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5

Of course $I-N$ is not always invertible. Clearly, if $A$ is a non-invertible function, then $N=I-A$ causes $I-N=I - (I-A) = A$ to be non-invertible. The thing is that the expression $$(I+N+N^2+\cdots)$$ makes no sense if $N$ is not some special matrix. For example, if $N=I$, what is $I+N+N^2+\dots$? It equals an infinite sum of identity matrices and it ...


2

A condition: If $N$ is nilpotent matrix, then there exists $(I - N)^{-1}$. A nilpotent matrix, for example: an upper triangle matrix with entries on diagonal are $0$.


2

For part (i), we don't even need right-continuity or stationary increments from $X$. Notice that if $t>s$, then for any (bounded-measurable) function $f: E \to \mathbb{R}$, we can write $$E[f(X_t)|\mathcal{F}_s] = E[f(X_t-X_s+X_s)|\mathcal{F}_s] = E[g(X_t-X_s,X_s)|\mathcal{F}_s]$$ where $g(x,y) = f(x+y)$. Now, we know that $X_s$ is ...


2

Let $Y_t$ be a Poisson process with rate $\mu$. Since $\mu_n = \mu$, we can take the deaths as following $Y_t$ until everyone has died. So given you start with $N$ people, and $n \ge 0$, the probability that $X(t) = n$ is the probability that there are exactly $N-n$ occurrences in the $Y_t$ process in the interval $[0,t]$. The number of such occurrences ...


1

Following your approach you can write : $E[1\{X_{t+s}=y,X_T\in A \}|\mathcal{F}_t]=(1)E^{X_t=x}[1\{X_{t+s}=y\}.1\{X_T\in A\}]$ $=(2)E^{X_t=x}[E^{X_{t+s}=y}[1\{X_{t+s}=y\}.1\{X_T\in A\}]]$ $=(3)E^{X_t=x}[1\{X_{t+s}=y\}.E^{X_{t+s}=y}[1\{X_T\in A\}]]$ $=(4)E^{X_t=x}[1\{X_{t+s}=y\}.P(X_T\in A|X_{t+s}=y)]$ $=(5)E^{X_t=x}[1\{X_{t+s}=y\}].P(X_T\in A|X_{t+s}=y)$ ...


1

Since $N(t)$ is Poisson, which is discrete (its pdf is $\frac{e^{-\lambda t}(\lambda t)^k}{k!}$), then you need summation and not integral: $$\mbox{E}(e^{-\alpha t} S(t)) = \sum_{k=0}^{\infty} e^{-\alpha t}s e^{(\alpha-\lambda \sigma)t} (\sigma + 1)^{k} \frac{e^{-\lambda t}(\lambda t)^k}{k!} =\\= se^{-\lambda\sigma t}\sum_{k=0}^{\infty} \frac{e^{-\lambda ...


1

I would do this. Show $$\begin{align}P_t (x,A) &{:=} \int_A g_t (x,y)dy \\ P_0 (x,A) &{:=} \delta_x (A), \end{align}$$ where $g_t$ density of the folded normal distribution, is a Markov kernel (Chapmann K. Equation, etc.). Then you know there exists a unique Markov process. That's the reflected Brownian motion. Have someone any comments? Is that ...


1

Hint: Use $$P_t f(x) = e^{-\lambda t} \underbrace{\sum_{y \in E} f(y) P^{(0)}(x,y)}_{f(x)} + t \underbrace{\sum_{n \geq 1} e^{-\lambda t} \frac{\lambda^n t^{n-1}}{n!} \sum_{y \in E} f(y) P^{(n)}(x,y)}_{\text{bounded (for e.g. $|t| \leq 1$)}}$$


1

Note that if $\langle 1 - \lambda, \vec u \rangle$ is an eigenpair for $N$, then: \begin{align*} N\vec u &= (1 - \lambda)\vec u \\ &= \vec u - \lambda \vec u \\ &= I_2\vec u - M \vec u &\text{since $\langle \lambda, \vec u \rangle$ is an eigenpair for $M$}\\ &= (I_2 - M) \vec u \end{align*} This suggests that $N$ should be the $2 \times ...


1

I believe this approach will work for your problem: Modify the state transition matrix as follows. Since both states $0$ and $6$ are absorbing states, and you start every new run at state $1$, identify states $0$ and $6$ as state $0$, and have state $0$ go to state $1$ with probability $1$. That is, the new matrix will look like this: $$ A = \left[ ...


1

To your first question, the answer is generally no. The matrix $$ N= \frac 23 \pmatrix{1&1\\1&-1} $$ will have (Euclidean) operator norm $4/3 > 1$. $(I - N)^{-1}$ will exists if $\rho(N)<1$ ($\rho(N)$ denotes the spectral radius of $N$). This in turn is true if and only if $\|N\| < 1$ for some multiplicative matrix norm $\|\cdot \|$.


1

With respect to the game, this is exactly the St. Petersburg paradox. reference: http://en.wikipedia.org/wiki/St._Petersburg_paradox The game has infinite expected value but is worth only a finite amount in practice if that makes sense?


1

Hence, since 0 < s < t, we know the distribution of arrivals in $[0,t]$ is Uniform. Since each occurrence is uniform and independent of all others, The formula you stated is correct. Basically, view the events $A_i$ = {arrival i occurs in the first s minutes of $[0,t]$} for $1 ≤ i ≤ 8$ $A_i$ occurs with probability $\frac st$ by uniformity. Each ...


1

Regular Markov chain means that all the entries of the transition matrix $P$ are positive. Let Markov chain have $n$ states. If $\pi _i$ is the equilibrium probability for state $i$, you can say: $$ \pi _i = \sum _{j=1}^n\pi _j P_{ji} $$ Assume by contradition that $\pi _i=0$. Since all the entries of $P$ are positive, from equation above you can conclude ...


1

Since $X_{\tau}$ is $\mathcal{F}_{\tau}$-measurable, we have $$\mathbb{E}_x(g(X_{\tau}) \mid \mathcal{F}_{\tau}) = g(X_{\tau})$$ for any measurable function $g: E \to \mathbb{R}$. Use this for $g(x) := \mathbb{E}_x(f((X_t)_{t \in I})$.


1

In order to make the entire process be Markov, you can raise the dimension, and consider a four state chain: the states are $(A,1),(A,0),(B,1),(B,0)$. (This is a common theme: you can increase the dimension to take away memory effects, or reduce the dimension at the cost of adding memory effects.) The first component represents "Alice will come in next/Bob ...


1

Set $\tau_j := (\lfloor 2^j \rfloor+1)/2^j$. Then $\tau_j$ is a discrete stopping time and $\tau_j \downarrow \tau$ as $j \to \infty$. Fix $F \in \mathcal{F}_{\tau+}$. Then, by the right-continuity and the dominated convergence theorem, $$\mathbb{E}(e^{\imath \, \xi (Y_t-Y_s)} 1_F) = \lim_{j \to \infty} \mathbb{E}(e^{\imath \, \xi ...


1

It is easier to calculate the expectation directly than it is to write down the distribution. This is usually the case. Since you haven't said that much about what you've tried yet, let me start with a hint. Try to calculate the expected time to hit $\{ 0,n \}$ starting from $x$ with $0<x<n$. Then send $n \to \infty$. You can set up a system of ...


1

Since $V(\tau) \in \{0,1\}$, we have: $$ S(t) = \int_0^t V(\tau)d\tau = \int_0^t 1\{V(\tau)=1\}d\tau $$ where $1\{V(\tau)=1\}$ is an indicator function that is 1 if $V(\tau)=1$, and 0 else. Then $\lim_{t\rightarrow\infty} S(t)/t$ is the limiting fraction of time being in state $1$, which you can work out via basic steady state theory (and the answer is ...


1

The statement is true in the limit of $\delta t \to 0$. In that case, the probability of two events goes down as $(\delta t)^2$ which is ignorable in comparison with $r\delta t$. Strictly speaking, the result is only true at some time $t$ if the rate per unit time at time $t$ does not go to infinity. Thus if $$r(t) = \frac{1}{(t-4)^2}$$ then at $t=4$ you ...



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