Continuity of Gaussian ProcessesAuthor(s): M. B. Marcus and L. A. SheppSource: Transactions of the American Mathematical Society, Vol. 151, No. 2 (Oct., 1970), pp.377-391Published by: American Mathematical SocietyStable URL: http://www.jstor.org/stable/1995502Accessed: 19/11/2010 15:23

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TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 151, October 1970

CONTINUITY OF GAUSSIAN PROCESSES

BY

M. B. MARCUS AND L. A. SHEPP

Abstract. We give a proof of Fernique's theorem that if Xis a stationary Gaussian process and a2(h) =E(X(h) - X(O))2 then X has continuous sample paths provided that, for some E > O, a(h) < 0(h), 0 < h < E, where 0 is any increasing function satisfying

M 0~~~~~f (h) dh <co (*) Jo h(log (1/h))112

We prove the partial converse that if a(h) ? 0(h), 0 <h< e and 0 is any increasing function not satisfying (*) then the paths are not continuous. In particular, if a is monotonic we may take 0 =a and (*) is then necessary and sufficient for sample path continuity. Our proof is based on an important lemma of Slepian.

Finally we show that if a is monotonic and convex in [0, e] then a(h)(log 1/h)112 -- 0

as h -- 0 iff the paths are incrementally continuous, meaning that for each monotonic bounded sequence t = t1, t2, ...,X( + - X(tn) -X(> 0, w.p.l.

1. Introduction. Let X be a zero-mean separable stationary Gaussian process with continuous covariance p(h)=EX(s)X(s+h). Xavier Fernique (1964) made considerable progress toward a solution of the well-known problem [6], [3], [1] of finding necessary and sufficient conditions on p in order that X have continuous sample paths. Define u2(h) = E(X(h) - X(O))2 = 2(p(O) -p(h)). Fernique obtained that if g(h) ? 0(h), 0 _ h < 1, where b is monotonic 4 (nondecreasing) and satisfies

(1.1) f' ~~~~~,b(h) dho lh(log / 0h)1

dh <

then X has continuous paths. He also obtained (under some additional conditions) that if 0 is monotonic t and (1.1) fails to hold then there exists a discontinuous process X (which he gave as a random lacunary Fourier series) with a(h) ? 0(h), O < h ? e. Modifying his method of random lacunary Fourier series, we obtain (?3) using a basic inequality of D. Slepian (1962) that if for some e>0, g(h) >0(h), O < h < e where b is monotonic t and (1.1) fails to hold then the paths of X are

discontinuous. As an immediate consequence we find that if g is monotonic t then

(1.2) oh(log I1h)112 dh < oo

is the necessary and sufficient condition for X to have continuous paths. In parti-

cular if the convariance of X is of Polya's type, the question of whether the paths

are continuous is settled completely.

Received by the editors January 10, 1969. Copyright ( 1970, American Mathematical Society

377

378 M. B. MARCUS AND L. A. SHEPP [October

M. Nisio [9] has obtained conditions for continuity of X based on the spectral distribution function F. As a corollary to the results of [9], it has been proved [10] that if F(2n + 1) - F(2n) = Sn is eventually decreasing then :2(sn)1"2 < x is a necessary and sufficient condition for continuity of X. We point out by examples (?5) that neither'(1.2) nor X;(sn)112 < oo is a nasc for continuity among all covariances.

Finally, we observe that a discontinuous process X may or may not possess the following property: call X incrementally continuous if for each monotonic (either increasing or decreasing) bounded sequence t= t1, t2, ...

(1.3) X(tn+1)-X(tn) >0 as n-c so

except for a null set (which may depend on t) of paths X. We show (?4) that a Gaussian process X with covariance of Polya's type is incrementally continuous if and only if

(1.4) g(h)(log (I/h))112 ->0 as h -O0.

As an immediate consequence we find that there exists a process X which is incrementally continuous yet not sequentially continuous. (A process X is se- quentially continuous if tn -* t implies X(tn) -* limit a.s.; for Gaussian processes sequential continuity is equivalent to sample path continuity.)

We would like to acknowledge the help of R. M. Dudley who provided an enlightening example bearing on ?4, of H. 0. Pollak who found the simple proof of Boas' inequality, and of N. D. Ylvisaker who had suggested the use of Slepian's inequality to one of the authors.

2. Fernique's sufficient condition for continuity of paths. The purpose of this section is to prove the following theorem of Fernique [5]. Alternate proofs can be found in [13]-[16].

THEOREM. Let X be a separable stationary Gaussian process with zero mean and let g2(h) = E(X(h) - X(0))2. Let 2, be any nondecreasing majorant of a, so that

(2.1) a(h) ?_ 0(h), 0 _ h _ 1,

(2.2) #(h) t , 0 _ h ? 1.

If (1.1) holds, then the paths of X are continuous. Note: it is clear that without loss of generality we could take b to be the least nondecreasing majorant of a,

0(h) = max [cr(u) : 0 ? u ? h].

Proof. By Belyaev's theorem [1] (see [4] for an alternate proof) discontinuous Gaussian processes are unbounded on every set S dense(') in some interval with probability one, so we need only prove that for some M and S

(2.3) P{lX(t)A _ M, teS} > 0

(1) See Appendix 3.

1970] CONTINUITY OF GAUSSIAN PROCESSES 379

where S is dense in some interval. Oddly enough, the choice of the subset S is the crucial part of the proof. We shall choose S={k/2: 0< k/2p ? 1} where k and p are nonnegative integers and 2, = 22, following Fernique [5] (see also [8]). S is sparse enough so that (2.6) (below) holds and yet dense enough so that (2.8) (below) also holds. If t E S, let k(p, t) be the largest integer k for which k/2p ? t, so that k(p, t) = [2pt ]. Writing r(p, t) = k(p, t)/2p we have

(2.4) 0 ? r(p+l, t)-r(p, t) < 1/2p

and 00

(2.5) X(t) = X(r(O, t))+ 2 (X('r(p+ 1, t))-X(r(p, t)) P=1

where the series converges because only finitely many terms are nonzero. To prove the right side of (2.5) bounded in t for X in a set of positive measure we proceed as follows. Forp=0, 1, 2, . . .; k=0, 1, 2,. .., 2p-1; q=0, 1, 2,. .., 2p -1, define

4(p, k, q) = X(k/2, +q/2,+ 1)- X(k/2,).

Let D2(4) = Eg2 and set c(p) = b2P2, p = 0, 1, 2,. . ., where b will be chosen later. Let A be the event that for some (p, k, q) in the range above, I (p, k, q)I> c(p)D(4(p, k, q)). Thus we have

X0 2p,-1 2p,-1

P(A) < :2 :2 : P{li-g > C(p)}, p=O k=O q=O

where 7 = ;/D(4) is standard normal. Since P{l-gi > c} ? exp (-_c2/2) for c > 1 we find choosing b large enough that

00

(2.6) P(A) ? < (2p exp (-b22P1) < 1. p=o

Thus comp (A) has positive measure and if X E comp (A) we have

(2.7) | X(r(p + 1, t))-X(r(p, t))I ? c(p)D(4),

where 4=g(p, k, q) and k=k(p, t), q=(r(p+ 1, t)-r(p, t))2p+1. We have

D2 = D2(4(p, k, q)) = u2(q/2p+ ) = u2(r(p+1,t)-(p,t)) < 02('r(p+l, t)-r(p, t)) < 02(1/2p)

by (2.1), (2.2), and (2.4). Thus from (2.7) and (2.5)

coo

(2.8) X(t)I < I X((O, t))I + : c(p)0(1 /2p).

Again by monotonicity of 0 we have

2PI20((1/2p) <- r 1 0(2-u) du

380 M. B. MARCUS AND L. A. SHEPP [October

and so by (1.1) the series in (2.8) converges:

(2.9) 2 c(p),(l/2p,) ? b I (2-u) . < x Finally, since r(0, t) has only three possible values, it follows directly from (2.8) and (2.9) that (2.3) holds for some value of M.

3. A sufficient condition for the paths to be discontinuous.

THEOREM. Let X be a stationary Gaussian process with zero mean and let au2(h)

=E(X(t + h) - X(t))2. Let b be any nondecreasing local minorant of g, that is for some e>0

(3.1) g(h) > +(h) > 0, 0 < h _ 6

(3.2) 0(h)t, 0<h?e.

If ( . 1) does not holdfor b then the paths of X are not continuous.

Proof. We will show that if the hypothesis holds for a there is a separable, zero mean, stationary Gaussian process Y for which, in the range s E I, t E I, I= [0, e]

(3.3) E( Y(t)-y(s))2 < Kg2? l t ),

(3.4) P(sup [Y(t) : te = I]o= ) = 1

where K is some constant. In other words we will obtain an unbounded process Y whose incremental variance is bounded by a constant times the incremental variance of X. Given such a Y the following inequality of Slepian shows that the paths of X are also unbounded and hence discontinuous.

LEMMA (SLEPIAN). Let X and Z be separable zero mean Gaussian processes such that EX2(t) = EZ2(t), EX(s)X(t) ? EZ(s)Z(t) for s, t E L Then

(3.5) P (sup [X(t) : t el] > M) > P (sup [Z(t) : t fe I] > M).

We include Slepian's proof for completeness, but defer it to the appendix. To see that X is discontinuous if Y exists satisfying (3.3) and (3.4), take Z(t)

=(aq + Y(t))/b, where q is a standard normal variable independent of Y, and a and b are constants to be chosen. If b is large enough so that EY2(t) < b2EX(t)2 and K< b2 we may choose a so that EX2(t) = EZ2(t). We then have E(Z(t) -Z(S))2

=E( Y(t) - Y(s))2/b2 ? U2(J t-s s) and the hypothesis of Slepian's lemma is satisfied. Since the right side of (3.5) is unity for every M, X is unbounded also.

To produce a process Y satisfying (3.3) and (3.4) we take Y to be a random lacunary Fourier series, the stationary Gaussian process defined by

00

(3.6) Y(t) = a7(q cos 2nt+ -ql sin 2nt n=O

19701 CONTINUITY OF GAUSSIAN PROCESSES 381

where {71n} and {n} are independent standard normal sequences and I a2 < oc. Note that

(3.7) EY(s)Y(t) = E a2 cos 2n(t-S), n=O

co

(3.8) E(Y(t)-Y(s))2 = 2 E a2(1 -cos 2n(t-s)). n=O

We now use the following theorem of Szidon [12, ?6.4] to show that Y(t) satisfies (3.4) if

(3.9) 2 lan = 00.

LEMMA (SZIDON). If E (b2 + C2) < 00, E (|bnl + 1CnI) =o0, and On+ l/On> A> 1 then the L2 function with Fourier series 2 (bn cos Ont + Cn sin Ont) is unbounded.

To show that (3.4) holds if (3.9) holds let bn = an'qn, en = anrg, and On = 2n so that the series of the lemma is Y. Note that E I (bn + c2) = 2 E an and so E (b 2 + c2) < 00

a.s. Similarly, E[exp (- b (Jbnl + ICnI) =7l (E exp- Ibnl)E(exp (- ln J))0 by (3.9) and so > (Ibnl + ICnl)=oo a.s. It follows from the lemma that Y(t) is a.s. unbounded on [0, 2Xm] and hence [1] on every interval [0, e]. Thus the problem reduces to finding an 12 sequence an satisfying (3.9) for which the function Y defined by (3.6) satisfies (3.3).

We shall assume first that U2 is monotonically increasing and that a2(2 - n)

_2U2(2 -n), n = 1, 2,.. ., (call this assumption A) because if A holds (which it does if the covariance is of Polya-type, U2 then being monotonic and convex) the construction of the an sequence is simpler. Later we give a construction which works in general. Define fn,= 2(2- ) and observe that fn2 eventually decreases. Define gn as the largest convex minorant of fl, so that gn =fn2 for certain values of n and between these values, gn2 is linear and lies strictly below f2. We claim first that g2+1 < gn2 2gn2+1. The first inequality holds because gn is convex and tends to zero. To prove the second inequality observe that either gn2+ 1 =fn2+ org+1 n n

In the first case, 2g n ]2 >f2n while in the second case g2+1-gn +2 =gn2

-gn+1 by the definition Of g2 and so 2 n n =gn2+2>0. Thus we have that

gn < 2g2+1 for all values of n. The hypothesis of the theorem requires that (1.1) not hold for zk. By A, =,b=

and by monotonicity of a it is easy to see that (1.1) is the same as

(3.10) ~~~00 g(2 -n) _00fn

(3. 10) 00= Vn El /n n=1 n=i

We claim next that ', gn/ Vn = o0 also. To see this let r and s, defined as above,

382 M. B. MARCUS AND L. A. SHEPP [October

be any two consecutive points where f and g are equal. Then if r ? n ? (r + s)/2 we have by linearity

2 2 g2+2(g-gr2)(n-r)/(s-r) > g2-g2(n-r)/(s-r) = gr2(S-n)/(s-r)

= fr2(s-n)/(s-r) > fn2(s-n)/(s-r) > fn2/2

and since fn/Vn decreases,

(3.11) fn < 2 f?n < 2\/2 < 2A/2 r_n<s \/n rv-n?(r + s)l2 VIn r?n<(r + s)2 \/ n r<n<s V\n

Now we define a2 =g 2-g 2+l. Since g2 is convex, an decreases monotonically. We now use the following inequality of Boas, proved in the appendix.

LEMMA. Let gn = n, aj2 where aj decreases. Then

g n <2 E an. V\n = n=

It follows from the lemma that (3.9) holds and we need only verify (3.3). Fix t < 1 and let N be the largest integer for which 2Nt ? 1. By (3.8) we see that

N 00 (3.12) E(Y(s+t)- y(S))2 < S a 22nt2+4 2 a .

n=1 N+1

The second term on the right in (3.12) is 4g+1 <4fN2 +- 42(2 - N1) < 4a2(t) since a is monotonic by A. The first term on the right of (3.12) is less than

N N N-1 n t g2222n = t2gN I 22j+t2 I (gn-g2+1) 1 2

(3.13) N=l j=l

< 2t2gN22N4 +t2 222n4 n=1

since g2_-g2+1 <gn2/2 as we proved above, and since IL= 1 22i ? 22n(4/3). From (3.13) we get upon subtracting the last term that

N

(3.14) t2 gn22n ? 2t2gN222N < 2g2. n=1

As before, 2g2<4g2+1<4fN2+j=4g2(2-ND-1)4u2(t) and so each term in (3.12) ?4U2(t) and (3.3) holds with K=8. This proves the theorem in the case when a

is monotonic and a2(2-n)<?22(2-n-1). We point out that the proof would work as well even if r2(2-n) < au2(2-n-1) as long as a <4. The case ao=4 is the general case because for any a and h we have

(3.15) a2(2h) = 2 ((1 -cos 2hA) dF(A) = 2 f(2-2 COS2 hA) dF(A) < 4r2(h).

1970] CONTINUITY OF GAUSSIAN PROCESSES 383

We now give another proof of the theorem, slightly more involved, but which works in general. We start from scratch at the expense of some repetition. Since the incremental variance u2(t) has a least positive zero there is an 6 > 0 for which

(3.16) 0 < 0(6/2) = min [a(u): 6/2 < u ? 6],

that is, the minimum of a on [6/2, 6] is taken at the left endpoint. To see that such an 6 exists, let 2z be the least positive zero of a, and let m(u) = min [a(x): u ? x ? z]. We now choose 6 as the largest number ? z for which 0(6/2) = m(z/2), which exists because 0 < m(z/2) ? a(z/2) and a tends to zero at zero. Then 0< a(6/2) < a(x) for 6/2 <x ? z/2 since 6 was largest and 0(6/2) ?a(x) for z/2 ? x <6 because 0(6/2)

=m(z/2) and 6<z. Thus (3.16) holds for 6.

Having chosen 6, we next define

(3.17) +(h) = min [a(u): h ? u ? 6].

t dominates any nondecreasing minorant t' of a on [0, 6] and so if the hypothesis of the theorem holds then (1.1) fails to hold for /, since

+'(h) = min [+'(u): h ? u ? 6] ? min [a(u): h ? u < 6] = +(h)

for 0 < h < 6. Fixing t as in (3.17) we have that if 2t < 6,

02(2t) = min [u2(2u) t ? u ? 6/2]

= 4 min [o2(U): t < u < 6] ? 4 min [r2(U): t < u < 6/2]

by (3.15) and so by (3.16) we have

(3.18) 02(2t) ? 402(t).

Let now nf= /2(2 ,) n= 1, 2,..., so that fn2 and let gn be the largest convex minorant offn2 defined as before. By hypothesis, f t (2 - U6)u - 1/2 du= oo and so

Efn/A/n=oo since t is monotonic. By (3.11) we see that gn/\/n=oo as well. We next define a convex subminorant h2 of g2 by extending certain edges of the

graph of g2, viewed as a convex polygon. Let so =0, and define 0< r1 <s1 <r2

<S2 <... as follows. Suppose that s, -1 has been defined. Let rj be the first n > sj_ _ for which gn>2gg+1 if there are such n. (Let rj=oo if gn_2g2+1 for all n>sj-1.) If rj < o, let sj be the first n > rj for which gn= 2gn+1, noting that s, < x since other- wise gn decreases exponentially contradicting E gn/\/n = o. Now define hn=gn for sj 1<_ n < rj + 1 and for rj + 1 < n < sj define h so that hn-hn+ 1 = h2(sj)-h2(sj + 1)

=g2(sj) - g2(sj+? 1) (where h(n) is written for hn to avoid double subscripts). If

rj = oo, set hn=gn for all n > sj, 1. Since h2 has been defined in r + I < n < sj by extending the edge between sj and sj+ 1 of the graph of g2 backwards throughout

rj + 1 < n < sj, and since g2 is convex, we must have hn _ gn for all n and we see that h2 is also convex. We next show that E hn/Vn = oc. For rj< n <sj, gn+ <gn//2 and so gn<g(rj)0\2rr-n for rj<n<sj. Hence with r=rj, s=sj, we have

s-1 i s-1

gn/V/n ? (g(r)/V/r) \ V/2T < 4g(r)/\r = 4h(r)/V\r ? 4 2 h(n)/V\n. n=r n=r n=r

384 M. B. MARCUS AND L. A. SHEPP [October

Since hn =gn for s<n<rj,+1, it follows that > gn/A/n<4 E hy/A/n and so

E h,N/n = oo.

Set a =h -h2+1 in (3.6). Since an decreases (3.9) holds by Boas' lemma. By Szidon's lemma the process Yof (3.6) satisfies (3.4). By Slepian's lemma the theorem follows from (3.3) and (3.4) and the only thing left to show is that the right side of (3.12) is bounded by Ku2(t) for t < e, where N is the largest integer for which 2Nt<_ e e as above. The second sum in (3.12) is 4h2+1 _4g +j<4fN2+= 42(2 N) =< 402(t) ?4u2(t). The first sum in (3.12) will be bounded by Ku2(t) if we can show that

N (3.19) 2 (h2_hn +j)22nt 2 < Kg2,

n=1

because g f - VN2(2 -Ne)402(2 - N - 1_) ? 402(t) $ 4u2(t), where we have used

(3.18). It is the construction of h that plays the crucial role in proving (3.19) and (3.19) is not necessarily true with the right side replaced by KhN.

We begin with the observation that

(3.20) 4gn+ 1 = gn for all n = 1, 2, .

To prove (3.20) observe that either gn+ 1 =fn2+ l or gn+l <fn,+ l. In the first case, 4g +1=4fn2+1 n while in the second case, g2+1-gn+2 =gn-g2+ by the definition of g2 and so 4g2+1>2gn2+=gn+g2+2>g2. Thus (3.20) holds.

Now suppose that N in (3.19) satisfies Sk < N< Sk+ 1* For j< k we have with

s=sj-,, r=rj, t=sj,

r r r (3.21) 2 (h2-h2 +1)22n = 2 (gn-g2+1)22n < 2 gn22n-

n=s+l n=s+l n=s+l

But summing by parts we get

r r r-1 n 2 g~22n =g2 2k + (gn-gn+ 1) 2k

(3.22) n=s+l n=s+l n=s+l1=+ (3. 22) n = s + 1

r-1

< 4g22 _g2 )22n

Since g2<2g2+1 for s<n<r-1, gn-g2+1<g2/2 for s+1<n<r-1 and sub- stituting in (3.22) we get

r r (3.23) 2 g 22 _3gr222r + 2 g222.

n=s+l n=s+l

Subtracting the last term in (3.23) from both sides and using (3.21) gives

(3.24) 2 (hn - h )22 < 2gr22r n=s+1

1970] CONTINUITY OF GAUSSIAN PROCESSES 385

Since h is linear in the range r + 1 < n _ t + 1 we have

t t

2 (h2-h2+1)22n = (h 2-h 2+1) 2 2n

(3.25) n=r+2 n=r+2

< (g2_-g2+ )22tW4 < 2gt222t.

Since (h2+ 1 -h 2+ 2)22(r + 1) < 4h 2+ 122r < 4h 222r = 4gr222r we have from (3.24) and (3.25) that

Si

E (h - h 2+1)22n < 6g(r1)222r, + 2g2(sj)2 2s,

(3.26) n=sj 1+ < 8g2(Sj)22s,

the last inequality being valid because by (3.20) gn < 4g2+ 1 for all n and so g(n)222n

is monotonically increasing in n. Now again by (3.20), gn < 4g2+ 1 for sj_-< n <sj but there is at least one value of n in the range sj__ 1<n <sj for which g < 2g +1

namely n = sj, 1. Hence

(3.27) g(sj 1)222s, - I g(Sj)222s,.

Summing the inequalities (3.26) and using (3.27) we get

(3.28) n (hnh 21)2 _ 8g(sk)22Sk (1+2+- +2

The same proof as in (3.24) shows that if Sk < N? rk that

N

(3.29) (hn2 -h 2n+1)22n < 2gN22 N fl = sk +1

and the same proof as in (3.25) shows that (3.29) is valid also for rk<N<Sk+ 1*

Thus we have from (3.28) and (3.29)

N

(3.30) E (h2 -h 1)22n < 1 8g222N n=l

again using the monotonicity of gn22n. Since t222N < e2 we see that (3.19) is valid

with K= 18e2. The theorem is proved.

4. Discontinuous processes which are incrementally continuous but not sequentially continuous. A process X will be called incrementally continuous if for every monotonic (either increasing or decreasing) bounded sequence t = t1, t2,...,

(4.1) X(tn + 1) - X(tn) -->-O as n -*> oo

except for a set A(t) of paths X with P(A(t)) = 0. Of course if X consists of a single

function then that function must be continuous, but the Poisson process is an

example of a discontinuous process which is incrementally continuous. On the

other hand, it seems to be difficult to find a process which is incrementally

386 M. B. MARCUS AND L. A. SHEPP [October

continuous without also being sequentially continuous, that is whenever tn -+ t, X(tn) -* limit, except for a null set A(t). Since it can be shown(2) that a sequentially continuous Gaussian process always has continuous sample paths, examples of processes with the desired properties can be constructed by using the theorem of ?3 and the following theorem.

THEOREM. If X has covariance p of Polya's type then X is incrementally continuous if and only if (1.4) holds.

Proof. The property of Polya-type processes that we make use of is that if s1 < s2 < t1 < t2, then

(4.2) E(X(s2)-X(s1))(X(t2)-X(t1))-< 0.

Indeed, if t2-S2 < t1-s1 we write the left side of (4.2) as

(P(t2-S2)-p(tl-S2))-(p(t2-S1)-p(tl-S1)) = (P'(02)-P'(01))(t2-t1)

= P "(03)(02-01)(t2-tl) < 0

since 0<02<t2-S2<t1-S1<01. On the other hand if t1-s1 < t2-s2 we write the left side of (4.2) as (p(tl - S) - p(t1 -s2))- (p(t2 - S) - p(t2 -s2)) and proceed analogously. Thus (4.2) holds.

Now we use two lemmas from [7], omitting their proofs.

LEMMA 1. Let Xand Ybe Gaussian variables with zero mean and suppose EXY< 0. Then if a>0, b>0,

(4.3) P(X > a, Y > b) < P(X_ a)P(Y _ b).

LEMMA 2. Let B1, B2, . .. be events with P(BjBk) <P(B,)P(Bk) for all jA k. If

, P(B,n) = c0 then P(Bn infinitely often) = 1.

Let t be a monotone sequence and e > 0. Define

Bn = {X(tn +1) - X(tn) _ e}, n = 1,2.

By (4.2), Lemmas 1 and 2, and the usual Borel-Cantelli lemma, we see that X is incrementally continuous if and only if : P(Bn) < oo for every monotonic sequence t. Supposing that tn t t which involves no loss of generality, we see that n =tn+ -tn t 0 must satisfy 7 an < 00. We have

(4.4) P(Bn) = P{n _ e/o(8n)}

where -1 is a standard normal variable. If (1.4) fails to hold then there exist un, 0

and 0>0 for which a2(un)> 0/log (1/uUn). If we let (81,82, .)=(ui, u,.. .,U,

U2, U2, .. ., U2, U3, . . .) where the block of u1's is of length A1 > 0, the block of u2's

(2) See Appendix 4.

1970] CONTINUITY OF GAUSSIAN PROCESSES 387

is of length A2 _ 0, and so on, we can make : an <cc and Z P(B) = 00 provided that we can choose the A's so that

(4.5) AnUn<cc

and

(4.6) 2 An exp (-e2/a2(Un)) = oo

hold. The sum in (4.6) is at least I Anun210. If we choose e2 = 0/2, it is clear that there can be found nonnegative integers A1, A2, for which : AnUn < oo and AnUn = c. Thus if (1.4) fails for X of Polya type then X is not incrementally continuous.

On the other hand, if (1.4) holds, then X is incrementally continuous even if X is not of Polya's type because (e2/2a2(8a))- = o(1/log (1/8n)) and it follows that 2 P(Bn) < oo since P(Bn) < Sn for sufficiently large n.

It is easy to find covariances which satisfy (1.4) but not (1.1) and thereby con- struct incrementally continuous processes which are not sequentially continuous.

5. Examples. We give first a continuous process with lim sup (log 1 /h)cr2(h) # 0 thereby showing that the condition that X be of Polya type cannot be dropped entirely in proving that X is not incrementally continuous in ?4. At the same time, a of the example also fails to satisfy (1.2) and thereby shows that (1.2) cannot be necessary for sample continuity in general. The example is

(5.1) X(t) = n 2 cos 22 t+n sin 22't)

where X and 71' are independent standard normal sequences. X is continuous be- cause 2 (IqnI + IJ- )/n2 < oo, a.s. On the other hand, we have

(5.2) (log 22k)ar2(1/22k) (1 - cos 1)(log 2)2k/k4 _o cc

and so lim sup (log 1 /h)oa2(h) #0. A short calculation shows that (1.2) fails to hold. Of course, a2is not monotonic.

We remark that an extension of the above example shows that there is no simple analogue of Slepian's lemma in terms of spectral distribution functions (s.d.f.). In fact if F is the s.d.f. of any discontinuous process then there is a s.d.f. G of a continuous process Y with G(x) < F(x) for all x > 0, G(oo) = F(oo). Indeed this is trivially seen by taking Y(t) = : an( cosg CO nt+9n,sinnOnt) where -q and 79' are as in (3.6). If 2 IanI < cc, Y is continuous no matter what values the 0's take and if On-+ cc sufficiently fast, it is clear that G will have uniformly fatter tails than F. Similarly, it is shown that if F is any s.d.f. with F* 0 then G1 and G2 may be found with G1 < F? G2 and Gl(oo) = G2(0c) where the corresponding processes are either continuous or discontinuous.

The next example shows that the condition

(5.3) :Sn12 < cc, Sn= F(2n+=1)

388 M. B. MARCUS AND L. A. SHEPP [October

where F is the spectral cumulative distribution function of X, fails to be sufficient for continuity of X in general. Our example, which is of Polya type, is defined as follows:

Let ,u be the measure having a point mass

(5.4) Am = (I/2m)1/2 atpm = 1/22m,

m= 1, 2,.... Define F(x) by setting F'(x) =f(x) for all x, where

(5.5) f(x) Jo X2t2 t d(t)

If Pm<X-' <Pm+i,

r1/X r1 2 (5.6) f(x) ? t d(t)+ (t) 2Am+Pm+ +4X-2Amp

and so, for 0 _ k < 2m,

2m+k+1) F(22m+ k) = S22m+k < 2Am+lpm+122m+k

+4AmP 12 2mk

It follows easily that (5.3) holds. To prove that X is not continuous it suffices to check that (1.2) fails to hold because X is of Polya type, as is seen by noting that

f(x) p(t) cos xt dt =-! J p'(t) sin xt dt (5.8) o

-Csx f Jl--c2?s2Xtt2dp(t)

where we have set tdp'(t)=du(t), p(l)=p'(l)=0, ,u as in (5.4). We have It It

p(O)-p(t) = J p'(u) du =- '(t)t + Jdu(u)

(5.9) 1 t t = { X!u du(u) + du(u) = du(u).

Thus if Pm+ 1 < t <Pm, P(O)-p(t) _ Am+ 1 and it is a simple matter to check that (1.2) fails to hold. Thus X is discontinuous in spite of (5.3). Of course, sn in this case is not a monotonic sequence.

Appendix. 1. Proof of Slepian's lemma. The intuitive explanation of Slepian's inequality

is that although X and Z are instanteously of equal variance, X oscillates more than Z because it is less correlated. It is enough to prove (3.5) for finite sets I= {t1, . . ., tj}. The left side of (3.5) is then 1 - Q where Q= Q(r) is given by

AM d M (A. 1) Q(r) = dxl .. dx.g(xl, x.n; r)

1970] CONTINUITY OF GAUSSIAN PROCESSES 389

and the Gaussian density g(x, r) is given in terms of its characteristic function by

(A.2) g(x, r) = { d*e** { den(21T)- exp (i '

xj{-(1/2) 2 2 rjk6j6k)

and rjk= EX(t)X(tk), r={rjk}. From (A.2) we obtain immediately that ag/lrjk - 8g/axiJaxk for j#, k and so differentiating (A. 1) with respect to r12 say we get

r Q(r) dxl dXn

___ (Xl_X2

Xn ,r)

Integrating we get

Q(r) = dx3. . dXng(M, M, x3,..., xn, r).

Thus aQ(r)/1rjk> 0 for all j =k. Now suppose that Sjk=EZ(tj)Z(tk) and let

rjk(A) = Ark + (1- A)Sjk.

Then r(A) ={rjk(A)} is positive definite and so q(A) = 1 - Q(r(A)) is defined. We see that

q'(A) =r- i (8/8rj)Q(r(A)) dr,k(A)IdA j k

= -E E ((lark) Q(r(A)) (rjk - Sik) ? 0

k

since rjj = sjj and for j =k, a QIarjk_0 and sjk> rk. Thus

q(l) = 1 - Q(r) = P{max X(tj) ? M} ? q(O) = 1 - Q(s) = P{max Z(tj) > M},

proving (3.5). 2. Proof of Boas' inequality. The following proof was shown to us by H. 0.

Pollak. We have since a1 ? a2 ? * *, for each m ? n

m 1/2 M-1 1/2 m ~~~1/2 rn-i 1/2 -1.

(B.1) (k=n k=n k=n k=n

< a2/(m + 1 2-n)"2am = am/(m + 1-n)1/2.

Adding on m, we get

00 m ~1/2 rn-1 \1/2\ 0

2 (( ak) -( k ak) 2

= ( a2) =gn

(B.2) m=n k=n k=n k=n

< am/(m + -n)"2. m=n

390 M. B. MARCUS AND L. A. SHEPP [October

Dividing by VIn and adding, we get

00 00 00

2g.l\n//<2 (11/\In) 2~ a.1(m +1-n)111 n=1 n=1 m=n

m m

(B.3) = am 1/(n(m+ 1-n)) 12 m=l n=1

? 2 2 am, m=1

proving the lemma. We remark finally that the reverse inequality, I an, ' 2 : gn/V/n is valid even

without assuming that the a's are monotonic. Indeed, we may write

00 cx0 a) I a) oo 1/2

(B.4) 2~~ an= , am_ < gn 2lm

n=1 n = Il m=n n=1 m=n

and the remark follows. 3. Footnote 1, ?2. Actually Belyaev's theorem shows only that a discontinuous

Gaussian process is a.s. unbounded on any interval I. To see that X is in fact a.s. unbounded on any set S dense in I, we proceed as follows. If S' c I is a countable set which acts as a set of separability for X, then, since X is unbounded on I, sup [I X(t)I: t E SUS'] = oo, a.s. But sup [I X(t)I : t E S] = sup [IX(t)I: t E SUS'] w.p.l. because every point t E S' is a limit of points tn E S and so X(tn,') -* X(t) a.s. along some subsequence {tnl} of {tn} since X(tn) -? X(t) in the mean. Since S' is countable we see that sup [IX(t)l : t e S]=sup [IX(t)l : t e SUS']=oc w.p.l. Thus the claim that (2.3) is sufficient for the theorem is proved.

4. Footnote 2, ?4. To see that a discontinuous Gaussian process is not sequen- tially continuous, note first that as a result of Appendix 3 above if X is discon- tinuous and a < b,

(D.1) P(jX(a+(b-a)j/n)l < M,j = 0, 1 ... . n) -O0, as n -s o.

Define t2 = t4 = ... = 0 and choose 0 < aj < b1, j= 1, 2, .. ., with bj 4 0. Define tl, t3 ... and n(O) =0< n(1) < ... inductively as follows: Choose for j= 1, 2, .... t2n(j-l)+ 1, t2n(j_1)+3, . . ., t2n(j)-1 so each belong to (aj, b1) and together satisfy for each j= 1, 2,. ..

(D.2) P(IA(t2i + 1)l I< j, n(j -1) _ i < n(j)) < I /j,

which can be done on account of (D. 1). We see that tn- 0 but X(tn) do not have a limit because P(X(t2,,+ 1)-X(t2n) -O) =P(X(t2n+ l)-X(O) -O) P(I X(t2n,+ l)lI n = 1, 2... . is bounded)= 0 by (D.2). Thus X is not sequentially continuous.

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