By Ayalvadi Ganesh, Neil O’Connell, Damon Wischik (auth.)

**Big Queues** goals to provide an easy and chic account of ways huge deviations idea might be utilized to queueing difficulties. huge deviations thought is a set of strong effects and common suggestions for learning infrequent occasions, and has been utilized to queueing difficulties in numerous methods. The strengths of enormous deviations idea are those: it truly is strong sufficient that you may solution many questions that are not easy to respond to differently, and it really is basic adequate that you'll be able to draw huge conclusions with no hoping on specific case calculations.

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**Sample text**

Bernoulli random variables Xi , with P (X1 = 1) = p = 1 − P (X1 = 0), and this made the calculation easy. Here is another example where the calculation is also easy: the average is of normal random variables. d. sequence of normal random variables with zero mean and unit variance, and let Sn = Y1 + . . + Yn . The sample mean Sn /n is also Large Deviations in Euclidean Spaces 25 normally distributed, with mean zero and variance 1/n. Thus, for any x > 0, P ∞ 2 Sn 1 e−nz /2 dz >x = √ n 2π x ∞ 1 1 2 2 √ e−nx /2 .

1), namely the term with k = qn, is suﬃcient to determine the correct exponential decay rate in n of this sum. Since it is only this decay rate that we are interested in, we can replace the sum by the largest term. It turns out this feature is characteristic of many situations where the theory of large deviations is applicable. 2 for more. d. Bernoulli random variables Xi , with P (X1 = 1) = p = 1 − P (X1 = 0), and this made the calculation easy. Here is another example where the calculation is also easy: the average is of normal random variables.

This satisﬁes the recursion and is stationary. However, in the calculations that follow, we will only be concerned with X(−t, 0], which by stationarity has the same distribution as X(0, t], so we could just as well set X0 ∼ Normal(0, σ2 ) and deﬁne X|(0,t] from this. What is Λ(θ)? Clearly ESt = μt, and Cov(A0 , At ) = at σ 2 , giving a|i−j| = Var St = σ 2 1≤i,j≤t σ2 t(1 − a2 ) − 2a(1 − at ) . (1 − a)2 And since St is normal, 1 Λt (θ) = θμt + σt2 2 2 where σt = Var St . Dividing by t and taking the limit, Λ(θ) = θμ + θ2 1 + a .