# Tag Info

1

Fix $\epsilon>0$. Since $(X_t)_{t \geq 0}$ satisfies the SLLN, there exists for almost all $\omega \in \Omega$ a constant $S=S(\omega)>0$ such that $$\left| \frac{X_s(\omega)}{s} - \mathbb{E}(X_1) \right| \leq \epsilon \quad \text{for all s \geq S}.$$ As $U_t \uparrow \infty$, we have $U_t(\omega) \geq S$ for $t \geq T=T(\omega)$ sufficiently ...

1

If there is someone at your university or in your area who is an applied statistician, you should begin by having some discussions on these topics with him or her. Someone who knows you personally can give you a level of advice you can't get here. I'm not sure what you might gain from participating in competitions because they are intentionally beyond the ...

0

What the book it hinting at is the knowledge that if $$X_t=X_0+\int_0^t X_s\,ds +B_t,$$ then $X$ is a continuous semimartingale and $$\int_0^t e^{-s}\,dX_s=\int_0^t e^{-s}X_s\,ds+\int_0^t e^{-s}\,dB_s.$$ The differential form of an SDE is more than just a way to save ink (or electrons); it can (as here) serve as a guide to deducing one integral ...

0

Edit: Maybe I should flesh out my answer a bit. Recall previous discussion on the matter. \begin{align*} E[Z] &= \sum_{n=0}^\infty E[Z|N = n]P(N=n)\\ &=\sum_{n=0}^\infty E[X_1+\dotsb+X_{n+1}|N=n]P(N=n)\\ &=\sum_{n=0}^\infty (n+1)E[X_1|N_n]P(N=n)\\ &=\frac{1}{2}\left[\sum_{n=0}^\infty nP(N = n)+\sum_{n=0}^\infty P(N=n)\right]\\ ...

0

Convert $A_1$ to $A$ and $A_2$ to $B$ for ease of use. As I understand the question, $A$ has drawn a single black ball and $B$ has drawn no black balls. The question asks what is the probability that $B$ will eventually draw a black ball. Using Ross's notation, $AAA$ is the only failure condition. $AAB, ABA, ABB$ are the possible win conditions. This ...

0

You can use conditional probability here. P(A2(B) given A1(B) is = P(A2BintersectionA1B)/P(A1B). From here you know P(A1B)=3/1000. This would simply come out to be 2/999. Do you have a way of checking this answer is correct or not?

0

Imagine each sequence of the balls laid out in a separate line. Since balls have no preference for positions, each black ball is equally likely to occupy odd/even positions. $A_2$ wins if they occupy (in order) odd-even-odd, odd-odd-even, or odd-even-even positions, and loses if they occupy odd-odd-odd positions Thus $P(A_2\text{wins}) = \dfrac34$ The ...

1

$$X_{n+1} = X_n + T_{n+1}$$ where $T_{n+1}$ is independent of $X_1,\ldots,X_n$ and $$\Pr(T_{n+1}\in A) = \int_A \frac{dt}{(2+t^2)^{3/2}}$$ for every measurable set $A$. (Thus $T_{n+1}$ has a t-distribution with $2$ degrees of freedom.)

5

Neither of them holds, in general, for stochastic integrals. The trouble already starts if you consider measures which need not to be non-negative, i.e. signed measures. For a signed measure $\mu: (\Omega,\mathcal{A}) \to \mathbb{R}$ we cannot expect that the triangle inequality $$\left| \int f(x) \, \mu(dx) \right| \leq \int |f(x)| \, \mu(dx) \tag{1}$$ ...

2

For an approximation, which is rather good because there are a lot of white balls, each black ball when drawn has $\frac 12$ chance to go to either player. The winning sequences are $AAB, ABB, ABA, BAB$, so the chance of your event is $\frac 12$. The correct value will be a little higher because $A$ is more likely than $B$ to get the first black ball.

1

So the thing is that what you are looking for is exactly the content of lemma 3. I report the claim here : Lemma 3 : Let $X$ be a càdlàg square integrable martingale and ${\xi}$ be a bounded predictable process. Then, $\int\xi\,dX$ is a square integrable martingale. In your case $W$ is a Brownian motion, so you fit the conditions for the lemma as a ...

0

Let $X_t = \int_0^t W_s^3\,dW_s$. Note that $[W,X]_t = \int_0^t W_s^3\,ds$ and $[X,X]_t = \int_0^t W_s^6\,ds$. Also define $N_t := W_t\sin(X_t)$. We will apply the two dimensional version of Ito's lemma to the process $N$. \begin{align}N_t = N_0 &+ \int_0^t \sin(X_s)\,dW_s + \int_0^t W_s\cos(X_s)\,dX_s\\ &-\frac{1}{2}\int_0^t W_s\sin(X_s)\,d[X,X]_s ...

0

Within the "usual" terminology every Doob martingale is a martingale but not vice versa. The Doob martingales are exactly the uniformly integrable martingales. Here for an integrable r.v. $X$ and a filtration $(\mathcal{F}_t)_{t\in T}$ the corresponding Doob martingale $D_t$ is defined via $D_t:=E(X | \mathcal{F}_t), t\in T$. A martingale (stochastic ...

1

Firstly, exponential distributions are continuous, which means you're dealing with probability densities rather than masses. $$f_X(x)=\lambda~\mathsf e^{-\lambda x}~\mathbf 1_{x\in(0;\infty)}\textrm{ and }f_Y(y)\textrm{ likewise.}$$ Secondly $Z=X+Y$ so $f_{X, Z}(x, z) = f_X(x)~f_Y(z-x)$ and $f_Z(z) = \int_\Bbb R f_X(x)~f_Y(z-x)\operatorname d x$ Thusly: ...

1

$E[X|\{ \emptyset, \Omega \}]$ is defined as the a.s. unique random variable in $\sigma(\{ \emptyset, \Omega \})$ such that for any set $A\in\sigma(\{ \emptyset, \Omega \})$, $E[E[X|\{ \emptyset, \Omega \}];A]=E[X;A]$. $\sigma(\{ \emptyset, \Omega \})$ is just $\{ \emptyset, \Omega \}$ and any r.v. measurable wrt this collection is a.s. constant. Take ...

1

First, it is not the same as the probability of having $k/0.65$ customers walk into the shop. For example, taking $k = 5$, it is really unlikely that exactly $2/0.65 \approx 7.692$ customers will walk into the shop. (Well, maybe less unlikely if the shop is located in King's Landing...) What you have is a rate of customers walking in and buying coffee of ...

1

Hints: 1) Law of total probability, in particular, the second formula. Note that since $Y$ is continuous, you will use integrals instead of sums. 2 and 3): Use what you get from #1. $N$ is poisson distributed, so use $$\mathbb{E}[g(N)] = \sum_{n=0}^{\infty}g(n)f_N(n)\text{.}$$

0

If the probability of the $j$-th map is $p_j$, and $X_j$ the variable which counts how many times the map $p_j$ appears in $N$ rounds, then $X_j \sim Bi(N,p_j)$ is a binomial distribution; the mean is $$E[X_j] = Np_j$$ Is this what you're looking for?

0

Maye does this work : We have $Y_t=\int_0^tY_sf(s)dW_s$ where W is a brownian motion, so $Y_t$ is a Wiener(ito) integral and thus it is a martingale

1

If you mean that $L = M$, then $$\operatorname{E}[Z] = \operatorname{E}[\operatorname{E}[Z \mid M]] = \operatorname{E}\left[\left. \sum_{k=1}^M \operatorname{E}[X_k] \;\right| M \right] = \operatorname{E}[M \mu] = \mu \operatorname{E}[M] = \mu \lambda.$$ The second moment is more difficult, since $\operatorname{E}[Z^2]$ is messy to deal with using the above ...

0

How about a process with only two paths: $X^{(1)}_t$ and $X^{(2)}_t$, each has probability of 1/2, and $X^{(1)} = 0$ at $t=0$, $X^{(2)} = 1$ at $t=0$, and both are $1/2$ on $(0\ , 1]$?

0

For $M$ of the form $M_t=\int_0^t f(s)\,dW_s$ with $f$ locally square integrable, the (local martingale) $K_t:=\int_0^t M_s\,dM_s$ satisfies $$E[K_t^2]=\int_0^t E[M_s^2f(s)^2]\,ds\le E[M_t^2]\int_0^t f(s)^2\,ds=\left(E[M_t^2]\right)^2<\infty,$$ and $K$ is indeed a martingale.

0

Fix a random variable $X$ (measurable at time zero if you like) which is finite but not in $L^\infty$, then define $\lambda_t = X$ for all $t$. Then $$\int_0^T \lambda_t^2\,dt = T X^2.$$ By assumption $T X^2 < \infty$ for all $T, \omega$, but there is no $C$ such that $TX^2 \leq C$ almost surely, as this would imply $X \in L^\infty$. The difference ...

0

No, in general $X_t$ is not a martingale; if $Y \in L^2_{loc}(M)$ then you can say that $X_t$ is a local martingale though, and then try to see if it's indeed a true martingale using a number of different criteria. In general it's not true though. If $Y_t = M_t$ then it's not guaranteed, either; since $$d(M_t^2/2) = M_t dM_t + \frac 12d[M]_t$$ you get that ...

2

Let the sample space be $[0,1]$ and let each element of the filtration be the borel $\sigma-algebra$. Let $I$ be some non-measurable subset and $\tau_x=\mathbb{1}(x), x\in [0,1]$. Then $\{\tau_x<t\}$ is the complement of a singleton for $x\le 1$ and $[0,1]$ otherwise so it belongs to each $F_t$, but $\{sup_{x\in I}\tau_x<t\}=\cap_{x\in I}\{\tau_x ... 1 Personally I would use a graph rather than a transition matrix. The first graph here corresponds to your transition matrix, with the blue numbers representing the number of coins to toss in the next round and the green numbers of probability of moving to the next state. But you are not interested in tosses which are all tails, i.e. do not change the ... 1 It might simplify slightly to say we just require the chain to reach state$4$at any point, starting at$0$, and we can forget about the variable$T$. So re-define$u_i$as $$u_i = P(X_n = 4 \text{ for some n}\mid X_0=i).$$ Our transition probability matrix: $$\begin{matrix} \qquad 0 \qquad & 1 \qquad & 2 \qquad & 3 \quad & 4 ... -1 Refer to this? Also$$M_{\Theta}(t) = E[\exp(t\Theta)]= E[\exp(t\lim \frac{B_n + 1}{n+2})]= E[\lim\exp(t \frac{B_n + 1}{n+2})]= \lim E[\exp(t \frac{B_n + 1}{n+2})]= \lim \frac{1}{n+1}[\exp(t \frac{1}{n+2}) + \exp(t \frac{2}{n+2}) + ... + \exp(t \frac{n+1}{n+2})]= \lim \frac{a(n)}{(n+1)(1-a(n))} (1-a(n)^{n+1}), \ \text{where} ... 1 For brevity, I'm going to say "we win" if the game stops with a sum of$5$, and "we lose" if the game stops with a sum of$7$. Let$p_k$be the probability that we win, given that the first roll is$k$. Suppose we roll a$1$and then a$2$. After the$2$, the probability that we will win is$p_2$; that is, all that matters after that roll is that we last ... 1 I'd consider$8$states, namely for$1\leq k\leq 6$the states$s_k:\ $"last roll was$k\>$, but game is not yet over", and the two end states$e_5$and$e_7$. Denote by$x(n)$the$(1\times8)$row vector giving the probabilities that after$n$rolls we are in the state$s_1$,$s_2$,$\ldots\ $,$s_6$,$e_5$,$e_7$respectively. It follows that ... 0 Well I am not sure this is true. I give you an argument to be discussed about which I'm not totally comfortable. Let's fix$T_n=T$, let's suppose it has a law$P_T(ds)$and that$T$is independent of$(M_t)_{t>0}$and finally that$M_t$positive for all$t>0$. Now for any$t>0$we have a fixed$p\in (1,\infty)$such that$M_t\in L^p$, and we ... 0 "f" is the limit$\sigma-algebra$, correct? I think your idea is right. Denote$E_n=\{N=n\}=\{\omega: \omega_1=T,\ldots,\omega_{n-1}=T,\omega_n=H\}$, writing an element of the sequence space as$\omega$. If$A$contains any element of$E_n$, it contains all of$E_n$. Any non-empty proper subset of$E_n$is of the form$E_n \cap \cap_{n\in ...

-1

(a) For any fixed $t$, we know from the condition that $X\mid_{[0,t-\frac{1}{N}]\times \Omega} \to (\mathbb R, \mathcal B(\mathbb R))$ is $\mathcal B([0 ,\ \ t-\frac{1}{N}]) \times \mathcal F_t$ measurable, therefore, for any $A\in \mathcal B(\mathbb R)$, $(X\mid_{[0,t)\times \Omega})^{-1}(A)= \mathop{\cup}_{N}\left ((X\mid_{[0,t-\frac{1}{N}]\times ... 0 I've taken a look at the book and this chapter is introductory and somewhat informal, so I imagine the authors are more specific about what they mean by a white noise in space or time and what they mean by the S(P)DE in your question in later chapters. Nevertheless, I have addressed aspects of your question below. A discussion of Definitions 2, 3 and 5 are ... 1 This is actually a big open question in Computational Complexity. There are (among many others) two big classes of problems called P and BPP. Without going into too much detail, P includes all the problems which can be efficiently solved deterministically, while BPP includes those which can be solved efficiently using a source of randomness with a small ... 1 Since $$X_s^2 = (X_{s-}+\Delta X_s)^2 = X_{s-}^2+ 2 X_{s-} \Delta X_s + (\Delta X_s)^2$$ we have $$\Delta (X_s^2) \stackrel{\text{def}}{=} X_s^2-X_{s-}^2 = 2 X_{s-} \Delta X_s + (\Delta X_s)^2.$$ Hence, $$\Delta (X_s^2) - 2 X_{s-} \Delta X_s - (\Delta X_s)^2 = 0$$ for all$s$. This means that the sum $$\sum_{s \leq t} \left( \Delta (X_s^2) - 2 X_{s-} ... 0 a) It is given that X_{|U=p} \sim Geo(p), so E[X|U=p] = \frac{1}{p} and Var[X|U=p] = \frac{1-p}{p^2}. b) Using the total expectation and total variance:$$ E[X] = E[E[X|U]] = E\left[\frac{1}{u} \right] = \int_{1/3}^1\frac{1}{u}\frac{3}{2}du = \frac{3}{2}(0 + \ln(3)) \approx 1.65 $$For the variance of X use the total variance formula, i.e., ... 0 Let me show a method for 1-d case first: dX = aX dN By Ito's formula ( In this special case, it is straightforward, I guess we don't even need Ito's formula) ln(X(t)) = ln(X(0)) + \sum_{s\in [0,t]}(ln(X_s)-ln(X_{s-})) Note that, whenever there is a jump at s, ln(X_s)-ln(X_{s-})=ln(1+a), so we have ln(X(t)) = ln(X(0)) + N(t)ln(1+a) X(t) = X(0) ... 2 If (X,Y) are jointly Gaussian with mean zero, then P(X>0,Y>0)={\arccos(-\rho)\over 2\pi} where \rho is the correlation of X and Y. Therefore, for Brownian motion and 0<s<t$$P(B_s>0, B_t>0)={\arccos(-\sqrt{s/t})\over 2\pi}.$$1 Consider the function f(x,t) = \frac {e^{r(T-t)}}{x}. Then X = f(t, S) and you can use Ito lemma to say that$$dX_t = \frac{\partial}{\partial t}f(t,x)\mid_{ t,S} dt + \frac{\partial}{\partial x}f(t,x)\mid_{ t,S} dS_t+\frac 12\frac{\partial^2}{\partial x^2}f(t,x)\mid_{ t,S} d \langle S \rangle _t$$Which results in$$dX_t = \frac ... 1 Let$S(t)$be governed by the SDE $$dS(t)=\mu S(t)dt+\sigma S(t)dW_t$$ Let$f(S)=S^n. Heuristically, we can write \begin{align} d(S^n)&=\frac{\partial f(S)}{\partial t}\,(dt)+\frac{\partial f(S)}{\partial S}\,(dS)+\frac12\frac{\partial ^2f(S)}{\partial S^2}(dS)^2\\\\ &=\frac{\partial S^n}{\partial t}\,(dt)+\frac{\partial S^n}{\partial ... 3 If the chain has an absorbing state, then the answer is no, because then P(X_0=x_0,...X_n=x_n)>0 but P(X_n=x_n,...,X_0=x_0)=0, where x_n is the absorbing state. On the other hand, suppose your markov chain can be broken down into two disjoint communicating classes, i.e. two disconnected Markov chains. Then if the individual markov chains are ... 0 (I repost here my partial answer from mathoverflow, http://mathoverflow.net/questions/229044/malliavin-derivative-under-change-of-measure/229874#229874) Here an answer for the case with determinist drift as mentioned in the edit. (Note: I fail to see why to use different notations for F and \tilde{F} as it is the same process) As dF_t = \mu_t \, dt ... 1 Using summation by parts, as suggested, it actually becomes quite clear. $$\sum_{n=1}^{N}t_n(W(t_{n+1})-W(t_n))=t_N W(t_N)-t_1W(t_1)-\sum_{n=1}^N W(t_n)(t_{n+1}-t_n)$$ Witht_N=T$and$t_1=0$. This gives: $$\int_0^TtdW(t)= TW(T)-\sum_{n=1}^N W(t_n)(t_{n+1}-t_n)$$ Taking$n \rightarrow \infty$gives the Riemann integral. Thus the results ends up being: ... 2 Hint: Use the summation by parts formula (also known as Abel's summation formula): $$\sum_{k=m}^n f_k (g_{k+1}-g_k) = (f_{n+1} g_{n+1}-f_m g_m) - \sum_{k=m}^n g_{k+1} (f_{k+1}-f_k).$$ 0 I don't understand your question that well. From what I understand it is akin to population distribution and sampling distribution for the process. If one is continuous, the other is continuous too. To further describe, let us say you have normal distribution with$N(\mu,\sigma^2)$where$\mu$and$\sigma^2$are true parameters, then sampling parameters ... 1 Formally : $$(\frac{\partial}{\partial z}R)(z,s)=\sum_{n=0}^{\infty}\sum_{i=-\infty}^{\infty}r(i,n)iz^{i-1}s^n$$ $$(\frac{\partial^k}{\partial s^k}(\frac{\partial}{\partial z}R))(z,s)=\sum_{n=k}^{\infty}\sum_{i=-\infty}^{\infty}r(i,n)iz^{i-1}\frac{n!}{(n-k)!}s^{n-k}$$ So that : $$(\frac{\partial^k}{\partial s^k}(\frac{\partial}{\partial ... 1 Your idea is correct, but you have some typo and some mistake. From the first equation we have:$$ \phi_1=\rho(1)(1-\phi_2) $$and, substituting, the second equation becomes:$$ \rho(2)=\frac{\rho^2(1)(1-\phi_2)^2+\phi_2[1-\rho(1)(1-\phi_2)]}{1-\phi_2} $$that, for 1-\phi_2 \ne 0, gives:$$ ... 0 This really is just a matter of notation. Usually, when considering random variables (or indeed, stochastic processes), you don't specify the dependency on$\omega\in\Omega$. That is, if$X$is a real-valued random variable, then$X$is a function$\Omega\to\mathbb{R}$, and for$\omega\in\Omega$,$X(\omega)\in\mathbb{R}$; but commonly we just write ... 0 There are two approaches. One way is to use the procedure for continuous-time markov chain analysis by finding an embedded markov chain which operates in discrete time. You can read details here. In this case, the embedded chain is simply$X_n$which is a simple random walk on$\mathbb{Z}/m\mathbb{Z}$. This is a finite state space, and$X_n\$ is aperiodic for ...

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