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Think about what happens the first time you get an H followed by a T. Case 1: you're looking for H-T-H, and you've seen H-T for the first time. If the next toss is H, you're done. If it's T, you're back to square one: since the last two tosses were T-T you now need the full H-T-H. Case 2: you're looking for H-T-T, and you've seen H-T for the first time. If ...

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The authors are providing a simple means for estimating the parameters of a mean-reverting Orstein-Uhlenbeck process via a regression on returns at discretized points in time. The model they are considering has a representation as a stochastic differential equation of the form [pg. 16, Eq. (12)] $$\newcommand{\rd}{\mathrm{d}} \rd X(t) = \kappa (m - X(t)) ... 11 I think what you're looking for is a copula. You've got two marginal distributions (specified by either parametric or empirical cdfs) and now you want to specify the dependence between the two. For the bivariate case there are all kinds of choices, but the basic recipe is the same. I'll use a Gaussian copula for ease of interpretation. To draw from the ... 10 Suppose you toss the coin 8n+2 times and count the number of times you see a "HTH" pattern (including overlaps). The expected number is n. But it is also n for "HTT". Since HTH can overlap itself and "HTT" cannot, you would expect more clumping with "HTH", which increases the expected time for the first appearance of HTH. Another way of ... 9 \sqrt{n}\sup_x|F_n-F|=\sup_x|\frac{1}{\sqrt{n}}\sum_{i=1}^nZ_i(x)|  where Z_i(x)=1_{X_i\leq x}-E[1_{X_i\leq x}] by CLT you have G_n=\frac{1}{\sqrt{n}}\sum_{i=1}^nZ_i(x)\rightarrow \mathcal{N}(0,F(x)(1-F(x))) this is the intuition... brownian bridge B(t) has variance t(1-t) http://en.wikipedia.org/wiki/Brownian_bridge replace t by F(x). ... 9 This problem can quickly be reduced to one of finding the quantile of a trapezoidal distribution. Let us rewrite the process as$$ P(x) = U_1 \cdot \frac12 \sin x + U_2 \cdot \frac12 \cos x + \frac{1}{2} (\sin x + \cos x) \>, $$where U_1 and U_2 are iid \mathcal U(-1,1) random variables; and, by symmetry, this has the same marginal distribution ... 8 I like to draw pictures. These diagrams are finite state automata (FSAs). They are tiny children's games (like Chutes and Ladders) that "recognize" or "accept" the HTT and HTH sequences, respectively, by moving a token from one node to another in response to the coin flips. The token begins at the top node, pointed to by an arrow (line i). After each ... 8 In a certain sense, the answer depends on how lazy you want to be. If you are willing to do some very mild bookkeeping, then it is easy to calculate on any particular day the exact probability of select a half-tablet based on your previous draws. This is a form of conditional probability calculation. If you don't like to do any bookkeeping at all, then a ... 8 I think most people will agree that successive outcomes of soccer matches (of the same team?!) are not independent of each other. Clearly there are factors, such as injured players, making matches that are close in time dependent. The exact nature of these invisible ties is nearly impossible to state correctly and in particular completely. People very ... 7 Probably the best idea is just to quote Ulam: The first thoughts and attempts I made to practice [the Monte Carlo Method] were suggested by a question which occurred to me in 1946 as I was convalescing from an illness and playing solitaires. The question was what are the chances that a Canfield solitaire laid out with 52 cards will come out successfully? ... 7 I am a novice in sports analyses as well as in time series but perhaps a quick empirical example: There is a package "vcd" which contains all soccer games in the German Bundesliga from 1963 to 2008. We can use this dataset to have a look whether we see some anecdotal evidence for a correlation between two consecutive games (it is already sorted). For ... 6 If n is large enough, your expected value should approach the mean of the distribution. Yes that's correct. So probability that value is greater than expected value should be 0.5. This would only be correct if the distribution is symmetric - which in your game isn't the case. You can see this easily if you think about what the median value ... 6 Additional conditions are needed. (A near-proof of this fact is that many incredibly smart individuals have been thinking deeply about these issues for over 100 years. It is highly unlikely that something like this would have escaped all of them.) First of all, note that the formula for V that you give is part of the conclusion of the associated central ... 6 It seems to me that you have enough data to model the dependence on space-time and meteorological influences of both the bias of forecast errors (i.e. tendency to systematically over-/underestimate [first moment]) and their variance [second moment]. For exploration of the bias, I'd just do a lot of scatterplots, heatmaps or hexbin plots. For exploration of ... 6 At its core, the use of simulation is exploiting the idea that if you repeatedly simulate a process, you can - with a sufficiently large sample size - start to get results that resemble an analytical solution even when that analytical solution is unknown. There are all kinds of questions that can be answered with that - Monte Carlo methods are a pretty ... 6 Yes. There is a very efficient (linear time) algorithm, and the intuition for it comes directly from the uniformly-sampled case. Suppose we have a partition of [0,T] such that 0=t_0 < t_1 < t_2 < \cdots < t_n = T. Uniformly sampled case In this case we have t_i = i \Delta where \Delta = T/n. Let X_i := X(t_i) denote the value of the ... 5 Existence proofs are notoriously difficult to justify and appreciate. One reason is that they don't seem to have any consequences. Whether or not you read and understand an existence proof does not change your subsequent work with a given stochastic process because that work relies on properties of the process and not the fact that it exists. The purpose ... 5 We (A colleague and I) finally wrote a paper on that one. To summarized things we proposed two solution to quantify and give a statistical summary of the (spatio-temporal) propagation of errors along Denmark and along look ahead times. In the first one we compute the correlation between all pairs of wind farms and for all pairs of look ahead times (this ... 5 To jump to the conclusion immediately, the "momentum" does not change the fact that the normal distribution is an asymptotic approximation of the distribution of the random walk, but the variance changes from 4np(1-p) to np/(1-p). This can be derived by relatively elementary considerations in this special case. It is not awfully difficult to generalize ... 5 In addition to the Kloeden and Platen books already mentioned, the book Simulation and Inference for Stochastic Differential Equations by Stefano Iacus is good. He is also the author of the sde package for R. 5 The response above by Xi'an addresses a different question than the one you are asking, I believe. He answers the question of the probability of 'HH' in a given pair of flips. This is slightly different than flipping until you get two heads in a row since the sequence 'THHT' would not register as two heads in a row in his MC. You were very close to having ... 5 This is a computational exercise, so think recursively. The current state of coin flipping is determined by the ordered pair (N,M) with N\ge M\ge 0. Let the expected number of flips to reach N consecutive heads be e(N,M): (1) There is a 50% chance the next flip will be heads, taking you to the state (N,M+1), and a 50% chance the next flip will ... 5 It is often the case in sports analytics that people ask questions about more ethereal concepts like momentum, clutch, or home-field advantage. At the surface it would sound silly to say that these things don't exist. However, whether or not they exist is a separate question from whether or not we can meaningfully use them in any sort of predictive ... 4 For Kolmogorov-Smirnov, consider the null hypothesis. It says that a sample is drawn from a particular distribution. So if you construct the empirical distribution function for n samples f(x) = \frac{1}{n} \sum_i \chi_{(-\infty, X_i]}(x), in the limit of infinite data, it will converge to the underlying distribution. For finite information, it will be ... 4 These two books may be a good starting point http://www.amazon.com/Numerical-Stochastic-Differential-Equations-Probability/dp/3540540628 http://www.amazon.com/Numerical-Solution-Computer-Experiments-Universitext/dp/3540570748/ref=pd_bxgy_b_img_b 4 I'd suggest starting with a quick read of the chapter of Law and Kelton's "Simulation Modeling and Analysis" textbook that discusses methods for selecting distributions to use in Monte Carlo simulations. This chapter discusses methods for selecting candidate distributions, fitting the distributions to your data, and then testing the goodness of fit. It's ... 4 To solve this problem, I will use stochastic processes, stopping times, and dynamic programming. First, some definitions:$$X_n \doteq \#\text{(of consecutive heads after the nth flip)} We also allow a value for $X_0$ to mean the number of consecutive heads before we start. So, for $X_0 = 0$ and the sequence of flips HHTHHHTHTTHH, the corresponding ...

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First question. Denoting by $({\cal F}_n)$ the filtration generated by the process $(X_n)$ we will prove that the event $A:=\{W_y= 1\}$ does not belong to the $\sigma$-field ${\cal F}_1$. This implies that $W_y$ is not a stopping time. One has $A=\{X_1 \neq y\} \cap \{X_2=y\}$. If $A$ did belong to ${\cal F}_1$ then one would have $\Pr(A \mid {\cal ... 4 This looks like homework so I'm trying to give a hint, not a solution. For part (b), you definitely want to use the structure of the graph. Without loss of generality suppose you start at$12$and your first step is to$1$. Can you say what the probability is that you hit$11$before you hit$12$? 4 It's a mixture of three distributions, and can be found pretty easily by brute force, if one allows oneself to handwave over some important details (e.g., "is$\lambda < \mu$"). Let$n$be the number of customers in the system;$n=0$means no-one is being served or waiting,$n=1$means one customer is being served but no-one is waiting, etc. Let$p_n\$ ...

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