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This problem relates to empirical distribution theory. Here, a complete treatment of the subject: EMPIRICAL PROCESSES: Theory and Applications https://www.stat.washington.edu/jaw/RESEARCH/TALKS/Delft/emp-proc-delft-big.pdf


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From your previous question, you know how to do this if you know $\mathsf E(Z\mid N)$ and $\mathsf {Var}(Z\mid N)$ So: $\mathsf E(Z\mid N) = \sum\limits_{i=1}^{N+1} \mathsf E(X_i)$ By linearity of Expectation. Similarly: $\mathsf E(Z^2\mid N) =\mathsf E\Big(\big(\sum\limits_{i=1}^{N+1} X_i\big)^2\Big) = ...


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I will explain the general case for the first moment. I will assume that by $Z$, you mean $$Z =\begin{cases} 0,& \text{if } N = 0\\ X_1+\dotsb+X_N,&\text{if } N>0.\end{cases}$$ (I am ignoring the $N+1$ thing.). Also, the $X_i$ are iid. $N$ non negative integer valued random variable. Let $E[X_i] = \mu$, and $E[N] = \nu$. Then to find $E[Z]$, ...


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Isn't the frequency just once per 20 minutes? "Hello" is printed only when the number is 20. That is with 1/20 probability. So, even if the rate varies each time, over an average of 7 hours, by 5% error margin, you can say the rate is 1 "hello" per 20 minutes.


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In every minute the probability that it will print hello is $\frac{1}{20}$ assuming that the "Random" is fair in JAVA Let $A$ denote the event that it will print hello per one minute so $\mathrm {P}[A]=\frac{|A|}{|\Omega|}=\frac{1}{20}$ Therfore the rate is one hello per $20$ minutes


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Additonnally, good properties of Sobol (uniformity) comes from the choice of Polynonme and initial integers for each dimension. Property A and A' should be satistified for better convergence.


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There is no "measure" like the one you are looking for. The proof establishes a measure of the randomness of a probability function on all permutations, not a measure of the randomness of an individual permutation/configuration of the deck. No individual deck configuration is "more random" than another in any sense that is meaningful for probability and ...


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I dont know if this is still relevant, but there is a paper about this problem: O'Donoughue, N.; Moura, J.M.F., "On the Product of Independent Complex Gaussians," in Signal Processing, IEEE Transactions on , vol.60, no.3, pp.1050-1063, March 2012 Best, Rick


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Hint: Suppose for a moment that the ideal number generator generates any integer between $ A $ and $ B $ instead of real number. Then the probability of $ X_1 + X_2 \le C $ can be found as $ \sum_{M} P(X_1 = M) P(X_2 \le C - M) $ For a concrete example, let say A = B = 6, then the probability that $ X_1 + X_2 \le 8 $ is $ P(X_1 = 1)P(X_2 \le 7) + P(X_1 ...


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It seems reasonable to hypothesize that the number of baserunners in each inning is not completely random. After all, the teams exercise strategy to try to win, by arranging the batting order, choosing when to replace the pitcher, and so forth, with the objective of scoring as many runs as they can (which requires getting baserunners) and preventing the ...


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Random distribution of 74 balls into 89 boxes, by simulation: $$\begin{array}{rr|rr} =0 & 39 & \ge0 & 89 \\ =1 & 32 & \ge1 & 50 \\ =2 & 13 & \ge2 & 18 \\ =3 & 4 & \ge3 & 5 \\ =4 & 1 & \ge4 & 1 \end{array}$$ I'd say that settles it, but feel free to draw your own conclusion.


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If 1 currency = 200 aluminium, and aluminium is 9800 times more likely to be chosen than platinum, it might make sense to make 1 platinum be worth 9800 aluminium. If so, 1 currency = 200 aluminium * (1 platinum/9800 aluminium) = 1/49 platinum. Then 1 platinum would be 49 currency. For the rest of the metals, just replace the 1/9800 with the appropriate ...


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After resesarch, this question relates to Kernel Density Smoothing and estimation from empirical distribution: https://en.wikipedia.org/wiki/Kernel_density_estimation KDE should converge to the theoritical when the number of sample increases... How this convergence is done, is related to the random generator structure. Another item is the consistency of ...


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There is an anecdote about the notion of stochastic processes. They say that when Khinchin wrote his seminal paper "Correlation theory for stationary stochastic processes", this did not go well with Soviet authorities. The reason is that the notion of random process used by Khinchin contradicted dialectical materialism. In diamat, all processes in nature are ...


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Sobol sequences belong to the class of Quasi Random Generators (by opposition of Pseudo Random Generators). Quasi Random Generators by construction minimize the discrepancy between the sub square (ie sub interval). Discrepancy is the (maximum) between 2 points inside sub-interval. Quasi Random Generators are deterministic generators of points. Sobol uses ...


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For the second part of your question, there are $10$ possibilities for the first position, of which $1$ qualifies, for a probability of $1/10$. There are $100$ possibilities for the fourth and fifth positions, of which $19+3+6 = 28$ qualify, for a probability of $18/100 = 9/50$. Thus, the probability that both qualify is $$ \frac{1}{10} \times ...


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The total number of sequences is $10^{9}.$ Use constructive counting. If $9$ is in the first position and $93$ takes up positions three and four, we now have only $10^{6}$ choices (we select the remaining $6$ digits). Our answer is thus $\frac{10^{6}}{10^{9}} = \boxed{\frac{1}{1000}}.$


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You can follow the approach presented in the paper Quasi-Random resamplings, with applications to rule-samplng, cross-validation and (su-)bagging. Additionally if feasible I would suggest keeping track of previous draws, as repetitions might arise before the whole set of permutations is exhausted. Depending on the form of your function (e.g. if there is a ...


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Here's a more general question: Partition a set $X$ of $N$ elements into disjoint non-empty subsets $X_1, X_2,\dots, X_m$ with $X_i=n_i$. Pick randomly from $X$ uniformly, with replacement. Let $T$ be the number of pairs picked when you've found at lest one element from each $X_i$. What is the expected value of $T$? In your question, if we have $n$ ...



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