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0

To my knowledge there aren't any great approaches to this. Most models "flatten" out the data and ignore patterns over time. Depends on the outcome. Most don't need the extra data/complex approach. This doesn't mean people aren't trying. Have you looked at the physioNet stuff? It's likely more esoteric than you're looking for but may be ...

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Not an answer, at least not to your question, but an example of how to use Ito's formula (http://en.wikipedia.org/wiki/It%C5%8D_calculus). Since $[B]_t=t$, we have $$B_t^n = \int_0^t nB_s^{n-1}dB_s + \frac{n(n-1)}{2} \int_0^t B_s^{n-2} ds$$ for $n\ge 2$. In particular, $$B_t^3 = 3 \int_0^t B_s^2 dB_s + 3 \int_0^t B_sds$$ so that $$\int_0^t B_s^2 dB_s ... 1$$E(X_{n+1}\mid \mathcal F_{n}) = E(\eta_{n+1}+X_{n}\mid \mathcal F_{n})=E(\eta_{n+1}\mid \mathcal F_{n}) +E(X_{n}\mid \mathcal F_{n}) =E(\eta_{n+1}) + X_n$$Now look up the definitions. Analogously for Y_n - if you understand its subscript (I don't). 0 Yes, the null hypothesis for PP and ADF tests is that the process is difference stationary. The order of the difference is one. If you suspect that the order is higher, you should test the differenced series. For AR(p) process (and generaly for ARMA(p,q)) unit root means first order difference stationary. Hence the notation ARIMA(p,d,q), which means that ... 0 Judging from the comments there appears to be a lot of confusion and lack of intuition over this question. A trivial Monte Carlo simulation will give (roughly) the correct answer that can be used to gauge the validity of the solutions. Here it is in R: > firstHour <- rpois(n=10000000,lambda=4) ; secondHour <- rpois(n=10000000,lambda=4) > ... 0 Thinking this through, I believe this should be calculated with a binomial distribution with n = 8 and p = 0.5 as follows: P = \binom{8}{5} \cdot 0.5^{5} \cdot (1-0.5)^{3}  Let me try to proof this: Let X_1 = number of calls that arrive in the first hour X_2 = number of calls that arrive in the second hour X_3 = number of calls that arrive ... 4 Simply write R_n=R_{n-1}+X_n and take the conditional expectations with respect to \mathcal{F}_n. Then exploit the fact that$$E[g(X_n)f(R_{n-1})|\mathcal{F}_n]=f(R_{n-1})Eg(X_n),$$for any measurable functions f and g (assuming that the expectations exist). Also recall the formula$$\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta.

0

Here's how I'd do it in R. Note, I have used your original data with self-transitions (non-transitions) included. You can do the same thing with the "compressed time series," however. Say I have already constructed an edge list from your example data in the form of the following matrix el <- structure(c(1, 2, 3, 1, 2, 1, 2, 3, 1, 1, 1, 2, 2, 3, 3, 3, 4, ...

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