LIMIT THEOREMS FOR STOCHASTIC PROCESSES

Transcription

1 LIMIT THEOREMS FOR STOCHASTIC PROCESSES A. V. Skorokhod 1 1 Introduction 1.1. In recent years the limit theorems of probability theory, which previously dealt primarily with the theory of summation of independent random variables, have been extended rather widely to the theory of stochastic processes. Among the work done in this connection, we mention the articles of A. N. Kolmogorov [1, 2], M. Donsker [3], I. I. Gikhman [4], Yu. V. Prokhorov [5, 6], the author [7, 8], and N. N. Chentsov [9]. Kolmogorov, Donsker, Gikhman and the author treat various more or less special important cases of this kind of limit theorems. Yu. V. Prokhorov indicates a general approach to limit theorems for stochastic processes, basing his work on compactness criteria of measures in a complete separable metric space. It seems to me that the restriction to a complete metric space is not very natural, since in every specific case it becomes necessary to find in the trajectory space of the random process a complete metric satisfying definite conditions. This is not always possible, and even when it is, it is not always simple. The present article suggests a new approach to limit theorems which can be used for many topological spaces in which in general no complete metric exists Limit theorems for stochastic processes contain primarily statements from which one can tell whether or not certain quantities f [ξ n (t)], which depend on a sequence of random processes ξ n (t), have the same distribution for n as the distribution of f [ξ (t)], where ξ (t) is the process which is the limit of the sequence ξ n (t) (assuming, usually, that the finite dimensional distributions of ξ n (t) converge weakly to the finite dimensional distributions of ξ (t)). Such limit theorems are useful for the following two reasons. In the first place, the limit process ξ (t) often has a simpler structure; moreover there often exists a good analytic apparatus for its analysis, as is true, for instance, in the case of convergence to processes with independent increments or to Markov processes, describable by differential and integro-differential equations. For these cases it is often much easier to analyze the distribution of f [ξ (t)] than it is to analyze that of f [ξ n (t)] and this can be done when we have the proper limit theorems (an example is Kolmogorov s well known statistical criterion). Secondly, a process ξ (t) can often be approximated by a sequence of processes ξ n (t) whose structure is simpler. In order to analyze the properties of ξ (t) in this way we must know which properties of ξ n (t) are maintained in the transition to the limit, and this can also be reduced to studying the limit distributions of f [ξ n (t)], where f is some function defined on the trajectory of the process. Although such treatments are still, very rare and often not entirely rigorous (an example is Feller s 1 Original paper published in Teorija veroyatn. i prim. 1 (1956), no.3, English translation: Theory of Probability and its Applications 1 (1956), no.3, Springer International Publishing Switzerland A. A. Dorogovtsev et al. (eds.), Selected Works

2 20 1 Limit Theorems for Stochastic Processes [10] attempt to analyze some Markov processes by approximating them by discrete ones), analysis of stochastic processes based on approximating them by simpler ones are quite powerful and may soon be very widely used. It therefore becomes even more important to have limit theorems for stochastic processes We shall obtain the limit theorems in this article in the following way. Consider a sequence of processes ξ n (t) whose finite dimensional distributions converge to the finite dimensional distributions of the process ξ (t). We shall assume that with probability 1 the trajectories of both ξ n (t) and ξ (t) belong to a certain set K of functions x(t), and it is natural to pose that the functions of K are determined by their values on a certain countable everywhere dense set N of values of t. It then turns out that one can construct processes ξ n (t) and ξ (t) so that ξ n (t) converges with probability 1 to ξ (t) for all t in N, so that ξ n (t) and ξ n (t) as well as ξ (t) and ξ (t) have the same distributions, and so that the trajectories of ξ n (t) and ξ (t) belong to K with probability 1. Now let F be a class of functions defined on K, which are continuous in some topology S defined in K. We will find the conditions which must be satisfied by a sequence x n (t) in K which converges to x(t) in K for t in N, in order that x n (t) converges to x(t) in the topology S. If these conditions are fulfilled by the sequence ξ n (t), then ξ n (t) converges with probability 1 to ξ (t) in the topology S, and this means that for any f F the sequence f [ξ n (t)] converges with probability 1 to f [ξ (t)]. Furthermore, the distribution of f [ξ n (t)] will converge to the distribution of f [ξ (t)], and since the distribution of f [ξ n (t)] coincides with that of f [ξ n (t)], while that of f [ξ (t)] coincides with that of f [ξ (t)], we find that the distribution of f [ξ n (t)] converges to the distribution of f [ξ (t)]. This method of obtaining the limit theorems is used when K is the set of all functions without second order discontinuities. This is perhaps the simplest class of functions which can be defined by its values on an everywhere dense set of values of t (except, of course, the set of all continuous functions; for them, however, the problem has been solved by Prokhorov [5]). In addition, there is a wide class of processes whose trajectories have, with probability 1, no second order discontinuities. The basic difficulty which arises here is in the choice of reasonable classes of functions, that is in the choice of the topology. In the space of functions without second order discontinuities the author has already suggested a topology [7] in connection with the study of the convergence of processes with independent increments. It was found later that the applicability of this topology is wider than was thought, and that the topology is a very natural one 2. In the present article some other topologies, occurring naturally from certain points of view, are also considered. Section 2 is devoted to a study of the convergence conditions in these topologies, convergence conditions which contain the above mentioned subsidiary conditions must be fulfilled by a sequence of functions converging on an everywhere dense set in order to have convergence in the desired topology. Section 3 contains the limit theorems for stochastic processes. I should like to note that although the suggested method is used only for a rather special choice of topology and a rather narrow class of processes, it can be used 2 See A. N. Kolmogorov, Probability Theory and Its Applications, this journal, Vol. 1, No. 2 (1956).

3 Limit Theorems for Stochastic Processes 21 for other topologies and for more complicated processes such as, for instance, for stochastic functions of many variables. If this has not been done in the present article, it is only because the theory of random functions of many variables is not yet sufficiently developed for one to be able to judge which classes of functions are the most natural to consider (except, of course, continuous functions, for which the problem has been solved by Prokhorov [5]). I take this opportunity to express my deep gratitude to E. B. Dynkin and A. N. Kolmogorov for valuable consultation during the many discussions of the results here presented. 2 The Space K X. Topologies in this Space 2.1. Let X be a complete metric separable space with elements x,y., and let ρ(x,y) denote the distance between x and y. We denote by K X the space of all functions x(t) which are defined on the interval [0,1], whose values lie in X, and which at every point have a limit on the left and are continuous on the right (and on the left at t = 1). Let us consider certain properties of the functions which belong to K X. A function x(t) will be said to have a discontinuity ρ(x(t 0 0),x(t 0 + 0)) at the point t Ifx(t) K X, then for any positive ε there exists only a finite number of values of t such that the discontinuity of x(t) is greater than ε. (This follows from the fact that if there exists a sequence for which t k t 0 such that ρ(x(t k +0),x(t k 0)) > ε, then at t 0 the function x(t) would have no limit either on the right or on the left.) Let t 1,t 2,,t k be all the points at which x(t) has discontinuities no less than ε. Then there exists a δ such that if t t < δ and if t and t both belong to the same one of the intervals (0,t 1 ),(t 1,t 2 ),,(t k,1), then ρ(x(t ),x(t )) < ε. To prove this, assume the contrary. Then there would exist sequences t n and t n which converge to some point t 0 and belong to the same one of the intervals (0,t 1 ),,(t k,1), and the sequences would have the property that ρ(x(t n),x(t n )) ε. Now the points t n and t n lie on opposite sides of t 0 (otherwise ρ(x(t n),x(t n )) ε would be impossible), so that ρ(x(t 0 + 0),x(t 0 0)) ε. Therefore t 0 is one of the points t 1,t 2,,t k, which contradicts the statement that t n and t n belong to the same one of the intervals (0,t 1 ),(t 1,t 2 ),,(t k,1) Ifx(t) K X, then for all η > 0 there is a δ > 0 such that every point t [0,1] satisfies one of the inequalities ρ(x(t),x(t 1 )) < η, 0<t 1 t<δ ρ(x(t 1 ),x(t)) < η. 0<t t 1 <δ (2.1) This assertion follows directly from It may be stated in the following more compact way Ifx(t) K X, then

4 22 1 Limit Theorems for Stochastic Processes min[ρ(x(t 1 ),x(t));ρ(x(t),x(t 2 ))] 0 for δ 0. (2.2) δ<t 1 t<0<t 2 t<δ Finally, the more or less converse assertion can also be made Ifx(t) satisfies (2.2) then, there exists a x(t) K X which coincides with x(t) at all of its points of continuity. We shall show that x(t) has, at every point, a limit from the left and from the right. If this were not so, there would exist an η > 0 and three points t 1 < t 2 < t 3 arbitrarily close to each other such that ρ(x(t 1 ),x(t 2 )) > η and ρ(x(t 2 ),x(t 3 )) > η. But this contradicts (2.2). We note also that x(t) must definitely coincide either with x(t 0) or x(t + 0). Setting x(1) =lim t J x(t) and x(t) =lim x(t ), we obtain the desired function We shall now go on to define the topologies in the space K X. A general analysis of this problem (that is, of classifying the possible topologies and studying their properties) is probably of independent interest. Here we shall discuss the five topologies which are simplest from our point of view, and to which we are led by considering functionals of trajectories of processes usually studied in probability theory. In addition, these are natural generalizations of the uniform topology on the space of continuous functions, generalizations in harmony with the whole development of mathematics DEFINITION. The sequence of functions x n (t) converges uniformly to x(t) at the point t 0 if for all ε > 0 there exists a δ such that lim n t t 0 <δ t t t >t ρ(x n (t),x(t)) < ε. (2.3) Obviously if x n (t) converges uniformly to x(t) at every point of some closed set, then x n (t) converges uniformly to x(t) on this whole set. A necessary condition for convergence in all the topologies considered here will be that the sequence x n (t) converges to x(t) uniformly at every point of continuity of x(t). Thus for continuous functions, all topologies reduce to ordinary uniform convergence. Our topologies begin to differ when we consider the behavior of x n (t) in the neighborhood of points of discontinuity of x(t). The simplest behavior we might suggest is that the convergence be uniform also at points of discontinuity, in which case we obtain ordinary uniform convergence (in our notation we shall call this topology U). Although we shall consider this topology in addition to others, we remark that it is not the most natural one for K X. This is because the uniform topology in K X requires that the convergence of x n (t) to x(t) imply that there exists a number such that for all n greater than or equal to this number the points of discontinuity of x n (t) coincide with the points of discontinuity of x(t). This means that if t is considered to be the time, we must assume the existence of an instrument which will measure time exactly, and physically this is an impossibility. It is much more natural to pose

5 Limit Theorems for Stochastic Processes 23 that the functions we can obtain from each other by small deformation of the times scale lie close to each other. We thus are led to propose the topology J DEFINITION. The sequence x n (t) is called J 1 -convergent to x(t) if there exists a sequence of continuous one-to-one mappings λ n (t) of the interval [0,1] onto itself, such that lim n t ρ(x n (t),x(λ n (t))) = 0; lim n t λ n (t) t = 0. (2.4) Let us analyze in more detail how we define the behavior of x n (t) in the neighborhood of a discontinuity point t 0 if we require that x n (t) converges to x(t) in one of our topologies. Since x(t) has a limit from the left and the right at every point t, we know that in a sufficiently small neighborhood of t 0 it will be approximately equal to x(t 0 0) for t < t 0 and to x(t 0 +0) for t t 0. Therefore on the left of t 0 there exist intervals where x n (t) is approximately equal to x(t 0 0), and on the right there are those in which x n (t) is approximately equal to x(t 0 + 0) (since the points of continuity of x(t) are everywhere dense, and at these points the convergence is uniform). Thus the question of how x n (t) behaves in the neighborhood of t 0 is equivalent to the question how x n (t) changes from x(t 0 0) to x(t 0 + 0). In the topologies U and J 1 this transition takes the form of a single jump. In both of these topologies, for values of t close to t 0 the function x n (t) can take on values which are either close to x(t 0 0) or to x(t 0 + 0). If we wish to keep this last property, but do not require that the transition be in the form of a single jump, that is that a function x n (t) may change back and forth between the values x(t 0 0) and x(t 0 +0) several times in the neighborhood of a point t 0, then we obtain topology J DEFINITION. A sequence x n (t) is said to be J 2 -convergent to x(t) if there exists a sequence of one-to-one mappings λ n (t) of the interval [0,1] onto itself such that t λ n (t) t 0 and ρ(x n (t),x(λ n (t))) 0 as n. (2.5) t We shall define the two following topologies for the case in which X is a complete linear normalized space. In this case, together with every function belonging to K X we shall consider its graph Γ x(t), constructed in the following way. Let X T be the space composed of the pairs (x,t), where x X and t [0,1]. In this space we define the metric R by R[(x 1,t 1 );(x 2,t 2 )] = ρ(x 1,x 2 )+ t 1 t 2. (2.6) We then define the graph Γ x(t) as the closed set in X T which contains all pairs (x,t), such that for all t the point x belongs to the segment joining x(t 0) and x(t + 0) (henceforth we shall denote the segment joining x 1 and x 2 by [x 1,x 2 ]). The graph Γ x(t) is a continuous curve in X T, and therefore we can define convergence for graphs in the same way as we usually do for space curves. We then obtain the topology M DEFINITION. The pair of functions (y(s), t(s)) gives a parametric representation of the graph Γ x(t) if those and only those pairs (x,t) belong to it for which

6 24 1 Limit Theorems for Stochastic Processes an s can be found such that x = y(s),t = t(s), where y(s) is continuous, and t(s) is continuous and monotonically increasing (the functions y(s) and t(s) are defined on the segment [0,1]). We note that if (y 1 (s),t 1 (s)) and (y 2 (s),t 2 (s)) are parametric representations of Γ x(t), there exists a monotonically increasing function λ(s) such that y 1 (s) = y 2 (λ(s)) and t 1 (s)=t 2 (λ(s)) DEFINITION. The sequence x n (t) is called M 1 -convergent to x(t) if there exist parametric representations (y(s),t(s)) of Γ x(t) and (y n (s),t n (s)) of Γ xn (t) such that lim R[(y n (s),t n (s));(y(s),t(s))] = 0. (2.7) n s We can characterize the topology M 1 in the following way from the point of view of the behavior at a point of discontinuity t 0 of the function x(t). The transition from x(t 0 0) to x(t 0 + 0) is such that first x n (t) is arbitrarily close to the segment [x(t 0 0),x(t 0 )] and second that x n (t) moves from x(t 0 0) to x(t 0 ) almost always advancing. More rigorously, what this last condition means is that if x(t 0 0) < x 1 < x 2 < x(t 0 ) (using the natural ordering for the segment), then having reached the neighborhood of x 2, we will not return to the neighborhood of x 1. If we drop the second requirement and assume only that x n (t) goes from x(t 0 0) to x(t 0 + 0) remaining always arbitrarily close to the segment [x(t 0 0),x(t 0 )], we obtain the topology M DEFINITION. The sequence x n (t) is called M 2 -convergent to x(t) if lim inf n (y 1,t 1 ) Γ x(t) (y 2,t 2 ) Γ xn(t) R[(y 1,t 1 );(y 2,t 2 )] = 0. (2.8) Let S be any one of our topologies. We shall denote convergence in the topology S by the symbol x n (t) S x(t). Let us consider the relation between our topologies. It is clear that U is stronger than J 1, and that this in turn is stronger than J 2. It is also clear that M 1 is stronger than M 2. We recall that a topology S 1 is stronger than S 2 if convergence in S 1 implies convergence in S 2. If X is a linear space, we can use any of our topologies. It is easily seen that convergence in M 2 follows from convergence in any of the other topologies, and that convergence in J 1 implies convergence in any of the other topologies except U. Examples can be used to show that the topologies M 1 and J 2 cannot be compared. (Indeed, let X be the real line. We set { 0;t < 1 x(t)= 2, 1;t 1 2, [ ) 0;t 0, n, ( ) [ ) x n(t)= n 2 t n ; t 12 1 n, n, [ ] 1; t n,1,

7 Limit Theorems for Stochastic Processes 25 0;t < 1 2 1/n, 1; 1 2 n(t)= 1/n t < 1 2, 0; 1 2 t < /n, 1; t /n, x so that x n(t) M 1 x(t) although this does not converge 3 to x(t) in J 2, while x n(t) J 2 x(t) although this does not converge to x(t) in M 1 ). Denoting the statement the topology S 1 is stronger than S 2 bys 1 S 2 all the above can be summarized by M 1 U J 1 M 2. (2.9) J 2 Let us also mention the relation between our topologies and functionals ordinarily used in probability theory. We shall consider X to be the real line. Consider the following functionals ( ) The upper and lower bound of a function on some segment [t 1,t 2 ], namely M [t1,t 2 ][x(t)] = x(t); m [t1,t 2 ][x(t)] = inf x(t). t 1 t t t 2 1 t t The number of intersections ν [a,b] [t 1,t 2 ] [x(t)] with the strip [a,b] on the segment [t 1,t 2 ] is equal to k if it is possible to find k + 1 points t (0) < t (1) < < t (k) on [t 1,t 2 ] such that either x(t (0) ) a,x(t (1) ) b,x(t (2) ) a,, or x(t (0) ) b,x(t (1) ) a,x(t (2) ) b,, and is impossible to give k+2 points with these properties The first value γ + [t 1,t 2 ],a [x(t)] above some level a on the segment [t 1,t 2 ]. Let us first define γ + [0,1],a[x(t)]. Let τ a [x(t)] = inf x(t) a t, if x(t) a and τ a [x(t)] = 1, if x(t) < a. Then γ + [0,1],a [x(t)] = { x(τ a [x(t)]) a for τ a [x(t)] < 1, 1 for τ a [x(t)] = 1. Further, we set x(t 1 + 0),t < t 1, γ + [t 1,t 2 ],a [x(t)] = γ+ [0,1],a[x(t)], where x(t)= x(t),t 1 t t 2, x(t 2 0),t > t 2. The following assertions are easily proved, but we shall not present the proofs here A necessary and sufficient condition for x n (t) M 2 x(t) is that M [t1,t 2 ][x n (t)] M [t1,t 2 ][x(t)] and m [t1,t 2 ][x n (t)] m [t1,t 2 ][x(t)], where t 1 and t 2 are points of continuity of x(t) A necessary and sufficient condition for x n (t) M 1 x(t) is that 3 Remark of the editors. We define the sequence (x n(t)) t [0,1] above in a way different from that given mistakenly by the author: 0;t < 1/2 1/n; x n(t)= n(t 1/2)+1/2; 1/2 1/n t < 1/2 + 1/n; 1; t > 1/2 + 1/n. One can easily reveal that this sequence does not converge on any of the topologies U,J 1,J 2,M 1,M 2.

8 26 1 Limit Theorems for Stochastic Processes ν [a,b] [t 1,t 2 ] [x n(t)] ν [a,b] [t 1,t 2 ] [x(t)] for all t 1 and t 2 which are points of continuity of x(t) and almost all a and b A necessary and sufficient condition for x n (t) J 2 x(t) is that γ + [t 1,t 2 ],a [x n(t)] γ + [t 1,t 2 ],a [x(t)] for almost all a and for all t 1 and t 2 which are points of continuity of x(t) A necessary and sufficient condition for x n (t) J 1 x(t) is that γ + [t 1,t 2 ],a [x n(t)] γ + [t 1,t 2 ],a[x(t)] and ν[a,b] [t 1,t 2 ] [x n(t)] ν [a,b] [t 1,t 2 ] [x(t)] for all t 1 and t 2 which are points of continuity of x(t) and almost all a and b. Let us now go on to a more detailed analysis of our topologies, starting with the weaker ones and going on to the stronger The topology M 2. We shall first find a condition for M 2 -convergence which is equivalent to the definition but which is given not in terms of graphs but of the functions x n (t) and x(t) themselves A necessary and sufficient condition for x n (t) M 2 x(t) is that lim lim Δ M c 0 n 2 (c;x n (t),x(t)) = 0, (2.10) where Δ M2 (c;y(t),x(t)) = H(x(t),y(t c ),x(tc )), t c = max[0,t c], tc = min[1,t + c], and H(x 1,x 2,x 3 ) is the distance of x 2 from the segment [x 1,x 3 ]. Proof. NECESSITY. Let x n (t) M 2 x(t). We shall show that this implies that by choosing c sufficiently small we can make lim n Δ M2 (c;x n (t),x(t)) arbitrarily small. Let c be such that if t 1 < t 2 < t 3,t 2 t 1 < 2c, and t 3 t 2 < 2c, then min[ρ(x(t 1 ),x(t 2 ));ρ(x(t 2 ),x(t 3 ))] < ε. Let us now denote by t 1,t 2,,t k all the points at which the discontinuities are greater than or equal to ε, and let c be so small that ρ(x(t ),x(t )) < ε so long as t t < c and t and t both belong to the same one of the intervals (0,t 1 ),(t 1,t 2 ),,(t k,1). We choose n so large that (y 1,t 1 ) Γ x(t) inf (y 2,t 2 ) Γ xn(t) R[(y 1,t 1 );(y 2,t 2 )] < c/2. (2.11) Now if the points t c,t,tc all belong to the same one of the intervals (0,t 1 ),(t 1,t 2 ),, (t k,1), then H(x(t c ),x n (t),x(tc )) c/2 + 2ε. For, it follows from (2.11) that the graph Γ x(t) contains a point (x,t) such that t t < c/2 and ρ(x,x n (t)) < c/2, and hence if t < t, then H(x(t c ),x n (t),x(t c )) < ρ(x(t c ),x(t 0)) + ρ(x(t 0),x)+ρ(x,x n (t)),

9 Limit Theorems for Stochastic Processes 27 while if t > t, then H(x(t c ),x n (t),x(t c )) < ρ(x(t c ),x(t + 0)) + ρ(x(t + 0),x)+ρ(x,x n (t)). Let us now find H(x(t c ),x n (t),x(t c )) when t c < t i < t < t c (the case t c < t < t i < t c is entirely symmetric). It follows from (2.11) that Γ x(t) contains a point ( x, t) such that ρ( x,x n (t)) < c/2 and t t < c/2. If t t, then either t i > t and H(x(t c ),x n (t),x(t c )) ρ(x(t c ),x( t 0)) + ρ(x( t 0), x)+ρ( x,x n (t)) 3ε + c/2, or t i < t and H(x(t c ),x n (t),x(t c )) ρ(x(t c ),x( t + 0)) + ρ(x( t + 0), x)+ρ( x,x n (t)) 3ε + c/2. If, however, t i = t, then H(x(t c ),x n (t),x(t c )) H(x(t c ), x,x(t c )) + ρ( x,x n (t)) ρ(x(t c ),x(t i 0)) + ρ(x(tc ),x(t i + 0)) + H(x(t i 0), x,x(t i + 0)) + ρ( x,x n (t)) 2ε + c/2, since H(x(t i 0), x,x(t i + 0)) = 0. This proves the necessity. SUFFICIENCY. Assume that (2.10) is fulfilled. This means that there exists a c sufficiently small so that for n sufficiently large we have Δ M2 (c;x n (t),x(t)) < ε, (2.12) and in addition that c can be chosen so small that if t and t belong to the same one of the intervals (0,t 1 ),(t 1,t 2 ),,(t k,1); t t < 2c, then ρ(x(t ),x(t )) < ε and 2c is less than the length of any of these intervals. Let us show that for every point of Γ xn (t) there is a point of Γ x(t) such that in the metric R the distance between these points is less than δ, and that this last quantity can be made arbitrarily small by appropriate choice of ε and c. From the fact that H(x(t c ),x n (t),x(tc )) < ε, if t c and tc belong to the same one of the intervals (0,t 1 ),(t 1,t 2 ),,(t k,1), it follows that ρ(x n (t),x(t c )) < 3ε, which means that ρ(x n (t),x(t)) < 4ε. (2.13) In this case, therefore, for a point (y,t) of Γ xn (t) we can find a point ( x,t) of Γ x(t) (with the same t) such that ρ( x,y) < 4ε. Now assume that t c < t i < t < tc. Then H(x(t c ),x n (t),x(t c )) < ε and H(x(t c ),x n (t 0),x(t c )) < ε.

10 28 1 Limit Theorems for Stochastic Processes For every point of the segment [x n (t 0),x n (t)], therefore, there is a point x on [x(t c ),x(tc )] such that ρ(x,x) < 2ε. Now since ρ(x(t c ),x(t i 0)) < ε and ρ(x(tc ),x(t i + 0)) < 2ε, it follows that on the segment [x(t i 0),x(t i + 0)] there is a point x such that ρ(x, x) < 2ε. Thus R[(y,t),( x,t i )] < 4ε + c. This proves the sufficiency REMARK. We may consider, instead of H(x 1,x 2,x 3 ), any other quantity such that ϕ 1 (H(x 1,x 2,x 3 )) H(x 1,x 2,x 3 ) ϕ 2 (H(x 1,x 2,x 3 )), where ϕ i (t) are nonnegative continuous functions which vanish only for t = 0. If X is a Hilbert space, we may set H(x 1,x 2,x 3 )=ρ(x 1,x 2 )+ρ(x 2,x 3 ) ρ(x 2,x 3 ) GENERAL CONDITIONS FOR M 2 -CONVERGENCE. Necessary and sufficient conditions for x n (t) M 2 x(t) are that (a) x n (t) converges to x(t) on a set of values of t which contains 0 and 1 and which is everywhere dense on [0,1] and (b) lim c 0 lim n Δ M2 [c;x n (t)] = 0, where Δ M2 [c;y(t)] = t [0,1];t 1 [t c,t c +c/2] t 2 [t c c/2,t c ] H(y(t 1 ),y(t),y(t 2 )). Proof. The necessity of (a) is implied by the fact that if x n (t) M 2 x(t), then at every point of continuity of x(t). NECESSITY OF (b). x n (t) x(t) H(x n (t 1 ),x n (t),x n (t 2 )) H( x 1,x n (t), x 2 )+ρ( x 1,x n (t 1 )) + ρ( x 2,x n (t 2 )). Since x n (t) M 2 x(t), there exist points t 1 and t 2 such that t 1 t 1, t 2 t 2,ρ( x 1,x n (t 1 )), and ρ( x 2,x n (t 2 )) are small, where the points x 1 and x 2 belong to the segments [x(t 1 0),x(t 1 )] and [x(t 2 0),x(t 2 )], respectively. Now let c be such that x(t) cannot have two discontinuities equal to or greater than ε in a distance 3c, and such that if t 2 t 1 2c and if there is no discontinuity between these points which is equal to or greater than ε, then ρ(x(t 2 ),x(t 1 )) < ε. Let us evaluate H( x 1,x n (t), x 2 ) assuming that t 1 t 1 < c/4 and t 2 t 2 < c/4. If in (t 1,t 2 ) and (t c,tc ) there are no discontinuities of x(t) equal to or greater than ε, then H( x 1,x n (t), x 2 ) H(x(t c ),x n (t),x(t c )) + 4ε. If, however, one of the intervals (t 1,t 2 ) or (t c,t c ) has a discontinuity equal to or greater than ε at some point t, we consider the following two possibilities.

11 Limit Theorems for Stochastic Processes 29 1) t 1 < t < t 2 ; t c < t < t c. Then ρ(x(t c ), x 1 ) 2ε and ρ(x(t c ), x 2 ) 2ε, so that H(x 1,x n (t), x 2 ) 4ε + H(x(t c ),x n (t),x(t c )). 2) t lies between t 1 and t c (and may coincide with t 1 ). Then in the interval (t c/4,tc/4 ) all the discontinuities of x(t) are less than ε, so that ρ(x(t c/4),x(tc/4 )) < ε. This means that ρ(x(tc/4 ),x n(t)) < ε + H(x(t c/4 ),x n (t),x(tc/4 )). In addition, since ρ(x(tc/4 ),x(t c )) < ε, and the discontinuity in x(t) at t 2 is also less than ε, we have ρ(x n (t), x 2 ) < 3ε + H(x(t c/4 ),x n (t),x(t c/4 )). Therefore H( x 1,x n (t), x 2 ) < 3ε + H(x(t c/4 ),x n (t),x(tc/4 )). This proves the necessity. SUFFICIENCY. We must show that if conditions (a) and (b) are fulfilled, then lim c 0 lim n Δ M 2 (c;x n (t),x(t)) = 0. Let us assume, on the contrary, that there exist sequences c k 0,n k j (where the arrow indicates that we are considering the limit as j ) and an η such that Δ M2 (c;x n (t),x(t)) > η. This means that there exist points t nk j 1,tnk j 2,tnk j 3, such that tnk j 3 tnk j 2 = tnk j 2 tnk j 1 = c k (since the t nk j i lie, for sufficiently small c k, within the segment [0,1] because (a) and (b) imply that x n (t) converges uniformly to x(t) at 0 and 1) and H(x(t nk j 1 ),x n k j (tnk j 2 ),x(tnk j 3 )) > η. Condition (a) implies that for all ε and δ there is an n 0 such that in any interval of length δ there exists a point t for which ρ(x n (t ),x(t )) ε when n > n 0. Let δ 1 be so small that δ 1 <t 1 t 2 <0<t 3 t 2 <δ 1 min[ρ(x(t 1 ),x(t 2 ));ρ(x(t 2 ),x(t 3 ))] < ε. (2.14) Further, (b) implies that there exist c 1,c 2, and n such that for n > n we have Δ M2 (c,x n (t)) < ε; Δ M2 (c 2,x n (t)) < ε; c 1 > 2c 2 ; c 1 > c k > c 2 ;2c 1 < δ 1. (2.15) We then have either ρ(x(t nk j 1 ),x(t)) < 2ε, t (tnk j 2 c 1,t nk j 2 c 1/2),

12 30 1 Limit Theorems for Stochastic Processes or ρ(x(t nk j 3 ),x(t)) < 2ε, t (tnk j 2 + c 1/2,t nk j 2 + c 1), ρ(x(t nk j 1 ),x(t)) < 2ε, t (tnk j 2 c 2,t nk j 2 c 2/2), ρ(x(t nk j 3 ),x(t)) < 2ε, t (tnk j 2 + c 2/2,t nk j 2 + c 2) (this follows from (2.14) and from the fact that the four intervals do not intersect in pairs). If we now choose δ < c 2 /2, then for n k j > max(n 0,n ) there exist three points t 1,tnk j 2,t 3 such that H(x n k j (t 1),x n k j (t nk j 2 ),x n k j (t 3)) > η 12ε, and either c 1 /2 t nk j 2 t j c 1 or c 2 /2 t nk j 2 t j < c 2,i = 1,3. Thus either Δ M2 (c 1,x n k(t)) > η 12ε or Δ M2 (c 2,x j n k(t)) > η 12ε, which contradicts (2.15) j if we take 13ε < η. We note that in proving the sufficiency we have proved the following stronger assertion A sufficient condition for x n (t) M 2 x(t) is that (a) be fulfilled and that (b ) lim c 0 lim n Δ M2 (c,x n (t)) = 0. Thus we can obtain various sufficient conditions by making different choices of sequences c k 0. In studying the compactness of sets of functions in the topologies J 1,J 2,M 1, and M 2, the following lemma plays an important role Consider a sequence x n (t) which converges to an arbitrary function x(t) for t N (some set, everywhere dense on [0,1], containing 0 and 1), and let lim lim (Δ M c 0 n 2 (c,x n (t)) + ρ(x n (t),x n (0)) + 0<t<c Then there exists x(t) K X such that x n (t) M 2 x(t). 1 c<t<1 Proof. Let t i be the points of N. Let us prove the existence of lim t i t,t i >t x(t i) and lim x(t i). t i t,t i <t ρ(x n (1),x n (t))) = 0. It is sufficient to prove that one of these limits, say the limit from the right, exists. First let t = 0. Let us assume that lim ti 0 x(t i ) does not exist, which means that there is a sequence t ik and an ε such that ρ(x(t ik ),x(t ik+1 )) > ε. Let us choose c so small that lim n Δ M2 (c,x n (t)) < ε/4, and k such that t ik and t ik+1 are both less than c/2. Then, if t r is a point such that t ik + c/2 < t r < t ik + c and t ik+1 + c/2 < t r < t ik+1 + c,

13 Limit Theorems for Stochastic Processes 31 we have H(x(t ik ),x(t ik+1 ),x(t r )) < ε/4 and H(x(t ik+1 ),x(t ik ),x(t r )) < ε/4. But this contradicts the fact that ρ(x(t ik ),x(t ik+1 )) > ε. Let us now assume that the limit lim ti t,t i >t x(t i ) does not exist, where t is an internal point of [0,1]. We then have the following two possibilities. 1) There exists an ε such that in any neighbourhood of a point t we can find three points t r1 < t r2 < t r3 lying to the right of t for which ρ(x(t ri ),x(t r j )) > ε (i j;i, j = 1,2,3). 2) There exist x 1 and x 2 such that every sequence x(t nk )(t nk > t,t nk t) can be separated into two sequences x(t n ) and x(t k n ) (the set {n k} is the union of {n k k } and {n k }) such that lim x(t t n t n )=x 1 and lim x(t k t n t n )=x 2. k k k We shall show that both of these possibilities contradict the conditions of the lemma. The first case. Choose c 0 so that lim n Δ M2 (c,x n (t)) < μ if c < c 0 (we shall choose μ later). Let t and t be points of N such that if 0 < t t < c 0 /2, then, t c 0 < t < t < c 0 /2 and t + c 0 /2 < t < t + c 0 /2. Then, for t < t i < t,x(t i ) will be at a distance equal to or greater than μ from the segment [x(t ),x(t )]. Now let A be the set of points x of the segment [x(t ),x(t )] for which lim ti t,t i >t ρ(x(t i ),x) μ, and let us write x 1 = inf x A x and x 2 = x A x (the order on the segment [x(t ),x(t )] is defined by the fact that we consider x(t ) < x(t )). Now consider a point x of A such that ρ(x 1,x) > ε 2μ and ρ(x 2,x) > ε 2μ. If t and t are points which, for some c < c 0, satisfy t c < t < t c/2, t + c/2 < t < t + c for 0 < t t < c/4 and ρ(x,x(t )) < μ, we find that if 0 < t t < c/4, the point x( t) lies at a distance no greater than 2μ from the segment [x(t ),x]. Therefore x 1 and x 2 must lie at a distance no greater than 3μ from [x(t ),x] which is impossible if ε > 5μ. In the second case we choose, for every ε, a δ such that for 0 < t 1 t < δ we have either ρ(x(t 1 ),x 1 ) < ε or ρ(x(t 1 ),x 2 ) < ε. If ε < ρ(x 1,x 2 )/4, only one of the possibilities is realized. Choose c 0 such that lim n Δ M2 (c,x n (t)) < ε for c < c 0. Then if t 1 is a point for which 0 < t 1 t < δ and 0 < t 1 t < c 0 and ρ(x 1,x(t 1 )) < ε, we find that ρ(x( t),x 1 ) < ε for (t 1 t)/3 < t t < 2 3 (t 1 t) and t N, since and therefore H(x 1,x( t),x(t 1 )) < ε, H(x 1,x( t),x 1 ) < 2ε, or ρ(x 1,x( t)) < 2ε, and therefore ρ(x 1,x( t)) < ε (there are only two possibilities, namely ρ(x 1,x( t)) < ε, or ρ(x 1,x( t)) > 3ε). Choosing t 2 from N such that 1 2 t 1 < t 2 < 2 3 t 1, we find that ρ(x( t),x 1 ) < ε for (t 2 t)/3 < t t < 2 3 (t 2 t). Continuing this process ad infinitum, we find that ρ(x( t),x 1 ) < ε for t < t < 2 3 t 1 and therefore lim t>t, t t x( t) = x 1. Thus x(t) has limits from the right and the left at every point.

14 32 1 Limit Theorems for Stochastic Processes Setting x(t)= lim t>t, t t, t N x( t), x(1)= lim t N, t 1 x( t), we obtain a function from K X. We now have to prove that x 1 (t) M 2 x(t). To do this it is sufficient to show that x n (t) converges to x(t) at all points of continuity of x(t). Let t 0 be a point of continuity of x(t). Then if we choose t and t from N such that t 0 c < t < t 0 c/2 and t 0 + c/2 < t < t 0 + c, where c is chosen so that lim n Δ M2 (c,x n (t)) < ε, ρ(x(t ),x(t 0 )) < ε, and ρ(x(t ),x(t 0 )) < ε, we find that lim n ρ(x(t 0 ),x(t 0 )) < 4ε. This proves the lemma The topology M CONVERGENCE CONDITIONS. A necessary and sufficient condition that x n (t) M 1 x(t) is that (a) be fulfilled and that (b) lim c 0 lim n Δ M1 (c;x n (t)) = 0, where Δ M1 (c;y(t)) = H(y(t 1 ),y(t 2 ),y(t 3 )). t 2 c<t 1 <t 2 <t 3 <t 2 +c Proof. THE NECESSITY OF (b). Let x n (t) M 1 x(t). This means that there exist parametric representations (y(s),t(s)) for Γ x(t) and (y n (s),t n (s)) for Γ xn (t) such that lim n s t n (s) t(s) = 0 and lim n s ρ(y(s),y n (s)) = 0. Let t 1 < t 2 < t 3 and s (n) 1,s(n) 2,s(n) 3 be chosen so that t n (s (n) i )=t i,(i = 1,2,3). Then H(x n (t 1 ),x n (t 2 ),x n (t 3 )) = H(y n (s (n) 1 ),y n(s (n) 2 ),y n(s (n) 3 )) H(y(s(n) 1 ),y(s(n) 2 ),y(s(n) 3 )) +3ρ(y n (s),y(s)). s It remains to prove that for s 1 < s 2 < s 3 we can make H(y(s 1 ),y(s 2 ),y(s 3 )) arbitrarily small if t(s 3 ) t(s 1 ) is sufficiently small. Let us consider three cases. 1. If t(s 1 )=t(s 2 )=t(s 3 ), so that H(y(s 1 ),y(s 2 ),y(s 3 )) = If t(s 1 )=t(s 2 ) < t(s 3 ), then either the discontinuity in x(t) at t(s 1 ) is less than ε, i.e. ρ(y(s 1 ),y(s 2 )) < ε and therefore H(y(s 1 ),y(s 2 ),y(s 3 )) < ε, or the discontinuity in x(t) at t(s 1 ) is greater than ε and then for sufficiently small t(s 3 ) t(s 2 ) we have ρ(y(s 3 ),x(t(s 1 ))) < 2ε. This means that H(y(s 1 ),y(s 2 ),y(s 3 )) < 2ε. 3. If, finally, t(s 1 ) < t(s 2 ) < t(s 3 ), then either there is no discontinuity at any of these points greater than ε and therefore H(y(s 1 ),y(s 2 ),y(s 3 )) < H(x(t(s 1 )),x(t(s 2 )),x(t(s 3 ))) + 3ε (2.16) (it follows from that (2.16) is small), or at t(s 1 ) the discontinuity is greater than ε. This means that at the other points the discontinuities are less than ε

15 Limit Theorems for Stochastic Processes 33 for sufficiently small t(s 3 ) t(s 1 ) so that ρ(x(t(s 2 )),x(t(s 3 ))) is small and so is H(y(s 1 ),y(s 2 ),y(s 3 )) (if the discontinuity greater than ε is at t(s 3 ), the considerations are similar), or the discontinuity greater than ε is at t(s 2 ) in which case ρ(y(s 1 ),x(t(s 2 0))) and ρ(x(t(s 2 )),y(s 2 )) are small, and therefore so is H(y(s 1 ),y(s 2 ),y(s 3 )). This proves the necessity. SUFFICIENCY. First we shall show that if Δ M1 (c;x(t)) < h and if (x 1,t 1 ), (x 2,t 2 ),(x 3,t 3 ) are points of Γ x(t) such that t 2 c < t 1 < t 2 < t 3 < t 2 + c, then H(x 1,x 2,x 3 ) < h. To do this it is sufficient to prove the following two assertions: 1) if H(x 1,x 2,x 3 ) < h, H(x 1,x 2,x 3 ) < h, and x 1 [x 1,x 1 ], then H(x 1,x 2,x 3 ) < h; 2) if H(x 1,x 2,x 3) < h,h(x 1,x 2,x 3) < h, and x 2 [x 2,x 2 ], then H(x 1,x 2,x 3 ) < h. Assertion (1) follows from the fact that if y [x 1,x 3] and y [x 1,x 3], where ρ(y,x 2 ) < h and ρ(y,x 2 ) < h, then any point y on the intersection of [y,y ] and [x 1,x 3 ] will also satisfy ρ(x 2,y) < h. Assertion (2) follows from the fact that if y and y are points of [x 1,x 3 ] such that ρ(y,x 2 ) < h,ρ(y,x 2 ) < h, and x 2 = αx 2 +(1 α)x 2,(0 < α < 1), then we know about y + αy +(1 α)y that it satisfies ρ(x 2,y)=ρ(αx 2 +(1 α)x 2,αy +(1 α)y ) ρ(αx 2,αy )+ρ((1 α)x 2,(1 α)y ) αρ(x 2,y )+(1 α)ρ(x 2,y ) h (it should be recalled that although in this case X is a Banach space, we are using the distance rather than the norm for uniformity of notation). To prove our assertion, it is sufficient to construct, for every ε and for all sufficiently large n, parametric representations (y ε n(s),t ε n (s)) for Γ xn (t) and (y ε (s),t ε (s)) for Γ x(t) such that s tn ε (s) t ε (s) < ε; ρ(yn(s),y ε ε (s)) < ε. s Let (y(s),t(s)) be some parametric representation of Γ x(t) and let 0 = s 0 < s 1 < s 2 < < s m = 1 be values of the parameter such that t(s i ) are either points of continuity of x(t) or points where the discontinuities are no less than ε, and further let t(s i ) t(s i 1) < η for i = 1,,m 1 and min[ρ(y(s i ),y(s));ρ(y(s),y(s i+1 ))] < ε/2 for s (s i,s i+1 ). If finally t(s k )=t(s k+1 ), then ε/4 < ρ(y(s k ),y(s k+1 )) < ε/2. Let us choose an n 0 sufficiently large and an η < ε such that if t(s i ) is a point of continuity of x(t), then ρ(x n (t(s i )),x(t(s i ))) < ε/4 and Δ M1 (η,x n (t)) < ε/8 for n > n 0. Then if the point t = t(s r )= = t(s r+ j ) is a point of discontinuity of x(t) at which the discontinuity is no less than ε, then there exist on Γ xn (t) points (y (n) r,t r (n) ),(y (n) r+1,t(n) r+1 ),,(y(n) r+ j,t(n) r+ j ) such that tn r+1 t < η and ρ(y (n) r+i,y(s r+i)) < 7 8 ε (because the segment [y(s r 1),y(s r+ j+1 )] lies at a distance no greater than ε/4 from [x n (t(s r 1 )),x n (t(s r+ j+1 ))] and at a distance no greater than ε/2 from [x(t(s r ) 0),x(t(s r+ j ))], while all the points of Γ x(t) with t(s r 1 ) < t < t(s r+ j+1 ) lie no further than ε/8 from [x n (t(s r 1 )),x n (t(s r+ j+1 ))].

16 34 1 Limit Theorems for Stochastic Processes Thus if (y n (s),t n (s)) is some parametric representation of Γ xn (t), then there exist points 0 = s (n) 0 < s (n) 1 < < s (n) m = 1 such that t n(s (n) i ) t(s i ) < ε and ρ(y n (s (n) i ),y(s i )) < ε. Let λ(s) be the monotonic continuous function which maps [0,1] onto itself, such that λ(s i )=s (n) i. Then (y n (λ(s)),t n (λ(s))) will also be a parametric representation of Γ xn (t), and it is easily shown that This completes the proof. t n (λ(s)) t(s) < 2ε, ρ(y n (λ(s)),y(s)) < 4ε The topology J A necessary and sufficient condition that x n (t) J 2 x(t) is that lim lim Δ J c 0 n 2 [c;x n (t),x(t)] = 0, where Δ J2 (c;y(t),x(t)) = [min{ρ(x(t c ),y(t));ρ(y(t),x(tc ))}] t (the points t c and tc are defined as previously). Proof. NECESSITY. Let x n (t) J 2 x(t), so that there exists a sequence λ n (t) such that lim ρ(x n (t),x(λ n (t)) = 0; lim t λ n (t) = 0. (2.17) Therefore n t n t ρ(x(t c ),x n (t)) ρ(x(t c ),x(λ n (t))) + ρ(x(λ n (t)),x n (t)), ρ(x n (t),x(tc )) ρ(x(λ n (t)),x(tc )) + ρ(x(λ n (t)),x n (t)). Hence min{ρ(x(t c ),x n (t)),ρ(x n (t)),x(tc ))} ρ(x(λ n (t)),x n (t)) +min{ρ(x(λ n (t)),x(t c ));ρ(x(λ n (t)),x(tc ))}. In view of (2.17) we have only to show that (2.18) lim c 0 lim n t min{ρ(x(t c ),x(λ n (t));ρ(x(λ n (t)),x(t c ))} = 0. (2.19) But if λ n (t) t < c, we have t c λ n (t) t c, so that (2.19) follows from SUFFICIENCY. Assume there exists such a sequence c k 0 and n k that Δ J2 (c k ;x n (t), x(t)) < ε k for n > n k and ε k 0. Let t 1,t 2,,t k be the points at which the discontinuities of x(t) are no less than ε. Let us surround these points by intervals Δ 1,Δ 2,,Δ k so small that ρ(x(t),x(t i 0)) < ε for t < t i,t Δ i, ρ(x(t),x(t i + 0)) < ε for t > t i,t Δ i. (2.20)

17 Limit Theorems for Stochastic Processes 35 Let us choose c k so small that if t [0,1] k i=1 Δ j we have ρ(x(t),x(t ck 0)) < 2ε, ρ(x(t),x(t c k )) < 2ε. (2.21) Then for n > n k and t Δ i we have either ρ(x n (t),x(t i 0)) < 3ε + ε k, or ρ(x n (t),x(t i + 0)) < 3ε + ε k. (2.22) If ρ(x(t i 0),x(t i + 0)) 6ε + ε k, then only one of (2.22) can be fulfilled. Let us now construct the function λ ε,ε k n (t), which maps the segment [0,1] onto itself in the following way: λ ε,ε k n (t) =t everywhere except on those intervals Δ i where ρ(x(t i 0),x(t i + 0)) 6ε + 2ε k, and on these intervals λ ε,ε k n (t) maps the set of values of t for which ρ(x n (t),x(t i 0)) < 3ε + ε k into the set t < t i,t Δ i, and the set of values of t for which ρ(x n (t),x(t i + 0)) < 3ε + ε k into the set t t i,t Δ i. Obviously λ ε,ε k n (t) t is equal to or less than the greatest of the Δ i intervals, and ρ(x n (t),x(λ ε,ε k n (t))) 4ε + ε k for n > n k. Now by choosing the sequences ε 0,k and δ = {maxlengthδ i } 0, we obtain the desired result. It is easily seen that in proving the sufficiency we have proved the following stronger assertion A sufficient condition for x n (t) J 2 x(t) is that lim lim Δ J n 2 (c;x n (t),x(t)) = 0. c CONVERGENCE CONDITION IN THE TOPOLOGY J 2. Necessary and sufficient conditions for the functions x n (t) to be J 2 -convergent to x(t) are that (a) x n (t) converge to x(t) on an everywhere dense set of [0,1] which contains 0 and 1, and (b) lim c 0 lim n Δ J2 (c,x n (t)) = 0, where Δ J2 (c,y(t)) = t c t 1 t c +c/2,t [0,1],t c c/2 t 2 t c min[ρ(y(t 1 ),y(t));ρ(y(t),y(t 2 ))]. Proof. NECESSITY OF CONDITION (b). Let x n (t) J 2 x(t). Then there exists a sequence λ n (t) such that t λ n (t) t and t ρ(x n (t),x(λ n (t))) 0asn. Therefore ρ(x n (t 1 ),x n (t)) ρ(x(λ n (t 1 )),x(λ n (t))) + 2ρ(x(λ n (t)),x n (t)), t Hence ρ(x n (t),x n (t 2 )) ρ(x(λ n (t),x(λ n (t 2 ))) + 2ρ(x(λ n (t)),x n (t)). t

18 36 1 Limit Theorems for Stochastic Processes min[ρ(x n (t 1 ),x n (t));ρ(x n (t),x n (t 2 ))] 2ρ(x n (t),x(λ n (t))) t +min[ρ(x(λ n (t 1 )),x(λ n (t)));ρ(x(λ n (t)),x(λ n (t 2 )))]. By choosing c sufficiently small and n so large that λ n (t) t < c/2, we can make the expression min[ρ(x(λ n (t 1 )),x(λ n (t)));ρ(x(λ n (t)),x(λ n (t 2 )))] t c t 1 t c +c/2;t [0,1],tc c/2 t 2 tc arbitrarily small, so that if t [c,1 c], then λ n (t 1 ) < λ n (t) < λ n (t 2 ) and λ n (t 2 ) λ n (t 1 ) < 3c. This means we can apply Now for t c and t 1 c, all the quantities x(λ n (t 1 )),x(λ n (t)),x(λ n (t 2 )) can be made, arbitrarily close to x(0) or x(1) by choosing c sufficiently small. The necessity is proven. To prove the sufficiency, we shall prove the following stronger assertion A sufficient condition for x n (t) J 2 x(t) is that condition 2.5.3(a) be fulfilled, and (b ) lim c 0 lim n Δ J2 (c,x n (t)) = 0. Proof. We have ρ(x(t ck ),x n (t)) ρ(x(t ck ),x(t )) + ρ(x n (t),x n (t )) + ρ(x n (t ),x(t )), ρ(x n (t),x(t c k )) ρ(x n (t),x n (t )) + ρ(x n (t ),x(t )) + ρ(x(t ),x(t c k )). Since x n (t) converges to x(t) on an everywhere dense set, there is for every t a t such that t t < δ, and as n, we have ρ(x n (t ),x(t )) 0; further ρ(x(t),x(t )) < ε. Then for δ < c k /4 Therefore if lim Δ J n 2 (c k,x n (t),x(t)) lim Δ J2 (c k + δ,x n (t)) + 2ε. n lim c k 0 lim n Δ J 2 (c k,x n (t)) = 0, we have lim lim Δ J c k 0 n 2 (c k,x n (t),x(t)) = 0, which, in view of 2.5.2, proves our assertion The topologies J 1 and U CONVERGENCE CONDITIONS IN THE TOPOLOGY J 1. Necessary and sufficient conditions for x n (t) J 1 x(t) are that (a) x n (t) converges to x(t) on an everywhere dense set containing 0 and 1, and (b) lim c 0 lim n Δ J1 (c,x n (t)) = 0, where Δ J1 (c,y(t)) = t c<t 1 <t t 2 <t+c min[ρ(y(t 1 ),y(t));ρ(y(t),y(t 2 ))].

19 Limit Theorems for Stochastic Processes 37 Proof. NECESSITY OF CONDITION (b). If x ( t) J 1 x(t), then there exists a sequence of monotonic continuous functions λ n (t), which map [0,1] onto itself such that lim n t λ n (t) t = 0 and lim n t ρ(x n (t),x(λ n (t))) = 0. Then min[ρ(x n (t 1 ),x n (t));ρ(x n (t),x n (t 2 ))] 4ρ(x n (t),x(λ n (t))) t +min[ρ(x(λ n (t 1 )),x(λ n (t)));ρ(x(λ n (t)),x(λ n (t 2 )))], and since (1) λ n (t 1 ) < λ n (t) < λ n (t 2,) and (2), as c 0 and n, we know that λ n (t 2 ) λ n (t 1 ) becomes arbitrarily small, it follows that Δ J1 (c,x n (t)) approaches zero as n and c 0. SUFFICIENCY. Assume that (a) and (b) are fulfilled. We must construct a sequence of monotonic functions λ n (t) mapping [0,1] onto itself such that t ρ(x n (t),x(λ n (t))) 0 and λ n (t) t 0asn. To do this it is sufficient to construct continuous monotonic functions λn ε (t) mapping [0,1] onto itself, such that for n > n ε we have ρ(x n (t),x(λn ε (t))) < ε and lim ε 0 lim n λn ε (t) t. Let us attempt to construct such functions. Let t 1,t 2,,t k be all the points where the discontinuities of x(t) are no less than μ. Let us choose c and N so that for n > N we have Δ J1 (c,x n (t)) < μ. (2.23) Further, let this c be so small that if t and t both belong to the same one of the intervals (0,t 1 ),(t 1,t 2 ),,(t k,1), then ρ(x(t ),x(t )) < μ so long as t t < c. The fact that x n (t) converges to x(t) on an everywhere dense set implies that there exist points 0 = τ (c) 0 < τ (c) 1 < τ (c) 2 < < τ m (c) = 1 such that τ (c) i τ (c) )) < μ for n > N. Let t l lie between i 1 < c/2,(i = 1,2,,m), and ρ(x n (τ (c) i ),x(τ (i) i τ (c) i l and τ (c) i l +1. Then if the discontinuity at t l is greater than 8μ, it follows that ρ(x(τ (c) i l ),x(τ (c) i l +1 )) > 6μ. This implies that ρ(x n(τ (c) i l ),x n (τ (c) i l +1 )) > 4μ for n > N. But if n > max{n,n } and t (τ (c) i l,τ (c) i l +1 ), we have min{ρ(x n(τ (c) i l ),x n (t)), ρ(x n (τ (c) i l +1 ),x n(t))} < μ. Therefore in the interval (τ (c) i l,τ (c) i l +1 ) there is a point t l such that t (τ (c) i l,t l ) implies ρ(x n (τ (c) i l ),x n (t)) < μ and t (t l,τ (c) i l +1 ) implies ρ(x n (t),x n (τ (c) i l +1 )) < μ. (If this were not so, there would be points t 1 < t 2 < t 3 (τ (c) i l,τ (c) i l +1 ) such that ρ(x n (t 1),x(τ (c) i l ) < μ, ρ(x n (t 2),x n (τ (c) i l +1 ) < μ, ρ(x n(t 3),x(τ (c) i l ) < μ, or ρ(x n (t 1 ),x n(t 2 )) > 2μ and ρ(x n(t 2 ),x n(t 3 )) > 2μ, which is impossible according to (2.23)). Now let t k1,t k2,,t kr be points where the discontinuities are greater than 8μ, and let the t ki be points constructed for the t ki in exactly the same way as t l was constructed for t l. Let λ ε n (t) be a function continuous and monotonic on each of the

20 38 1 Limit Theorems for Stochastic Processes intervals (0,t k1 ),(t k1,t k2 ),,(t kr,1) of the form λ ε n (t)=a i t + b i, λ ε n (t ki )=t ki. Obviously λn ε (t) t < c/2. Let us evaluate ρ(x n (t),x(λ n (t))) for t (τ (c) j,τ (c) j+1 ). If in this interval there is not a single point t ki (and therefore no point t ki ), then ρ(x n (t),x(λn ε (t))) μ + min{ρ(x n (t),x n (τ (c) j ));ρ(x n (t),x n (τ (c) j+1 ))} +max{ρ(x(λn ε (t)),x(τ (c) j ));ρ(x(λn ε (t)),x(τ (c) j+1 ))} < 12μ. If t (τ (c) j,τ (c) j+1 ), however, and t k ν and t kν also belong to (τ (c) j,τ (c) j+1 ), then either t (τ (c) j,t kν ),λn ε (t) < t kν or t (t kν,τ (c) j+1 ),λ n ε (t) > t kν. This implies that ρ(x n (t),x(λn ε (t)) < 3μ. If we set ε = 12μ, we obtain the desired result CONVERGENCE CONDITIONS IN THE TOPOLOGY U. Necessary and sufficient conditions for x n (t) U x(t) are: (a) x n (t) J 1 x(t) and (b) if t 1,t 2,,t k denote points where the discontinuities of x(t) are greater than ε, where ε is arbitrary except that x(t) has no discontinuities whose magnitude is ε, and if t (n) 1,t(n) 2,,t(n) k n denote the points where the discontinuities of x n (t) are greater than ε, then for n greater than a certain value we have k n = k,t i = t (n) i. The necessity of both conditions is obvious. The sufficiency is seen if we note that by constructing λ ε n (t) (see the proof of the sufficiency in 2.6.1), we obtain λ ε n (t)=t Compactness conditions in the topologies J 1,J 2,M 1 and M 2. We note the following conditions necessary for compactness If a set of functions K K X is to be compact in one of the topologies J 1,J 2,M 1,M 2, it is necessary that for all t [0,1] and x(t) K the values of x(t) belong to a single compact set A of X. Indeed, if we have a sequence of points x k (t k ), then by choosing a sequence n k, such that x nk (t) M 2 x 0 (t),t nk t 0, we find that the distance between x nk (t nk ) and the segment [x 0 (t 0 0),x 0 (t 0 )] approaches zero, which means that x nk (t nk ) is compact, so that the segment is compact Sufficient condition for compactness. The set of functions K is compact in a topology S, where S is J 1,J 2,M 1 or M 2, if is fulfilled and if lim lim (Δ S (c,x(t)) + ρ(x(0), x(t)) + c 0 x(t) K 0<t<c 1 c<t<1 ρ(x(1),x(t))) = 0 (2.24) (by lim of a certain numerical set we mean its maximum limit point).

21 Limit Theorems for Stochastic Processes 39 Proof. Choosing some everywhere dense set N of values of t which contains 0 and 1, we may take from any sequence x n (t) a subsequence x nk (t) such that if t N then lim nk x nk (t) exists. Since Δ M2 (c,x(t)) Δ S (c,x(t)), we have lim lim (Δ M c 0 n k 2 (c,x nk (t)) + ρ(x(0), x(t)) + 0<t<c 1 c<t<1 ρ(x(1),x(t))) = 0. According to 2.3.5, therefore, there exists an x(t) K X such that x nk (t) M 2 x(t). This means that x nk (t) converges to x(t) on an everywhere dense set containing 0 and 1, and that (2.24) is fulfilled. Hence x nk (t) S x(t). Bearing in mind that Δ J1 (c,x(t)) and Δ M1 (c,x(t)) are monotonic functions of c, it is easy to obtain the following: Necessary and sufficient condition for compactness in the topologies J 1,M 1. 4 Let S 1 denote either J 1 or M 1. Conditions and lim (Δ S1 (c,x(t)) + ρ(x(0), x(t)) + c 0 x(t) K 0<t<c 1 c<t<1 ρ(x(1),x(t))) = 0 are necessary and sufficient for K to be compact in S 1. The next example shows that condition (2.24) is not necessary either for J 2 or for M Example. Consider the set K consisting of the functions x s (t),s [ 1 4, 1 2], given by 0, 0 t < s, 1, s t < 1 x s (t)= 4 + s/2, 1 0, 4 + s/2 t < 1 2, 1 1, 2 t 1. The set is J 2 -compact and therefore also M 2 -compact, but lim Δ J2 (c,x s (t)) = 1, s lim Δ M2 (c,x s (t)) = 1, for c < 1 s A certain inequality. The inequality we are about to derive hardly has any independent meaning, but will be needed in the proof of the limit theorems. Let us write R(δ,x(t),y(t)) = 0 t 1 δ inf ρ(x(t),y(t)), t t 1 t+δ R(δ,x(t),y(t)) = max{ρ(x(0),y(0));ρ(x(1),y(1));r(δ,x(t),y(t))}. Further, we set 4 The compactness conditions for the topology J 1 have been obtained by A. N. Kolmogorov.