you try by yourself before. Compute the expression of the CDF $F^{(n)}_X$ of $X\stackrel{\rm def}{=} \max(X_1,\dots, X_n)$, using independence. Recall that for $t\in\mathbb{R}$, $$ F^{(n)}_X(t) = \mathbb{P}\{ X \leq t \} = ...


1

Let $\varphi(\theta)=\mathbb E[e^{\theta X_1}]$ be the moment generating function and $\psi(\theta)=\log\varphi(\theta)$ the cumulant generating function of $X_1$. We compute $$ \varphi(\theta) = e^{-\theta}\mathbb P(X_1=-1)+e^{\theta}\mathbb P(X_1=1)=\frac12\left(e^{-\theta}+e^{\theta}\right)$$ and $$\psi(\theta) = \log\left(\frac 12 ...


0

Elaborating on @Did's comment, from $\mathbb E[f(\xi_1)\mid S_0=k]=f(k)$ we have $$f(k)= f(k-1)\mathbb P(\xi_1=-1)+f(k+1)\mathbb P(\xi_1=1)=q\cdot f(k-1)+p\cdot f(k+1), $$ and hence $$f(k+1) - \frac1p f(k) +\left(\frac qp\right) f(k-1)=0.$$ This recurrence relation has characteristic polynomial $$\lambda^2-\frac1p\lambda +\frac qp, $$ with roots ...


0

Collecting all mentioned arguments. 0*. The function $t\mapsto Y_t = X_{t+s}-X_s$ is continuous a.s. (I believe that this is also necessarily). $Y_0 = X_{0+s}-X_s = 0$. Observe that $Y_{t_2} - Y_{t_1} = X_{t_2+s} - X_{t_1+s}$ holds for any $0\leq r_1\leq r_2$. Then for all $0\leq r_1\leq r_2\leq \dots\leq r_n$ $Y_{t_2} - Y_{t_1}, Y_{t_3} - Y_{t_2}, ...


0

Suppose $(\mu,\sigma)$ obeys the linear growth condition $$|\mu(t,x)|+|\sigma(t,x)|<C(1+|x|),\ \forall t\in[0,T],\, x\in\mathbf R$$ for some positive constant $C$. By the Cauchy-Schwarz inequality and the Gronwall inequality, $$\mathbf E[r^2]<3\mathbf E[r(0)^2]e^{a(1+T)t}, \forall t\in[0,T]$$ for some positive constant $a$. $$\mathbf E ...


-1

UPDATE: Here is the short answer: The process is not necessarily Markov. Limiting or stationary distributions need not exist. (Original answer below with some corrections and edits noted.) Part I: Markov-ness. In general, this is not a Markov process. The exponential distribution is the only continuous distribution with the memory-less property. See ...


1

Hint: The density $f(a)=\frac12e^{-|a|}$ belongs to the Laplace distribution. The Laplace distribution arises when you subtract two iid exponential(1) variables. Second hint: If $U$ has uniform distribution on $[0,1]$ then $X:=-\ln (U)$ has exponential(1) distribution.


1

Just for your reference as an alternative solution, though more complex. Let $X$ be the number of coin tosses until we get a success. Then $$ \Pr(X = i) = q^{i-1}p $$ The corresponding generating function is $$ f(x) = \sum_{i=1}^\infty q^{i-1}p\cdot x^i = \frac{p}{q}\cdot\sum_{i=1}^\infty (qx)^i = \frac{p}{q} \cdot \frac{qx}{1-qx}=\frac{px}{1-qx} $$ Now ...


1

$\{Y=n+i\}$ is the event that among $n+i-1$ experiments there were $i$ failures, and secondly also that the $n+i$-th experiment was a success. There are $\binom{n+i-1}{i}$ ways of selecting $i$ experiments out of $n+i-1$. Each of these selections goes along with a probability of $p^{n-1}q^i$ of occurring. The $n+i$-th experiment has probability $p$ to ...


0

Of course, it depends on what high means when you say $\ldots$ "When $m$ and $V$ are very high, we get ridiculously high number." Nonetheless, I could not replicate your concerns. I did two simulations ($N = 10\,000$ each) using MATLAB, one with $m$ varying from $-70$ to $70$, and another with $v$ varying from $0.5$ to $30$. The two figures show the ...


1

Here is an example in discrete time explaining why the Markov property is crucial to guarantee that solutions do not cross, in the sense that if two solutions coincide at some given time then they coincide at any later time. Consider the AR2 process $(x_n)_{n\geqslant0}$ defined by some initial conditions $(x_0,x_1)$ and by the recursion ...


1

I realize this is a late answer to this post, but it still makes the top two to three results on Google for "ensemble average" and an answer has not yet been officially accepted. For posterity, I figured I would try to answer it to the best of my ability in the way that the question has been phrased. First, it is important to have a broad understanding of ...


1

Express $Y$ as an Ito process: $$ dY_t=X_t\,dt = X_t\,dt +\ 0\,dB_t. $$


1

You're certain to take exactly one step each from $1$ to $2$ and from $3$ to $4$; that's the constant $2$. At $2$, you have a Bernoulli experiment with success probability $\frac23$ that you perform until it succeeds, and at $4$ you have a similar Bernoulli experiment with success probability $\frac35$. The expected time until a Bernoulli experiment succeeds ...


2

If $X$ has generator $G$ and $b>0$ is a scalar, then $X(bt)$ is a Markov process with generator $bG$. If $G_1$ and $G_2$ are generators on the same state space, and (writing $D(G_i)$ for the domain of $G_i$) $D(G_1)\cap D(G_2)$ is large enough, then $G_1+G_2$ is the generator of another Markov process. Under the appropriate conditions, the semigroup $S$ ...


2

I think you misunderstood the meaning there. They talked about "contingent" claims, these are random variables given some other process. A simple example will be something like this: you have a simple SDE, $dX = dB$, with $X_0 = a$, the (strong) solution is, of course, $X_t = a+ B_t$. Now, you have another process on the same Brownian Filtration: $dY ...


1

Let's assume we are in the context of a filtered probability space $(\Omega,(\mathcal F_t),\Bbb P)$ satisfying the usual conditions. Here are a few features of optionality that make it important. If $T$ is a stopping time and $A\in \mathcal F_T$ then there is a bounded optional process $Z$ such that $Z_T=1_A$ on $\{T<\infty\}$. If $Z$ is an optional ...


1

"Progressive" and "optional" are refinements of "adapted" involving some degree of joint measurability of $(\omega,t)\mapsto X_t(\omega)$. The three notions coalesce in discrete time. Consider, for example, the notion "optional" in a discrete-time setting. So let $(\Omega,(\mathcal F_n)_{n\ge 0},\mathcal F,\Bbb P)$ be a filtered probability space. The ...


1

I think you've got the terminology mixed up a little. You have properties of a process (such as being adapted to a filtrartion, being progressively measurable, or being predictable), and then you have stopping and optional times (where the difference is, roughly speaking, merely that optional processes cannot answer questions regarding the "current time"). ...


0

So, you have a Poisson point process and under the condition that $n$ events have happened in the time interval, $(0;t]$, you want the joint probability density function of times for the sequence of times of these events. $\{\tau_i\}_{i\in\{1,..,n\}}$. Under this condition, the time of each event will be independently and uniformly distributed over the ...


2

The number of offspring of a cell is $X$. The probability that $X=k$ is given by: $~\mathsf P(X=k) = qp^k$ for $k\in\{0, 1, 2,\ldots \}$ and $q=(1-p)$. The expected value of $X$ is given by : $~\mathsf E(X) = \sum_{k=0}^\infty k~\mathsf P(X=k)$ Thence, using Geometric Series closed form (when $\lvert p\rvert<1)$: $$\begin{align}\mathsf E(X) ~=~& ...


1

$$X = \sum_{i=1}^n X_i$$ where $X_i \in \{ -1, 0, 1 \}$ is the change in $x$ at step $i$. Since the variables $X_i$ are independent and each has mean 0, this implies that $$E(X^2) = \sum_{i=1}^n E(X_i)^2 = n E(X_1^2)$$. From here, it should be easy to calculate $E(X_1^2)$ explicitly.


1

This might be helpful: Theorem 29 on page 173 of Protter's "Stochastic integration and differential equations": Let $M$ be a local martingale and let $H\in \mathcal P$ be locally bounded. then the stochastic integral $H\cdot M$ is a local martingale. I'm reading that book, but haven't got that far yet, will be interested to see if someone can provide a ...


2

Check out George Lowther's blog: https://almostsure.wordpress.com/2010/01/03/the-stochastic-integral/ Especially Lemma 3, answers your question "why cadlag?"


1

$X_1 X_2=0$ so $E[X_1 X_2]=0$ and thus, since $E[X_1]=E[X_2]=0$, the covariance and correlation must be zero. But, for example, $X_1=1 \implies X_2=0$ so they are not independent


2

The pdf of $X=(X_1,\,X_2)$ is $$ f(x_1,x_2)=\begin{cases} \frac{1}{4} & \text{for } (x_1,x_2)\in Q=\{(-1,0), (1,0), (0,-1), (0,1)\}\\ 0 & \text{otherwise} \end{cases} $$ and the variable $X=(X_1,X_2)$ can be represented in tabular form $$ \begin{pmatrix} (X_1,X_2)\\ f(x_1,x_2) \end{pmatrix}= \begin{pmatrix} (-1,0) & (1,0) & (0,-1) & ...


2

In addition to the hints, we need the following identities related to derangements: (1) $\displaystyle\sum_{i=0}^{n}\frac{!i}{i!~(n-i)!}=1,~(n\ge0),$ (2) $\displaystyle\sum_{i=0}^{n}\frac{!i~(n-i)}{i!~(n-i)!}=1,~(n\ge1),$ (3) $\displaystyle\sum_{i=0}^{n}\frac{!i~(n-i)~(n-i-1)}{i!~(n-i)!}=1,~(n\ge2),$ where I have adopted the subfactorial notation. The ...


1

There are at least two reasons. I am certain of the second, only 60% of the first one, so please keep this in mind. 1. Look at equation (2.7); $F_t$ is a stopped version of the integral of the progressively measurable process. It should follow that (at least restricted to locally bounded/integrable progressively measurable processes) that the integral ...


0

This is a good question. I do not have a full response to it, but rather a partial answer. Maybe one approach would be to try to make use of the conditional expectation $\mathbb{E}(T_X \mid \alpha)$, or more specifically its factorized form $\mathbb{E}(T_X \mid \alpha = a)$. If $T_{X^a}, T_{Y^a}$ denote the first entry time in some set $A_X, A_Y \in ...


5

In discrete time at least, the definitions I'm familiar with are fairly straightforward. Given an increasing filtration $\{\mathcal{F}_n\}_{n=0}^{\infty}$, a process $\{X_n\}_{n=0}^{\infty}$ is adapted if each $X_n$ is $\mathcal{F}_n$-measurable. For predictable processes, the random variables are measurable with respect to slightly smaller ...


3

(i) and (ii) are correctly reasoned and the calculations are okay. (iii) use conditional probability, and the independence of Poisson counts in disjoint intervals. That is: $(N_4{-}N_2)$ is i.i.d. to $N_2$ . $$\begin{align} \mathsf P(N_4\leq 5\mid N_2\leq 3)~=~&\dfrac{\mathsf P(N_4\leq 5, N_2\leq 3)}{\mathsf P(N_2\leq 3)} \\[1ex] =~& ...


4

The distribution of $\int_0^t W^2_s\,ds$ was found by Cameron and Martin in the 1940s. Mark Kac (1949, Transactions of the AMS) applied his method to find the Laplace transform of this integral: $$ \Bbb E\left[\exp(-u\int_0^t W^2_s\,ds)\right]={1\over\sqrt{\cosh(t\sqrt{2u})}}. $$


2

Regarding the indicator functions, just use their definition, i.e. they are equal to 1 on the event they indicate, and $0$ on its complement. Then $$ 1_{\{S\ge K_1\}}1_{\{S\ge K_2\}}=1_{\{S\ge K_1\}\cap{\{S\ge K_2\}}}=1_{\{S\ge \max\{K_1,K_2\}\}}. $$


1

As Wikipedia points out, there are several results called Riesz (representation) theorem. A common feature of some of them is that on some spaces, every linear functional is integration against something. The result that the author uses here is the representation of linear functionals on $L^q$: they are integrals against $L^p$ functions. More precisely, ...


1

It's enough that $f$ be Lebesgue measurable and $\int_0^t [f(s)]^2\,ds<\infty$ for each $t$. Indeed, in this case your stochastic integral is well defined. Moreover, if we set $A_t:=\int_0^t[f(s)]^2\,ds$, then $Y$ is a (mean zero) continuous martingale with quadratic variation $\langle Y\rangle_t=A_t$. Let $\tau(t):=\inf\{s:A_s>t\}$. Then ...


0

Introduce the Ito process $dZ_t=tdW_t$, in that case, $Y_t=e^{Z_t}$. By the way, to prove that it is a martingale, you have to provide an additional information. A driftless process is not necessarily a martingale.


0

If a function is continuous, you can connect its Riemann integral with a limit of a sum of elements. In your case, they are normally distributed. EDIT : let n an integer, and we define the a subdivision of the interval $[0,t]$ , as follows $t_i=\frac{t}{n}i$ with $i \in${ ${0,...,n}$} Since $W$ is a.s continuous, as well as h, we have $t \rightarrow ...


1

In the case of equality in law then the answer is clear from the independence hypothesis you have formally : $P(X^1_t=S_t.Y_t^1<a)=P(Y_t^1<a/s).P(S_t<s)=P(Y_t^2<a/s).P(S_t<s)=P(S_t.Y_t^2<a)$ and as $P(X^1_t<a)=P(X_t^2<a)$ this leads to the conclusion, $P(X_t^2<a)=P(S_t.Y_t^2<a)$. For the case of almost sure equality (or ...


0

for analog continous signals, we have time average.Time average is averaged quantity of a single system over a time inetrval directly related to a real experiment. or discrete signals, we have ensemble average. Ensemble average is averaged quantity of a many identical systems at a certain time.


1

Yes, your calculation is correct. It doesn't matter that the density can be bigger than $1$; it's just a density, not a probability. Now you just have to take into account that the events are uniformly distributed over $[0,t]$, not $[0,1]$. Yes, the conditional expectation of $W_1$ is different from the unconditional one. The unconditional one depends on ...


2

Yes, this results holds for any Lévy process which is integrable. It follows from the independence and stationarity of the increments that $$\mathbb{E}X_{k/n} = \mathbb{E} \left( \sum_{j=1}^k (X_{j/n}-X_{(j-1)/n}) \right) = k \mathbb{E}(X_{1/n})$$ and, similarly, $$\mathbb{E}(X_1) = \mathbb{E} \left( \sum_{j=1}^n (X_{j/n}-X_{(j-1)/n}) \right) = n ...


1

In the basic Doob-Meyer decomposition, we have $Z=M-A$ with $M$ a martingale and $A$ a predictable integrable increasing process. Comparing left limits $Z_-=M_--A_-$ with predictable projections ${}^pZ={}^pM-A$ and using the fact that ${}^pM=M_-$ because $M$ is a martingale, we have $Z_--{}^pZ=A-A_-$. Thus $A$ is continuous if and only if $Z_-$ is ...


1

The event $\{t(x)>2\}$ means the sum of 2 iid uniform RVs is less than $x$, and $P(X_1+X_2<x)=\int_0^{min(1,x)}\int_0^{min(1,x-x_1)}dx_2dx_1.$


0

As suggested by Jay.H, the solution follows from Jensen's inequality: $$ \mathbb{E}\left[\left|X_{t}\right|^{p}\right]=\mathbb{E}\left[\mathbb{E}\left[\left|X_{t}\right|^{p}\mid\mathcal{F}_{s}\right]\right]\geq\mathbb{E}\left[\mathbb{E}\left[\left|X_{t}\right|\mid\mathcal{F}_{s}\right]^{p}\right]=\mathbb{E}\left[\left|X_{s}\right|^{p}\right]. $$


1

1) Yes, check if they have the same characteristic function $\forall t$. Else, generally not. But you may be able to do that by some different characterization theorems for specific processes (e.g. Brownian Motion). $\\$ 2a) From a theorem on weak convergence, you have $$X_n\overset{D}\to X \iff \lim_{n\to \infty} \mathbb{E}[g(X_n)]=\mathbb{E}[g(X)]$$ for ...


1

If $k\leq (n-1)$ you're done. Otherwise, $k\geq n$. You know that it's possible to reach state $j$ from $i$ in at most $k$ steps with positive probability. By pigeon-hole principle, you have to visit at least one state twice in those $k$ steps (since $k\geq n$). If you now delete the loop to that path, you've necessarily reduced the number of steps to a ...


1

I got the answer! Suppose there is no $r \in \{1, \dots, n-1\}$ such that $P^r(j, i) > 0$. So $r \geq n$ (we know there is some $r$ that it happens). Then there are $i_1, \dots i_r$ such that $0 < \leq P(j, i_1)P(i_1, i_2)\dots P(i_r, i)$. As $r \geq n$, there is some $m_1$ and $m_2$ such that $i_{m_1} = i_{m_2}$. Then you can "cut" the path ...


2

See §10.6. Intermission: Harmonic Functions and Brownian Motion of Ulrich's Complex Made Simple.



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