An instructive toy model: two paradoxes

Transcription

1 Tel Aviv University, 2006 Gussin rndom vectors 27 3 Level crossings... the fmous ice formul, undoubtedly one of the most importnt results in the ppliction of smooth stochstic processes..j. Adler nd J.E. Tylor 1 3 An instructive toy model: two prdoxes b Mesures on the grph of function c The sme for rndom function d Gussin cse: ice s formul e Some integrl geometry An instructive toy model: two prdoxes We strt with very simple rndom trigonometric polynomil (even simpler thn 1c12): (31) X(t) = ζ cost + η sin t where ζ, η re independent N(0, 1) rndom vribles. Its distribution is the imge of γ 2 under the mp (x, y) (t x cost+y sin t). Time shifts of the trigonometric polynomil correspond to rottions of 2, (32) X(t α) = ζ(costcosα + sin t sin α) + η(sin t cosα costsin α) = = (ζ cosα η sin α) cost + (ζ sin α + η cos α) sint; the process (31) is sttionry, tht is, invrint under time shifts; (33) EX(t) = 0, EX(s)X(t) = cos(s t). The rndom vrible (34) hs the density M = mx t X(t) = ζ 2 + η 2 (35) f M (u) = ue u2 /2 for u > 0 ; 1 e u2 /2 du u 1 See Prefce (pge vi) to the book ndom fields nd geometry (to pper).

2 Tel Aviv University, 2006 Gussin rndom vectors 28 it mens tht P ( < M < b ) = b f M(u) du. Consider lso rndom vribles ( k ) (36) M n = mx X ; k Z n clerly, M n M.s. The density of M n is (37) f Mn (u) = 1 u tn π/n /2 n e u2 e x2 /2 dx for u > 0. utn π/n (n=3) u tn π n We see tht f Mn (u) f M (u) = ue u2 /2 s n. On the other hnd, (38) f Mn (u) n 1 e u2 /2 s u. A prdox! Think, wht does it men. Another prdox ppers if we condition X to hve the mximum t given t. By sttionrity we restrict ourselves to t = 0. The condition becomes η = 0 nd ζ > 0; thus, X(t) = ζ cost, nd the conditionl distribution of ζ is the sme s the unconditionl distribution of ζ (since ζ nd η re independent). The conditionl density of M is u 2 e u2 /2 for u > 0. This holds for t = 0, but lso for every t; we conclude tht the unconditionl density of M is lso u 2 e u2 /2, in contrdiction to (35)! Here is nother form of the sme prdox. For ech t the two rndom vribles X(t) nd X (t) re independent (think, why), distributed N(0, 1) ech. Thus, given X(t) = 0, the distribution of X (t) is still N(0, 1), nd the density of X (t) is u 2 e u2 /2 (for u > 0). On the other hnd, X( ) vnishes t two points, nd X (t) = ζ 2 + η 2 = M t these points (think, why). This rgument leds to nother density, u ue u2 /2 (for u > 0). Here is n explntion. The phrse given tht X(0) = 0 hs (t lest) two interprettions, known s verticl window nd horizontl window. Verticl window: we condition on X(0) < ε nd tke ε 0. η ζ

3 Tel Aviv University, 2006 Gussin rndom vectors 29 We get X(t) = η sin t with η distributed N(0, 1) (conditionlly). Horizontl window: we require X to vnish somewhere on ( ε, ε) nd tke ε 0. η ζ We get X(t) = η sin t, but now the conditionl density of η is u const u e u2 /2 (nd const = 1/2). Think bout the two interprettions of the phrse given tht X hs mximum t 0. Here is still nother mnifesttion of the prdox. Let us compre X = ζ cost+η sin t with Y = ζ cos 10t+η sin 10t. For ech t the two rndom vribles X(t), Y (t) hve the sme density t 0 (just becuse both re N(0, 1)). Nevertheless, Y ( ) hs 10 times more zeros thn X( ). Think, wht does it men in terms of the horizontl nd verticl window. 3b Mesures on the grph of function Before treting rndom functions we exmine single (non-rndom) function f C 1 [, b]; tht is, f : [, b] is continuous, nd there exists continuous f : [, b] such tht f(x) = f() + x f (t) dt for x [, b]. The number #f 1 (y) of points x [, b] such tht f(x) = y is function {0, 1, 2,... } { }. 3b1 Lemm. #f 1 (y) dy = b f (x) dx. 3b2 Exercise. Prove 3b1 ssuming tht f is () monotone, (b) piecewise monotone. This is enough for (sy) trigonometric polynomils. In generl, f C 1 [, b] need not be piecewise monotone, 1 but 3b1 holds nywy. Proof of Lemm 3b1 (sketch). We consider the set C = {x : f (x) = 0} of criticl points nd the set f(c) of criticl vlues; both re compct sets, nd mes f(c) = 0 by Srd s theorem (even if mesc 0). If y / f(c) then the set 1 Try x 3 sin(1/x).

4 Tel Aviv University, 2006 Gussin rndom vectors 30 f 1 (y) is finite (since its ccumultion point would be criticl), nd f 1 (y+ε) is close to f 1 (y) for ll ε smll enough; in prticulr, #f 1 (y+ε) = #f 1 (y). The sets B n = {y \ f(c) : #f 1 (y) = n} nd A n = f 1 (B n ) re open, B 1 B 2 = \ f(c), nd A 1 A 2 = f 1 ( \ f(c)) is full mesure subset of \ C. It is sufficient to prove tht #f 1 (y) dy = f (x) dx B n A n for ech n. We consider connected component of B n nd observe tht f is monotone on ech of the n corresponding intervls. Here is n equlity between two mesures on 2 concentrted on the grph of f (below δ x,y stnds for the unit mss t (x, y)). 3b3 Exercise. For every f C 1 [, b], (3b4) dy δ x,y = tht is, (3b5) dy 1 A (x, y) = b b dx f (x) δ x,f(x), dx f (x) 1 A (x, f(x)) for ll Borel sets A 2. Hint: first, consider rectngles A = (x 1, x 2 ) (y 1, y 2 ); second, recll the hint to 2c5. 3b6 Exercise. For every f C 1 [, b] nd every bounded Borel function g : [, b], b dy g(x) = dx f (x) g(x). Hint: first, indictors g = 1 A ; second, their liner combintions. eplcing g(x) with g(x) sgnf (x) we get n equivlent formul 1 b (3b7) dy g(x) sgn f (x) = dxf (x)g(x). 1 for > 0, 1 sgn = 0 for = 0, 1 for < 0.

5 Tel Aviv University, 2006 Gussin rndom vectors 31 We hve lso (seemingly) more generl equlities between (signed) mesures, (3b8) (3b9) dy dy g(x)δ x,y = b g(x) sgnf (x)δ x,y = dx f (x) g(x)δ x,f(x) ; b dxf (x)g(x)δ x,f(x). Tking g( ) 0 such tht both sides of (3b8) re equl to 1 we get probbility mesure on the grph of f. Treting it s the distribution of pir of rndom vribles X, Y we see tht the conditionl distribution of Y given X = x is δ f(x), nd the conditionl distribution of X given Y = y is const g(x)δ x, which does not men tht g is the unconditionl density of X. ther, the density is equl to f g. This is nother mnifesttion of the distinction between horizontl window nd verticl window. 3c The sme for rndom function Let µ be probbility mesure on C 1 [, b]. Two ssumption on µ re introduced below. Given x [, b], we consider the joint distribution of f(x) nd f (x), where ( f is distributed µ; in other words, the imge of µ under the mp f f(x), f (x) ) from C 1 [, b] to 2. The first ssumption: for ech x [, b], this joint distribution is bsolutely continuous, tht is, hs density p x ; (3c1) ϕ(y, y )p x (y, y ) dydy = ϕ ( f(x), f (x) ) µ(df) for every bounded Borel function ϕ : 2 (recll the hint to 3b6). 1 The function (x, y, y ) p x (y, y ) on [, b] 2 is (or rther, my be chosen to be) mesurble, since its convolution with ny continuous function of y, y is continuous in x. For exmple, µ cn be n rbitrry bsolutely continuous mesure on the (finite-dimensionl liner) spce of trigonometric (or lgebric) polynomils of degree n (except for n = 0). 1 The mening of df in µ(df) nd df(x)/dx is completely different...

6 Tel Aviv University, 2006 Gussin rndom vectors 32 The second ssumption: 1 (3c2) [,b] C 1 [,b] f (x) dxµ(df) <. (A simple sufficient condition: f µ(df) < ; here f = mx f + mx f or something equivlent.) For exmple, µ cn be n rbitrry nondegenerte Gussin mesure on the (finite-dimensionl liner) spce of trigonometric (or lgebric) polynomils of degree n (except for n = 0). 3c3 Exercise. The function f #f 1 (0) is Borel function on C 1 [, b]. Hint: first, {f : f 1 (0) = } is open; second, #f 1 (0) = lim n #{k : f 1 (0) [(k 1)2 n, k 2 n ) }. 3c4 Exercise. The function (y, f) #f 1 (y) is Borel function on C 1 [, b]. Hint: do not work hrd, consider (y, f) f( ) y. For given y we consider the expected (verged) #f 1 (y), (3c5) E ( #f 1 (y) ) = #f 1 (y) µ(df) [0, ]. 3c6 Exercise. dy E ( #f 1 (y) ) = Hint: 3b1 nd Fubini. C 1 [,b] Similrly, (3b5) gives (3c7) dy E 1 A (x, y) = b b dx dydy p x (y, y ) y. 2 dx dydy p x (y, y ) y 1 A (x, y) 2 for ll Borel sets A 2. More generlly, 2 (3c8) dy E b g(x, y) = dx dydy p x (y, y ) y g(x, y) 2 1 The function (x, f) f (x) on [, b] C 1 [, b] is continuous, therefore, Borel. 2 In 3b6 we use g(x), but g(x, y) cn be used eqully well (which does not increse generlity).

7 Tel Aviv University, 2006 Gussin rndom vectors 33 for every bounded Borel function g : 2. Substituting g(x)h(y) for g(x, y) we get (3c9) dy h(y) E g(x) = dy h(y) b dx g(x) dy p x (y, y ) y for ll bounded Borel functions g : [, b], h :. Therefore (3c10) E g(x) = for lmost ll y. Especilly, (3c11) E ( #f 1 (y) ) = b b dx g(x) dy p x (y, y ) y dx dy p x (y, y ) y for lmost ll y. Some dditionl ssumptions could ensure (continuity in y nd therefore) the equlity for every y. 3d Gussin cse: ice s formul Let γ be (centered) Gussin mesure on C 1 [, b] such tht for every x [, b] (3d1) f(x) 2 γ(df) = 1, C 1 [,b] (3d2) f (x) 2 γ(df) = σ 2 (x) > 0 C 1 [,b] for some σ : [, b] (0, ). 1 Ech x [, b] leds to two mesurble (in fct, continuous) liner functionls f f(x) nd f f (x) on (C 1 [, b], γ). The former is distributed N(0, 1) by (3d1); the ltter is distributed N(0, σ 2 (x)) by (3d2). 3d3 Exercise. The function σ( ) is continuous on [, b]. Hint: f (x+ε) f (x) 0 (s ε 0) lmost sure, therefore in probbility, therefore (using normlity!) in L 2 (γ). 1 See lso 3d14.

8 Tel Aviv University, 2006 Gussin rndom vectors 34 The joint distribution of f(x) nd f (x) is Gussin mesure on 2. It ppers to be the product mesure, N(0, 1) N(0, σ 2 (x)); in other words, the two rndom vribles f(x) nd f (x) re independent. It is sufficient to prove tht they re orthogonl, (3d4) f(x)f (x) γ(df) = 0. Proof: by (3d1), C 1 [,b] f(x + ε) 2 f(x) 2 0 = γ(df) = 2ε f(x + ε) f(x) f(x + ε) + f(x) = ε 2 γ(df) ε 0 f(x)f (x) γ(df). (Once gin, convergence lmost sure implies convergence in L 2 (γ)...) We see tht γ stisfies (3c1) with ) (3d5) p x (y, y 1 ) = ( σ(x) exp y2 2 y 2. 2σ 2 (x) Condition (3c2) boils down to σ( ) L 1 [, b], which is ensured by (3d3). Thus, we my use the theory of 3c. 3d6 Exercise. (ice s formul) 1 For lmost ll 2 y, Hint: (3c11) nd (3d5). E ( #f 1 (y) ) = 1 π e y2 /2 b σ(x) dx. Let us try it on the toy model (31). Here [, b] = [0, ], σ( ) = 1, nd we get E ( #f 1 (y) ) = 2e y2 /2. In fct, #f 1 (0) = 2.s., nd #f 1 (y) = 2 if M > y, otherwise 0; therefore E ( #f 1 (y) ) = 2P ( M > y ) = 2 f y M (u) du = 2e y2 /2 by (35). 3d7 Exercise. Clculte E ( #f 1 (y) ) for the rndom trigonometric polynomil of 1c12, X(t) = ζ 1 cost + η 1 sin t ζ 2 cos 2t η 2 sin 2t. 1 Kc 1943, ice 1945, Bunimovich 1951, Grennder nd osenbltt 1957, Ivnov 1960, Bulinsky 1961, Itô 1964, Ylvisker 1965 et l. See [1, Sect. 10.3]. 2 In fct, for ll y, see 3d12.

9 Tel Aviv University, 2006 Gussin rndom vectors 35 Integrting ice s formul in y we get (3d8) dy E ( #f 1 (y) ) 2 = π b σ(x) dx, which cn be obtined simpler, by verging (in f) the equlity 3b1 (nd using Fubini s theorem). Bsiclly, ice s formul sttes tht E ( #f 1 (y) ) = const e y2 /2, where the coefficient does not depend on y; its vlue follows esily from 3b1. We my rewrite ice s formul s (3d9) E ( #f 1 (y) ) = 1 e y2 /2 E b f (x) dx for lmost ll y ; note tht E b f (x) dx is the expected totl vrition of f. For exmple, on the toy model (31), #f 1 (0) = 2.s., the totl vrition is equl to 4M, nd EM = uf 0 M (u) du = π/2. The right-hnd side of ice s formul is continuous (in y); in order to get it for ll (rther thn lmost ll) y we will prove tht the left-hnd side is lso continuous (in y). First, we do it for one-dimensionl non-centered Gussin mesure. 3d10 Exercise. Let g, h C 1 [, b] nd h(x) 0 for ll x [, b]. Then the function y 1 du e u2 /2 #fu 1 (y), where f u ( ) = g( ) + uh( ), is continuous on. Hint: the integrl is equl to the totl vrition of the function x Φ ( y g(x) ) ( h(x) where Φ(u) = γ 1 (, u] ). 3d11 Lemm. The function y E ( #f 1 (y) ) is continuous on. Proof (sketch). It is sufficient to prove it on smll subintervls of [, b], due to dditivity. Assume tht γ is infinite-dimensionl (finite dimension is similr but simpler). We hve γ = V (γ ) for some V : S 1 E, S 1 being liner subspce of full mesure. Thus, f = g + uh, where u N(0, 1), h = V ((1, 0, 0,...)) nd g = V ((0,,,...)). Conditionlly, given g, we my pply 3d10 provided tht h does not vnish. Otherwise we do it on neighborhood of ny given point, picking up n pproprite coordinte of. 3d12 Corollry. Formuls 3d6, (3d8), (3d9) hold for ll y (not just lmost ll).

10 Tel Aviv University, 2006 Gussin rndom vectors 36 Especilly, (3d13) E ( #f 1 (0) ) = 1 E b f (x) dx. 3d14 Exercise. Formuls 3d6, (3d8), (3d9), (3d13) still hold if σ( ) my vnish. Hint: pss to the new vrible x new = x 0 σ(x 1) dx 1. 3e Some integrl geometry We consider curve on S n 1 = {z n : z = 1} prmeterized by some [, b]; Z C 1( [, b], n), Z ( [, b] ) S n 1. It leds to Gussin rndom vector in C 1 [, b], f(x) = Z(x), ξ, where ξ is distributed γ n. (Thus, f(x) = ζ 1 Z 1 (x) + + ζ n Z n (x) where ζ 1,...,ζ n re independent rndom vribles distributed N(0, 1) ech, nd Z 1,..., Z n C 1 [, b].) The function f( ) vnishes when the curve Z( ) intersects the hyperplne {z n : z, ξ = 0}. The ltter is just rndom hyperplne distributed uniformly, since ξ/ ξ is distributed uniformly of S n 1. Thus, E ( #f 1 (0) ) is the men number of intersections. On the other hnd, σ 2 (x) = E f (x) 2 = E Z (x), ξ 2 = Z (x) 2, thus, b σ(x) dx is nothing but the length of the given curve. By ice s formul (for y = 0), (3e1) the men number of intersections the length of the curve = 1 π. For exmple, the toy model (31) corresponds to Z : [0, ] S 1, Z(t) = (cos t, sin t). The curve is the unit circle, of length. The number of intersections is equl to 2 lwys. eferences [1] H. Crmér, M.. Ledbetter, Sttionry nd relted stochstic processes, Wiley 1967.

11 Tel Aviv University, 2006 Gussin rndom vectors 37 Index expected totl vrition, 35 horizontl window, 29 ice formul, 34 verticl window, 28 C 1 [, b], spce of smooth functions, 29 E ( #f 1 (y) ), expected number of level crossings, 32 #f 1 (y), number of level crossings, 29 p x (y, y ), 2-dim density, 31 σ(x), men squre derivtive, 33