1 Studies in the History of Statistics and Probability ... - Sheynin, Oscar

P{max ξ(t) ≥ x} = 0.01, 0 ≤ t ≤ 1where t is measured in years and the left part of the inequality is themaximal yearly wind velocity.Suppose that we know the values ξ 1 , ξ 2 , ..., ξ n of the maximalvelocity during the first, the second, ..., n-th year during whichmeteorological observations were made. However, wind velocities hadnot been recorded continuously but only several times a day, so thatthose maximal yearly velocities are in essence unknown. For the timebeing, let us nevertheless abstract ourselves from this extremelyessential difficulty.And so, we have those observations of the random variable ξ, themaximal yearly wind velocity, and we wish to assign an x such thatP{ξ ≥ x} = 0.01. (4.11)Had the number n been very large, we would have been obliged toselect such an x that about a hundredth part of the ξ i will be larger thanit. The trouble, however, is that n, the number of years during whichobservations are available, is much less than 100. Then, if x is suchthat (4.11) is fulfilled, that is,P{ξ i ≥ x} = 0.01 for each i,the number of variables ξ i larger than x will obey the Poisson law withparameter λ = 0.01n < 1. It will follow that most likely all of our ξ i willbe less than x so that we are only able to say that x should be largerthan each of the ξ i ′s with no upper boundary available.Therefore, we are tempted to smooth our ξ 1 , ..., ξ n by some law, forexample by the normal law N( x; ξ, s ) and determine x from equationN( x; ξ, s ) = 1− 0.01 = 0.99.Or, we will propose to identify the tail areas of the unknown functionF(x) with those of the normal law.We turn the readers’ attention to the fact that such a procedureshould not be trusted either when applying the normal, or any otherlaw, and that there exist both theoretical grounds and considerationsbased on statistical experiments for that inference. Theoretical groundsconsist in that the central limit theorem only states that the difference*between the exact distribution function P{ sn< x}and the normal lawis small:P s x x*{n< } − N( ) → 0.For example, if that probability P = 0.95, N(x) = 0.99 and thedifference is only 0.04 which is sufficiently small. However, therelative error*[1 − P{ sn< x}] ÷ [1 − N( x)] = 400%43