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0

There must have been $3$, $4$, $5$, or $6$ clients in five minutes, for otherwise the total deposited could not be $6000$. The number of clients has Poisson distribution with parameter $5$, so the probability of each of $3$, $4$, $5$, and $6$ are easily calculated. Call the probability there were $k$ clients $p(k)$. For $3$ clients, all must have ...


0

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


0

Strictly speaking, in probability theory you are free to excise null sets from the probability space. So you could define Brownian motion with a.s. continuous paths and then excise the null set where the paths are not continuous. Now the paths are always continuous. The process we've made is actually indistinguishable from the a.s. continuous process: their ...


0

***Interpreting you problem to mean that the fishermen don't leave until each has caught $3$ fish let $T_i$ be the time until fisherman $i$ catches $3$ fish. $$\Bbb E[T] = \sum_{i=1}^3\Bbb E[T_i] = 3\Bbb E[T_i]$$ by symmetry. Additionally, the time between catches is exponential with rate $2.5$ so, we exploit the independent increments property to solve ...


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Just seen this - probably too late, but if not..: re. (1): You are in $t$, i.e. $t$ is now, and $V_t$ and $S_t$ are real market prices observable right in this moment ($t$). re (2): same answer as to (1). $S_t$ and $V_t$ are here not symbols for random variables, but current realizations which you see now (in time $t$). Of course, you don't know future ...


0

In general, we cannot expect this. Just consider e.g. a Brownian motion $(M_t)_{t \geq 0}$ and the stopping time $$\tau := \inf\{t>0; M_t = 1\}.$$ Then $$\mathbb{E}(f(\tau) M_{\tau}) = \mathbb{E}f(\tau)=0 \iff f=0,$$ i.e. the claim holds only true for the (trivial) function $f:=0$.


1

It is probably marketed as an entry point into financial mathematics, since people in finance have realized that engineers and mathematicians can help them make computer programs to make money. However, lots of other fields exist: Signal processing, Image processing, Machine Learning, "Big Data" are just the ones I can think of straight away.


1

Mathematical biology (ranging from population dynamics to newer systems biology) makes wide use of both deterministic and stochastic models. See, for instance, http://www.math.wisc.edu/~anderson/RecentTalks/2010/CIBM.pdf by David Anderson for some readable slides.


1

Well, when it's known what is constant and what it's not, the key to solving this equation is observation that it's first order inhomogeneous linear equation: $$ \frac{d\omega}{dt} + c \left (\frac{be^{-bt}}{ae^{-bt}-C_1} \right) \cdot \omega = f(t) $$ General solution of this ODE is a linear combination of any solution of it and general solution of ...


1

By definition, $$U_t(X)-U_t(Y) = (X_0-Y_0) + \int_0^t (b(s,X_s)-b(s,Y_s)) \, ds + \int_0^t (\sigma(s,X_s)-\sigma(s,Y_s)) \, dW_s$$ and therefore $$\begin{align*}&|U_t(X)-U_t(Y)|^2 \leq \\ & \color{red}{3 |X_0-Y_0|^2} + 3 \left( \int_0^t |b(s,X_s)-b(s,Y_s)|ds\right)^2 + 3\left(\int_0^t |\sigma(s,X_s)-\sigma(s,Y_s)|dW_s\right)^2. \end{align*}$$ The ...


2

Let $E^I$ denote the set of all functions from $I$ to $E$. Note that $X$ can be viewed as a mapping from $\Omega$ into $E^I$, with $[X(\omega)](t) = X_t(\omega)$. Also, $X$ is measurable with respect to the $\sigma$-algebra $\mathcal{F}$ on $E^I$ which is generated by all of the projections $\pi_t: E^I\to E$, given by $\pi_t(f) = f(t)$. Hence there is a ...


0

The recurrent classes do not affect each other. Once we get into one recurrent class, we never leave, and the structure of the other recurrent class is irrelevant. Just take an example, say with 4 states and transition probability matrix $(P_{ij})$ given by: $$ (P_{ij}) = \left[ \begin{array}{cccc} 0 & 1/2 & 1/2 & ...


1

This is actually a necessary and sufficient condition (in the case of the joint generating function). Let $X$ and $Y$ have joint generating function $$\phi_{X,Y}(s,t) = \sum_{i,j=0}^\infty \mathbb P(X=i,Y=j)s^i t^j.$$ The product of the marginal generating functions is $$ \begin{align*} \phi_X(s)\phi_Y(t) &= \left(\sum_{i=0}^\infty\mathbb ...


0

Yes, if you fix $t$, $c1_{\{X_t<U\}}$, is a valid Radon-Nikodym derivative in the sense that you can define $\mathbb{Q}$ through the formula $$\mathbb{Q}(A)=\int_Ac1_{\{X_t<U\}}d\mathbb{P}=c\mathbb{P}(A\cap\{X_t<U\}).$$ That is, $\mathbb{Q}$ is defined to be restriction of $\mathbb{P}$ to $\{X_t<U\}$ (times the constant $c$). Now, $\mathbb{Q}$ ...


2

Determining the stable (long term) state probabilities The solutions of the following system of equations are the stationary state probabilities: $$\begin{matrix}[\pi_A\ \pi_B \ \pi_C]\\ \\ \\ \end{matrix} \begin{bmatrix} 0.2 & 0.4 & 0.4 \\ 0.4 & 0.3 & 0.3 \\ 0.6 & 0.2 & 0.2 \\ \end{bmatrix} ...


2

The transition matrix must take into account not just the current position (that is to say, whether she is at $a$ or $b$ right now), but also the position she was at immediately before. This is because in order to know what the probability of switching positions is going to be, we must know both her current position and the position immediately preceding ...


1

The answer you provided can't obey the condition that being in a state for two consecutive steps requires a different probability from when you are only in a state for one step. To describe this situation, you need to set conditional probabilities which depend on the two previous states. If the last two states were $[\dots aa]$ you transfer to $[\dots aab]$ ...


1

First a remark concerning the definition of $X^{(n)}$: Instead of the closed interval $A_k^{(n)} := [t_{k-1}^{(n)},t_k^{(n)}]$ you should use the half-open interval $A_k^{(n)} := (t_{k-1}^{(n)},t_k^{(n)}]$ (otherwise you might run into trouble at the boundary points $\{t_k^{(n)}\}$). Then $$X^{(n)}(s,\omega) := X_0(\omega) 1_{\{0\}}(s) + \sum_{k=1}^{2^n} ...


2

It holds that $$\mathbb{E}^x(F) = \int F(w) \, d\mathbb{P}_x(w) = \int F(x+w) \, d\mathbb{P}(w) \tag{1}$$ for any measurable function $F: (C[0,\infty),\mathcal{B}(C[0,\infty)) \to [0,\infty)$. This follows from the fact that $(1)$ holds for simple functions and we can extend the equality to all measurable non-negative functions (by the monotone convergence ...


1

Integrating any locally bounded predictable process (in particular a continuous adapted process) with respect to a local martingale (resp locally square integrable martingale) results in a new process which is a local martingale (resp locally square integrable). A proof of this can be found in Jacod & Shirjaev (Chapter 1; Theorem 4.31). (It basically ...


0

There are three exponential time 'contests', and each is independent of the others. First, A must arrive before B. Second (no memory), B must arrive before A finishes being served. Third (no memory), B must finish service before A finishes service. Thus, the desired probability is the product of three ratios: $$\frac{\lambda_a}{\lambda_a + \lambda_b} ...


1

First of all, note that the strong Markov property of Brownian motion does not state that $(B_t)_{t \in [0,\tau]}$ and $(B_{t+\tau})_{t \geq 0}$ are independent, but that $(B_t)_{t \in [0,\tau]}$ and $(B_{t+\tau}-B_{\tau})_{t \geq 0}$ are independent. This independence yields in fact $$\mathbb{E}(1_A \cdot 1_B) = \mathbb{E}(1_A) \mathbb{E}(1_B)$$ for any ...


6

From the stochastic integral analogue of integration by part, which is a corollary of Ito's Lemma, $$d(W_t\ln t) = \frac{W_t}{t}dt+(\ln t)dW_t,$$ so $$I:=\int_0^t \frac{W_s}{s}ds=W_t\ln t-\int_0^t (\ln s)dW_s=\int_0^t \ln\frac{t}{s}dW_s$$ as $W_t\ln t\to 0$ in probability, as $t\to 0^+$. Since the random variable $I$ is a linear combination of independent ...


-1

Here's my take on this, Let $S_t(s) = \int_0^t \mu(t,S_t,s)dt + \int_0^t \sigma(t,S_t,s)dW_t$ Then, $S_t(s)$ is a normally distributed random variable parametrized by it's expectation and variance. To see that $S_t(s)$ is normal, $S_t(s) = \int_0^t \mu(t,S_t,s)dt + \int_0^t \sigma(t,S_t,s)dW_t$, which is equal to a constant $\int_0^t \mu(t,S_t,s)dt = ...


0

Since the process is irreducible, we can assume without loss of generality that $X_0=0$, and it suffices to show that $\mathbb P(N_0<\infty)<1$, where $$N_0=\inf\{n>0: X_n=0\}$$ is the time until the first return to $0$. Let $$F(s) = \mathbb E\left[s^{N_0}\right]$$ be the generating function of $N_0$. It can be shown through some computation that ...


0

pi P = pi , and pi_0+pi_1+ ... +pi_n=1 where P is your matrix and pi=(pi_0 pi_1 ... pi_n) Here's what I started off with pi_0=kM(pi_1) pi_1=nL(pi_0)+kM(pi_2) pi_2=(n-1)L(pi_1)+kM(pi_3) ... pi_0+pi_1+ ... +pi_n=(pi_0)(1+1/kM{1+(n-1)/kM-(n-1)nL-(n-2)(n-1)L-...2*3L-1*2L})=1 Hopefully I didn't make any mistakes there with the simplification. Sorry ...


2

Sure assume $dy = \mu dt + \sigma dZ$ is an Ito process. $f(y) = e^y$ Normally by the chain rule, $df = \frac {\partial f}{\partial t}dt + \frac {\partial f}{\partial y}dy$ However, because $(dy)^2$ ~ $O(dt)$ we cannot neglect $(dy)^2$ terms. Taylor expand $df$ to order $2$ in $dy$ $\Rightarrow df = \frac {\partial f}{\partial t}dt + \frac {\partial ...


2

Recall Ito's formula, written in differential form, $$ df(X) = f'(X)dX + \frac12f''(X)d\langle X\rangle. $$ Assuming $w(t)$ is a semimartingale, since $y(x):=e^x$ is a $C^2$ function, we can compute $$ dy(w(t)) = e^{w(t)}dw(t) + \frac12e^{w(t)}d\langle w \rangle(t). $$ If by $w(t)$ you meant a brownian motion, then the quadratic variation term is just ...


0

The sigma algebra $F_t$ is generated by the variables $X_s,x≤t$. As a sigma algebra, you can consider it generated by the elements $$ \sum_{i=1}^p a_i X_{t_i}, p\in\Bbb N_{>0}, a_i\in\Bbb R, t_i\le t $$ Hence to prove the equation $$ \forall G\in F_t \ \ \forall f\ \ \ \ E(G\times f(X_{t+h} - X_t)) = EG\times E f(X_{t+h} - X_t)) $$ you can consider ...


1

If $$S(t) = X_1 + X_2 + \cdots + X_{N(t)}$$ is the aggregate claims and $N(t)$ is the claim count at time $t$, then $$\operatorname{E}[S(t)] = \operatorname{E}[\operatorname{E}[S(t) \mid N(t)]] = \operatorname{E}[N(t) \operatorname{E}[X_i]] = \operatorname{E}[X_i]\operatorname{E}[N(t)].$$ Since a single claim $X_i$ has expected size $\operatorname{E}[X_i] = ...


1

You're intuition's dead on with using Conditional variance. What you're looking for is right here: http://www.columbia.edu/~ww2040/3106F14/lec1023.pdf


0

The characteristic function of $(W_{t_1}^n,\dots,W_{t_d}^n)$ can be computed by the following date: the mean of this vector, and the covariance matrix. The convergence of $W_t^n\to W_t$ in $L^2$ shows that $\mathbb E[W_{t_i}^n]\to \mathbb E[W_{t_i}]$ and that $$\lim_{n\to \infty}\operatorname{Cov}\left(W_{t_i}^n,W_{t_j}^n ...


1

Hi it appears to me that you have done the hardest part. To finish the job you only have to apply of the Continuous Mapping Theorem. So you have already proven that $(W_t,Z_t)\to (X,Y)$ almost surely where $W_t =\frac{X_{Y_t}}{Y_t^p}$ and $Z_t=\frac{Y_t^p}{t}$ Now take $g(w,z)=w.z$ and apply Continuous Mapping Theorem to obtain almost sure convergence of ...


1

No, your proof is not correct There does in general not exist $\omega_t$ such that the equality $$\mathbb{E}(f(X_t)^2) = |f(X(t,\omega_t))| |\Omega|$$ holds. Hints: Use the estimate $$\begin{align*} |f(X_t)|^2 &= |(f(X_t)-f(0))+f(0)|^2 \\ &\leq 2 |f(X_t)-f(0)|^2 + 2 |f(0)|^2 \\ &\leq 2 C^2 |X_t-0|^2 + 2 |f(0)|^2 \end{align*}$$ to obtain ...


1

Look at $P(N(t)-N(s)>0) \sim \mathrm{Poisson}(\lambda(t-s))$ which is related to the independent increments property. if $M \sim \mathrm{Poisson}(\lambda(t-s))$, $P(M>0) = 1-P(M=0) = 1-e^{\lambda(t-s)}$


0

In your system of PDEs in the term: $ \frac{1}{2} b^{2} \frac{\partial^{2}h}{\partial w^{2}} (1-h^{2}) $ Don't you have an extra $b^{2}$ there?


0

Hint: Write $$X_t^2 = \bigg( \big[ (N_t-N_s)-(t-s) \big]+ \big[ N_s-s \big] \bigg)^2$$ for $s \leq t$ and expand the square. In order to calculate the conditional expectation $\mathbb{E}(X_t^2 \mid \mathcal{F}_s)$, consider the terms separately and use that $(N_t-N_s)-(t-s)$ is independent of $\mathcal{F}_s$ $N_t-N_s \sim N_{t-s}$


0

Since increments belonging to disjoint time intervals are independent, and assuming $s<t$, $$ E[X_s·X_t]=E[X_s^2]+E[X_s]·E[X_t-X_s] $$ and the expectation of the geometric Brownian motion is $E[X_s]=e^{μ·s}$. The square can be written as $$ X_s^2=\exp[(2μ−σ^2)·s+2σ·W_s]=\exp[((2μ+σ^2)−0.5·(2σ)^2)·s+2σ·W_s] $$ thus is also a geometric Brownian motion and ...


3

To show U.I., \begin{align} &\mathbb{E}[\exp(M_s);\exp(M_s)\geq K] && \\ &\leq \mathbb{E}[\mathbb{E}[\exp(M_\infty)|\mathcal{F}_s]; \exp(M_s)\geq K] && \text{(submartingale property})\\ & \leq\mathbb{E}[\exp(M_\infty); \exp(M_s)\geq K] \end{align} Hence you need only to show that $\sup_{0\leq s\leq ...


1

$\mathbb{E}\left(X_{1}+\cdots+X_{N}\mid N\right)=\mu N$ so that: $$\mathbb{E}\left(X_{1}+\cdots+X_{N}\right)=\mu\mathbb{E}N$$ $\mathbb{E}\left(\left(X_{1}+\cdots+X_{N}\right)^{2}\mid N\right)=\sigma^{2}N+\mu^{2}N^2$ so that: $$\mathbb{E}\left(X_{1}+\cdots+X_{N}\right)^{2}=\sigma^{2}\mathbb{E}N+\mu^{2}\mathbb{E}N^{2}$$ Working ...


1

Consider the application $$ F(\xi) : F(\xi)(T)=\xi_0+\int_0^Tf(\xi(t))dt+W(T) $$ $$ \left[ F(\xi) - F(\zeta)\right] (T) = \int_0^T \left[f(\xi(t)) - f(\zeta(t))\right] dt $$ under the same assumptions as in the Cauchy Lipschitz theorem, you can prove that (for the right space and right norm) $F$ is Lipschitz with constant $<1$ (the proof is exactly ...


0

It can be shown that the integral is a time-changed Brownian Motion and from there we can deduce the distribution from looking at this BM. Define $f(u) = \sigma(T-u)$ and $X_t = \int^t_0 f(u) dW_u$. Define the stopping time $$ \tau_t = \inf\lbrace u \geq0 : [X]_t > t \rbrace $$ and look at the process $B_t := X_{\tau_t}$. We see that by definition of the ...


1

Step 1: The assumption holds for all $f \in C_b^2$, i.e. functions which are twice differentiable and have bounded continuous derivatives. To this end, pick a function $\chi \in C^2_b$ such that $0 \leq \chi \leq 1$, $\chi(x) = 1$ for $x \in (-1,1)$ and $\chi(x)=0$ for all $|x|>2$. Then the function $$f_n(x) := f(x) \cdot \chi \left( \frac{x}{n} ...


1

The step you are asking about is just algebra: $$W(\sigma)=1-\int_0^\sigma e^{W(s)} 1_{|W(s)| \leq 1} ds$$ Substitute that into the definition of $\tilde{W}$: $$\tilde{W}(\sigma)=1-\int_0^\sigma e^{W(s)} 1_{\{ |W(s)| \leq 1\}} ds + \int_0^\sigma e^{W(s)} 1_{\{ |W(s)| \leq 1 \}} ds = 1.$$


2

Do you know what a $\sigma$-algebra is? It is a collection of subsets of the sample space $\Omega$. In particular, it is a collection of subsets that a) contains $\Omega$ as an element, b) is closed under set complement, and c) is closed under countable unions. The purpose of a $\sigma$-algebra is to act as the domain of a measure $P$. So the measure ...


1

A random varible is defined on a probability space which consists of a sample space $\Omega$, a $\sigma$-algebra $\mathcal{F}$ on $\Omega$ and a measure $P:\mathcal{F}\rightarrow [0,1]$. In this case, $X$ is a random variable if for ever $x\in\mathbb{R}$, $[X\le x]\in \mathcal{F}$. To wit, $X$ is a measurable function on the measure space $(\Omega, ...


0

A martingale is characterized by its characteristic function. apply the property with $f(x) = x$ proves that $W$ is a martingale fix $u$ and apply with $f(x) = \exp(iux)$; define $g(u, t) = \mathrm E(\exp(iuW_t)); M(t) = f(W_t) - \frac 12\int_0^t f''(W_s) ds$ and you get $$ 1 = M(0) = \mathrm EM(t) = g(u,t) + \frac 12 u^2\int_o^t g(u, s) ds $$ whose ...


2

Since $\mathbb{E}W_{\tau}=0$ and $W_{\tau} \in \{1,-1\}$, we have $$0 = \mathbb{E}(W_{\tau}) = \mathbb{E}(1 \cdot 1_{\{W_{\tau}=1\}} + (-1) \cdot 1_{\{W_{\tau}=-1\}}) = \mathbb{P}(W_{\tau}=1) - \mathbb{P}(W_{\tau}=-1).$$ Hence, $$\mathbb{P}(W_{\tau}=1) = \mathbb{P}(W_{\tau}=-1) = \frac{1}{2}.$$ This implies $$\mathbb{E}f(W_{\tau}) = \frac{1}{2} ...


0

Hints: (a) Conditioning on $Y$ will simplify the problem: \begin{eqnarray*} E(X(t)) &=& E(X(t)\mid Y=1)P(Y=1) + E(X(t)\mid Y=-1)P(Y=-1) \\ &=& \ldots \end{eqnarray*} (b) \begin{eqnarray*} Cov(X(s),X(t)) &=& E(X(s)X(t)) - E(X(s))E(X(t)). \end{eqnarray*} Use part (a) to evaluate the second term. For the first term, note that ...


1

Hints: Let $\arg\min_{{b_0,b_{-0}}}E\left[(X_{n+1}-b_0-b_{-0}'X)^2\right]=(\beta_0,\beta_{-0}')'=\beta\in\mathbb{R}^{n+1}, \quad X=(X_1,\dots,X_n)'.$ $\beta_0=E[X_{n+1}]-\beta_{-0}'E[X], \quad \beta_{-0}=Var(X)^{-1}Cov(X,X_{n+1}).$ $Cov(X, X_{n+1}-\beta_0-\beta_{-0}'X)=0.$ Use normality of $(X_1,\dots,X_{n+1})'.$ Show that ...



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