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Professor K. L. Chung 3 WF:vn 2e&Zl William Feller e4. 1C~r U94 4 L <^ r7 ? ci~t <^rry o < o )7 54 2 October 9, 1957 The m in (8.17) should be m + Thus (8.17) does not contain the m + La, 1xj limiting case m = 0, x = 1. This can be obtained by analogy with the corres ponding form for the diffusion equation, cf. the forthcoming paper in the Illinois Journal. p.549. The error in (8.13)(8.17) carries naturally over, and even here m should be replaced by m + [(1 lx]. p.549. (Footnote) From (9.7) it follows, as is wellknown, that m. is completely monotonic if and only if (~)n (n)(a ) < n =1, 2, ... (note the < 0). That this is so follows trivially from (9.13) using the resolvent relation (3.7). Best wishes. A4 t) . *' y y ^ Cordially, 6J^ PRINCETON UNIVERSITY PRINCETON, NEW JERSEY Department of October 18, 1957 Address reply to MATHEMATICS FINE HALL BOX 708 PRINCETON, N.J. Professor K. L. Chung Department of Mathematics Syracuse University Syracuse 10, N. Y. Dear Chungius: You sure know how to tease a poor creature. Formula (9.18) indeed contains a misprint, the l should be f: naturally, in the semigroup for P its gener ator a = XP I should occur rather than the generator Q of pmi. It is a pitty that just this formula contains the erroneous constant, for the calculation is trivial and the f would be clear from the meaning and by analogy with section 8, as whose counterpart (9.18) serves. I did not at first notice the missing bar since I thought that your mysterious query refers to the constant. Being in possession of the correct form all the time you would have acted kinder by telling me about the missing bar. Instead you cun ningly mislead me by implying that something goes wrong only when a is unbounded. I expect that you have similar teasers up your sleeve, e.g. when you talk about a gap on p. 542 but I don't mind giving you a written proof of my incompetence by admitting that I am unable to find it. Again you masterfully hide the real point. All I say is that the proof for u proceeds as the proof for 1, and this being trivially true on the basis of a general isomorphism there can hardly exist a gap in this place. I assume therefore that said gap appears earlier. However, each of the indicated steps seems so simple that I am completely puzzled about the possibility of a gap it seems to me that the explicitly given proofs are hardly required. The steps are: (1) ru = u implies Pi < u (2) This implies Pmin(t)u < u (p. 534) (3) u(t) = Pmin(t)u is monotone because of u(t + s; i) = in(b; i, j) u(s; j) < E mi(t; i,j) u(j) = (t; i). This also shows, putting t = s, that u(t) = u implies u(2t) = u. (p. 535) You have really stirred my curiosity and I shall be very grateful if you will kindly tell me at which of the steps the gap occurs and, if possible, wherein it consists. With kindest regards. Sincerely r, WF:vn William Feller PRINCETON UNIVERSITY PRINCETON, NEW JERSEY Department of October 29, 1957 Address reply to MATHEMATICS FINE HALL BOX 708 PRINCETON, N.J. Professor K. L. Chung Department of Mathematics Syracuse University Syracuse 10, New York Dear Chungius: Thank you for your third letter and for abandoning the mysterious component in favor of the preceptorial. Since you were so positive about "the gap" in the proof of Theorem 6.3 on p. 542, I spent hours in trying to fathom what was on your mind. Since it is tri vial that u and 1 differ only notationally, I had to break down the whole preceding part of the paper. And now it turns out that, after all, their is nothing to be added and all you wanted to know is how one passes from the Laplace transform relation XAPu < u to P(t)u < u. If you had asked that question, I would have from the beginning referred you to the basic section 3 in which the thing is treated apparently to your satisfaction. Although McKean has not read my paper he is so much at home in Laplace transforms that he would have answered the question if you had put it to him instead of playing your tricks on him too. I do not know who the other student or students are who flunked your test. That no student of mine, past or present, has read my paper is irrelevant to this incredible hoax: it should not affect your surprise at our not being able to answer a simple question that has not been put to us. As to p. 541, I had not realized that the question concerns the argument that S + s < implies + s )= Being perhaps too familiar with the U v tA U v subject matter, all steps appeared equally simple to me. However, I readily concede that the "it follows" of the text is too stingy and I am now glad to know about it. The point is that in the Markov chain with transition probabilities IL the differ ence = s) s = probability of an indefinite continuation of the pro cess outside of U U V. Since IL < 1 it follows that z < 1 s s = x and, by definition there does not exist a solution z. < 1 x of HIz = z except z. = 0. Probabilistically the whole assertion (b) is trivial and can be proved simpler; this together with the fact that the assertion is irrelevant for the paper induced Professor K. L. Chung me to hurry. I cannot tell you how comforted I am by the assurance dervied from your letters that these passages are the darkest in the paper. It must be well written! As for (9.18) I was not surprised to learn that your attempts of confusing me finally led to your being confused. The idea that you had the correct version was first stated by you. At any rate, you might have shown me the common courtesy of telling me (a) that my constant m was miscalculated instead of (b) that (9.18) seems to contain an additional error. I would at once have noticed the slip in writing f where the true infinitesimal generator nf (with P f* = Xp I) should figure and I would have given both corrections at once. With f* instead of the silly a the relation is a trivial identity. How ever, since you demand it, here is a detailed "derivation." We have to prove that P (i,) 8(i,w) = m+[,)lx] P(i,w) + (%P(i,j) &(i,j)]x(j) P =1 where [a, z] = z a(k)z(k). For i = 1, 2, ..., w we have from the definition 1 of P (i,) (P(i,j) 6(i,j)}x(j) j=1 = x(i) E in (i,j)x(j) x(i)m ZE a(k)Pin(k,J)x(j) j,k = x(i) (x(i) x,(i)) x((i)m[(a, xx ] = x,(i) x7(i)m,[a, xx ] = x(i)m~ [a, xX]) = x,(i)mX(Xp + m + [a, 1x,] [(, xx ]) = xx(i)m({p + m + [a, 1x]) m + [(, lx] PX(i,w)(A + m+ I intend to publish a correction stating that my m is wrong and giving the above correct version. If it is O.K. with you, I shall state that I am indebted to you for drawing my attention to these slips. Cordially yours, WF:vn William Feller October 29, 1957 May 21# 1964 Dear Feller: Thank you for giving me trademark for "i. o." I used it in my papers just as an abbreviation, sometimes even "i. o. a. e." as a linguistic trick. it is not hard to add "u." there for uni formity. As evidence of BSP, I was just thinking last night of Levy's strange theorem (p. 14/ of his L'Addition7 that if S is the partial sum of essentially divergent independent r. v.'s thee \VA : n(Sn>An i. o.) with pr. 0 or 1. I never checked his proof but here is a much simpler one which you may use as an exercise if you can. Replacing each summand Xn by Xn(AnAnl) Ao=0. one sees there is no loss of generality if one takes An=0. Now put for every real x f(x)=P(S >x i.o.) then it is i,:mediate that f=f*Pn where P is the distribution of Sn. Since f is monotone (though not necessarily continuous) and bounded it follows by a familiar lemma (which you generously attributed to ChoquetDeny) that f is a constant. The theorem then. follows from Kolmogorov's 01 lawy a absevve4 ky Levy Let o aalso take this opportunity to say that I have now done the boundary theory of iMarkov chains for a finite number of exits (no hypothesis on entrance boundary nor ztransience of states as I fool ishly assumed iA my first paper) with very simple probabilistic methods. (This was done analytically by David Wiilliams.) In particular there is a nice si:"le interpretation of your mysterious doing in the ease where u( o)= oaD and an explanation why only the diagonal elements can be OD. It is a. complete vindication of your foresight. Sincerely yours. 7 V.4 V> ,,, ., 0 y 4>'(f 