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0

We are given that $X,Y\sim \mathrm{Exp}(1)$ so that the probability density function (pdf) of $X$ is $f(x)=e^{-x}$ and the pdf of $Y$ is $f(y)=e^{-y}$ For all $z_1,z_2>0$ we need to evaluate the joint density of $X$ and $Y$ with the appropriate limits as follows:- ...


1

In general, the sum of two compensated poisson processes is not a martingale. Example Let $(N_t^1)_{t \geq 0}$ a Poisson process with intensity $\lambda^1 =1$. If we set $N_t^2 := N_{t/2}^1$, then $(N_t^2)_{t \geq 0}$ is a Poisson process with intensity $\lambda^2 = 1/2$. For the joint filtration $\mathcal{F}_t := \sigma(N_s^1,N_s^2; s \leq t)$ it holds ...


0

$$g(x,t) = ∫_0^t \theta(u) \exp(-\alpha(t-u) ) dB(u) \\ = e^{-\alpha t} ∫_0^t \theta(u) \exp(\alpha u ) dB(u) $$ Now use the fact that $t \to e^{-\alpha t}$ is of finite variation and the Ito formula for a product: If $X,Y$ are two Ito processes then so is $XY$: $$ dX_t = a_t dt + \sigma_t dB_t \\ dX_t = b_t dt + \sigma'_t dB_t \\ d(XY)_ t = X_t dY_t ...


2

$\newcommand{\E}{\mathbb E}$ I will assume here that $S$ is independent of $N$. (Very often people writing about this neglect to state that assumption. And so it is here, if in fact that was intended, and if it wasn't then the nature of the dependence would need to be mentioned at least implicitly.) You can use the law of total expectation: \begin{align} ...


1

Yes, it's the delta function. The formula means that for every $t$ the variable $f(t)$ has variance $R$ and for every $s\neq t$ variables $f(s)$ and $f(t)$ are uncorrelated. This corresponds to the more intuitve discrete-time white noise.


1

Actually, there is nothing left to do. From $$dZ_t = b \exp \left( \frac{c^2}{2} t - c W_t \right) \, dt$$ it follows that $$Z_t = \underbrace{Z_0}_{0} + b \int_0^t \exp \left( \frac{c^2}{2} s - c W_s \right) \, ds.$$ Hence, $$X_t = Y_t \cdot Z_t = b \exp \left(- \frac{c^2}{2} t+c W_t \right) \cdot \int_0^t \exp \left( \frac{c^2}{2} s - c W_s \right) ...


0

Some hints to (b): In an incomplete market $\Pi(C)$ is an interval because we put lower and upper bounds on the arbitrage free prices. For example the upper bound on the price of a non-replicable $C$ is $$ \inf(E_{\mathbb{Q}}[X/(1+R)]:X\geq C, \: X \text{ is replicable}) $$ because if it would trade at a higher price than this, you could do arbitrage by ...


1

What have you done so far, where are you stuck? The transition rate matrix for the standard M/M/1 queue model is given by $$Q=\begin{pmatrix} -\lambda & \lambda \\ \mu & -(\mu+\lambda) & \lambda \\ &\mu & -(\mu+\lambda) & \lambda \\ &&\mu & -(\mu+\lambda) & \lambda &\\ &&&&\ddots \end{pmatrix}.$$ ...


1

Hint It follows from Itô's formula that $$\exp(\theta t) \cdot B_t = \int_0^t \exp(\theta s) \, dB_s + \theta \int_0^t \exp(\theta s) B_s \, ds,$$ i.e. $$ \int_0^t \exp(\theta s) \, dB_s = \exp(\theta t) \cdot B_t - \theta \int_0^t \exp(\theta s) B_s \, ds.$$


0

Hmmm ... it seems that your book is confusingly using $X$ to denote both the process and the state space. Let me change the state space to $S$ to make things simpler. This is the description of a stochastic process; it is just a collection of random variables. Instead of just one or two, you have a random variable for every $t$ in some index set $T$: $$ ...


0

Apply Ito's lemma to $f(x,t) = t - x^2$ $$ \mathrm{d} f(W_t,t) = \mathrm{d}t - 2 W_t \mathrm{d}W_t - \frac{1}{2} 2 \mathrm{d}t = 2 W_t \mathrm{d}W_t$$ It is a property of an Ito integral $I_t = \int_0^t a(s, \omega) \mathrm{d}X_s$ where $X_t$ is a martingale and $a(t,\omega)$ is adapted that $I_t$ is a martingale. QED EDIT: you need a few more technical ...


3

Let $$W_t := \sigma B_t - \mu \cdot t$$ a Brownian motion with drift $\mu>0$ and $$T_x := \inf\{t \geq 0; W_t \geq x\}.$$ Using the stationarity and independence of the increments of the Brownian motion $(B_t)_{t \geq 0}$, it is not difficult to see that $\left(\exp\left[\frac{2\mu W_t}{\sigma^2}\right]\right)_{t \geq 0}$ is a martingale. In ...


1

Recall that an SDE of the form $$dX_t = b(X_t) \, dt + \sigma(X_t) \, dW_t$$ has a unique solution whenenver the coefficients $b$, $\sigma$ are globally Lipschitz continuous. In particular, we see that the SDE $$dX_t = (1+X_t) \, dt+ X_t \, dW_t$$ has a unique solution. Now let $(Y_t)_t$ the (unique) solution of the SDE $$dY_t = Y_t \, dt + Y_t \, dW_t ...


1

The conditional distribution of $(S_k)_{1\leqslant k\leqslant n}$ is uniform, that is, its density is $$n!\,t^{-n}\,\mathbf 1_D,$$ where $$ D=\{(s_k)_{1\leqslant k\leqslant n}\mid 0\leqslant s_1\leqslant\cdots\leqslant s_n\leqslant t\}. $$ One is interested in the conditional distribution of $(T_k)_{1\leqslant k\leqslant n}$ where $T_1=S_1$ and ...


0

The probability that 11 is a subsequence of $\sigma$ is the probability that a Markov chain on the state space $\{0,1,2\}$ starting from the state $0$ visits the state $2$ at or before time $k$, if the transition probabilities are $1-x$ for the transitions $0\to1$ and $1\to2$, and $x$ for the transitions $0\to0$ and $1\to0$, where $x=1-1/n$. The usual ...


0

The results follow from the following identities, using the structure of the process $(V_i)$. Note that $$X_i=\sum_{n\geqslant1}\beta^{n-1}V_{i-n} $$ hence $$ \mathrm{var}(X)=\sum_{n\geqslant1}\beta^{2(n-1)}\mathrm{var}(V)=\frac{\mathrm{var}(V)}{1-\beta^2}$$ and that, for every $j\geqslant0$, $$X_i=\beta^jX_{i-j}+\sum_{n=1}^j\beta^{n-1}V_{i-n} $$ hence $$ ...


2

Yes, your approach is correct. Your calculation simplifies (a bit) if you use the symmetry, i.e. the fact that $$\text{var} \left( \sum_{i=1}^n W_{t_i} \right) = \sum_{i=1}^n \text{var}(W_{t_i}) + 2 \sum_{i=1}^n \sum_{j=1}^{i-1} \text{cov}(W_{t_i},W_{t_j}).$$ An alternative approach is the following: The vector $(W_{t_1},\ldots,W_{t_n})$ is a Gaussian ...


1

$$Y_n=1/(1+X_n)\implies P(|Y_n|\geqslant2)\leqslant P(|X_n|\geqslant\tfrac12)$$


0

I have never heard of bounded in probability before this so I googled and am going off of this pdf. Here is a screenshot of the relevant definition(I assume the bent equals sign is meant to be $Θ$?): It felt unlikely that $\frac{1}{1+X_n}$ will be converging to 0 in probability so I decided to try to prove the other case.$\newcommand{\prob}{\mathbb{P}}$ ...


0

EDIT: Sorry I don't know why I was so sloppy with capital versus lowercase letters, I intended to make capitals random variables but I was sloppy and did it inconsistently in some places. I might edit it at some point. This follows the structure of the derivation of the regular B-S equation by Thayer Watkins. I guess the risk-free interest rate is $0$? Let ...


2

For simplicity (of notation), we assume $n=2$, i.e. that $(B_s)_{s \geq 0}$ is a $2$-dimensional Brownian motion, and $t=1$. For $k,j \in \mathbb{Z}$ and $m \in \mathbb{N}$ set $$A_{k,j}^m := \left[ \frac{k}{2^m}, \frac{k+1}{2^m} \right) \times \left[ \frac{j}{2^m}, \frac{j+1}{2^m} \right)$$ and $$B_{k,j}^m := \left[ \frac{k-1}{2^m}, \frac{k+2}{2^m} ...


1

Note that $$ G(z_1,z_2)=E(z_1^{X_1-X_2}z_2^{2X_1+X_2-N})=z_2^{-N}E((z_1z_2^2)^{X_1}(z_1^{-1}z_2)^{X_2}). $$ Since $X_1$ and $X_2$ are independent, $$ G(z_1,z_2)=z_2^{-N}g_1(z_1z_2^2)g_2(z_1^{-1}z_2). $$ This does not use the constraints.


2

If $(X_1,X_2,X_3)$ is independent, $$ G(z_1,z_2)=E(z_1^{X_1-X_2}z_2^{X_1-X_3})=E((z_1z_2)^{X_1}z_1^{-X_2}z_2^{-X_3}), $$ hence $$ G(z_1,z_2)=g_1(z_1z_2)g_2(z_1^{-1})g_3(z_2^{-1}). $$


2

In general, a point process is a random variable $N$ from some probability space $(\Omega,\mathcal{F},P)$ to a space of counting measures on ${\bf R}$, say $(M,\mathcal{M})$. So each $N(\omega)$ is a measure which gives mass to points $$ \ldots < X_{-2}(\omega) < X_{-1}(\omega) < X_0(\omega) < X_1(\omega) < X_2(\omega) < \ldots $$ of ${\bf ...


0

The integral to solve is: $E$$[S_s]$=$\int_{-\infty}^{\infty}$$S_s$$exp((\mu-\frac{1}2\sigma^2)s+\sigma W_s)$$exp$$(W_s^2/2t)$$dW$$/(\sqrt(2\pi t))$ From this integral the expression $S_s$$exp((\mu-\frac{1}2\sigma^2)s$ can be factored out. The remaining expression inside the integral is $exp(\sigma W_s)$$exp$$(W_s^2/2t)$=$exp(\sigma W_s+$$W_s^2/2t)$ This ...


1

From Wikipedia: For any disjoint bounded subsets $B_1,\ldots,B_n$ and non-negative integers $k_1,\ldots,k_n$ we have that $$\Pr[\xi(B_i) = k_i, 1 \leq i \leq n] = \prod_i e^{-\lambda \|B_i\|}\frac{(\lambda \|B_i\|)^{k_i}}{k_i!}.$$ The constant $\lambda$ is called the intensity of the Poisson point process. Note that the Poisson point process is ...


7

I will complement saz's answerby proving that $L$ (to use saz's terminology) is a random variable. EDIT: A simpler and more powerful proof was given later by saz here (see his remark at the end). Introduction & proof outline 0 Following is a proof that the measure of the path of an $n$-dimensional ($n \geq 1$) Brownian motion is itself a measurable ...


0

If $\mu$ is predictable and Lebesgue integrable, this converges to $0$. This is an Ito process with $\sigma =0$. Quadratic Variation of Ito process


1

If you start in j, then you automatically get one count of being in the target state. Now pick your next transition k. Do a weighted average of the expected number of visits from k to j, where you weigh by the probability of going to a particular k first. Let $S_{j}$ be a random variable that counts the number of times one visits state $j$. $$ S_{j} = ...


1

Suppose that $$dX_t = - \mu X_t \, dt + \sigma \, dW_t \tag{1}$$ and set $Y_t := e^{\mu t} X_t$. Itô's formula states that $$df(t,X_t)= \frac{\partial}{\partial x} f(t,X_t) \, dX_t + \frac{\partial}{\partial t} f(t,X_t) \, dt + \frac{1}{2} \frac{\partial^2}{\partial x^2} f(t,X_t) \, d\langle X \rangle_t. \tag{2}$$ Here, we have $f(t,x) = e^{\mu t} \cdot ...


0

Is this the way the system was written in your source or was it something like this for for example the second equation? $$\mathrm{d}r_2(t) = \omega r_1 \mathrm{d}t - \epsilon \mathrm{d}B(t) r_3(t)$$ Or I guess sometimes equivalently (although less correctly) above could be written as $$\dot{r}_2(t) = \omega r_1(t) - \epsilon \dot{B}(t)r_3(t)$$ The ...


2

Since $X_t$ is $N(0,t)$ for every $t$, one knows that $X_0=0$. Since the distribution of $X_t-X_s$ depends only on $t-s$, one knows that, for every $n$, $X_{n\Delta}-X_{(n-1)\Delta}$ is distributed like $X_{\Delta}-X_0=X_\Delta$, which is $N(0,\Delta)$. In particular, $X_{n\Delta}-X_{(n-1)\Delta}$ is centered with variance $\Delta$, QED.


2

$$E[ (W_{n\Delta} - W_{(n-1)\Delta})^2] = E[W_{n\Delta}^2] - 2 E[W_{n\Delta}W_{(n-1)\Delta}] + E[W_{(n-1)\Delta}^2]$$ Use what we know about the second moment of $W_t$, and rewrite the second term. $$ =n \Delta - 2 E\left[\left(W_{(n-1)\Delta} + (W_{n\Delta}-W_{(n-1)\Delta})\right)W_{(n-1)\Delta}\right] + (n-1)\Delta$$ Expand second term and use ...


4

If $(X_t)_{t \geq 0}$ is a process with independent and stationary increments and cadlag paths, then it is stochastically continuous. Indeed: Let $s<t$, then by the stationarity of the increments $$\mathbb{P}(|X_t-X_s|> \varepsilon) = \mathbb{P}(|X_{t-s}|>\varepsilon)$$ for any $\varepsilon>0$. Since $(X_t)_{t \geq 0}$ has cadlag sample paths ...


0

By your definition $h(t,b) = f(t)b$ . Just plug right in. Basically the $X_0$ term goes away because $B_0 =0$ and the second derivative term goes away because h is linear in b. Then it's just a matter of isolating the Ito integral on the left hand side.


0

For the second part: E[X] = E[n/Nk] = n/E[Nk] <-Property of expected value operator E[Nk] = n * Pr(Ai = k) = n * 1/n = 1 recall: E[X] = n/E[Nk] = n/1 = n Maybe I made some logical mistake though because andre has a really high reputation and our answers are different.


1

Let's say you're in the 211 state. Based on your assumptions the two singular opinion holders are twice as likely to switch. So two thirds of the time one of those guys switches. If they switch then you're equally likely to go to 310 or 220. So $$p_{211 \to 310} = p_{211 \to 220} = (2/3)(1/2)$$ If one of the other guys switches you have to go back to 211 so ...


3

Please note that this does not hold. @Stephen Montgomery-Smith already mentioned the "trivial" counterexample $$X_t := X, \qquad t \geq 0$$ for some $X \notin L^2$. Another counterexample is a one-dimensional Brownian motion $(X_t)_{t \geq 0}$; it is a martingale with continuous sample paths and satisfies $\mathbb{E}(X_t^2)=t$, hence $$\sup_{t \geq 0} ...


1

Here is to get you started. Let $N_t$ denote the number of X components used between times $0$ and $t$. The number of Y components used between times $0$ and $t$ is $N_t$ or $N_t-1$ or $N_t+1$. When $t\to\infty$, $N_t\to\infty$. By the strong law of large numbers, the total length of the $n$ first X components is $nE[L_X]+o(n)$, where $L_X$ denotes the ...


1

For every $a_1$ and $a_2$, $$ Y(t)=a_1X_1(t)+a_2X_2(t) $$ is $$ Y(t)=\mu t+Z(t), $$ where $$ \mu=a_1\mu_1+a_2\mu_2,\qquad Z(t)=a_1\sigma_1W_1(t)+a_2\sigma_2W_2(t). $$ (This is algebra.) Furthermore, if $W_1$ and $W_2$ are two independent standard Brownian motions, then $$ Z(t)=\sigma W(t) $$ for some standard Brownian motion $W$ and $$ ...


1

Is it obvious to you that the waiting time generated by this procedure will still be memoryless? The exponential distribution is the only continuous distribution that is memoryless. So we can completely characterize the waiting time by just computing it's mean. $$\mu = \left(\frac{\nu_i}{\nu}\right)\frac{1}{\nu} + \left(1 - ...


2

Because $Y_{t-k}$ is measurable with respect to the family $(\epsilon_{t-k-j})_{j\geqslant0}$, which is independent of $\epsilon_t$ since $\{t\}\cap\{t-j-k\mid j\geqslant0\}=\varnothing$, and because this implies that $Y_{t-k}$ and $\epsilon_t$ are independent.


1

Define $Y = \sup_{t\ge0} X_t$. We'll prove $P(Y> b)=a/b$ for $0\le a < b$. If we can prove this, it will follow by a continuity argument that $P(Y\ge b)=a/b$ as well. To do this, we first rewrite the probability in terms of a stopping time. Define $$ T = \inf\{t\ge0\mid X_t >b\}. $$ The variable $T$ is a stopping time. If $T<\infty$, then ...


1

As explained in the comments, $p_0(t)\mathrm e^{m(t)}$ does not depend on $t$ and $p_0(0)=1$ hence $p_0(t)=\mathrm e^{-m(t)}$ for every $t\geqslant0$. Since $p_0(t)=P(T_1\gt t)$, differentiating the function $p_0$, one sees that the density $f_1$ of $T_1$ is such that, for every $t$, $$ f_1(t)=\lambda(t)\cdot\mathrm e^{-m(t)}\cdot\mathbf 1_{t\gt0}. $$ More ...


1

Both answers are found in the same way. Let's denote the number of events in the first t hours(resp. in the last t hours) by $N_t$, that in the first $T$ hours by $N_T$. Then we know that $N_t$ is a Poisson variable with parameter $\lambda t$ and $N_T$ with $\lambda T$. What's more $N_T - N_t$ is Poisson with parameter $\lambda (T-t)$ and it is independent ...


3

No. Try $X_t=S_t-B_t$, where $S_t=\sup\limits_{0\leqslant s\leqslant t}B_s$. Another class of examples: $X_t=\displaystyle\int_0^tg(s)\,\mathrm dB_s$.


1

To make your tutor happy two things. Firstly, $u$ has no time dependence so the derivative of $u$ wrt $t$ is zero. Also with $f(z)=a(z)+ib(z)$ you have \begin{equation} \frac{1}{2}\frac{\partial^{2}f}{\partial z^{2}}d(Z_{t})^{2}+\frac{1}{2}\frac{\partial^{2}f}{\partial \bar{z}^{2}}d(\bar{Z_{t}})^{2}= \frac{1}{2}\left( \frac{\partial^{2}a}{\partial ...


6

Unfortunately, I'm not aware of an alternative proof (or a more detailed version of the proof by Mörters/Peres). So, instead of providing you with a source, I'll follow the lines of Paul Lévy; hopefully filling the gaps in his proof. Let $(B(t))_{t \geq 0}=((B^1(t),B^2(t)))_{t \geq 0}$ be a two-dimensional Brownian motion and $L := \lambda(B([0,1]))$. ...


1

Let $(X_t)_{t \geq 0}$ and $(Y_t)_{t \geq 0}$ simple predictable processes, i.e. $$\begin{align*} X_t &= 1_{\{t=0\}} A_0 + \sum_{k=1}^m 1_{\{S_k<t \leq T_k\}} A_k \\ Y_t &= 1_{\{t=0\}} B_0 + \sum_{j=1}^n 1_{\{U_k<t \leq V_k\}} B_k \end{align*}$$ where $S_k<T_k$, $U_k<V_k$ are stopping times and $A_k$ are $\mathcal{F}_{S_k}$, $B_k$ are ...



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