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und ve rTdte Gebiete SpringerVe 4g Berlin eidelberg New York Schriftil ung L. Schmetterer, Wien Herausgegeb von K. L. Chung, Sta rd R. Fortet, P i D. G. Ke all, Cambridge Kri eberg, Heidelberg Professor Dr. K. L. Chung Kunzi, Zurich Department of Mathematics J. Ne an, Berkeley Stanford University Ju. V. P chorov, Moskwa Stanford, California 94305, USA A. R6nyi, dapest H. Richter, M chen L. Schmetterer, i n F. Spitzer, Ithaca arch 15, 1995 Dear irof. Takacs: There is a disturbing mistake in p. 4 of your Cominatorial Methods etc. 1967. You wrote "Evidently o =gi+ gr. Ig = pJ This is false r t+1 r. [g = psi'] when n=*, the three k's are: 1L, O 1. For then u1=l, but gl=g2=l. Fortunately I found that o, r=grgr. Then your formula (10) becomes Z6r g g =nk, which is OK. Please confirm this correction unless it has already been corrected, because I may cite ti .Your proof is the nicest of the general ballot problem that I have seen. I was led to it because I used the result to prove an 51year old result of mine, see p. 560 of the inclosed incompletee) reference, by Guy. If you are interested in trying to solve the still unsolved pro blem, I can show you my solution in the case a=l, using a new probabili ty lemma to reduce it to the general ballot problem(on p. 2 of your bool. I do not know if this reduction is possible for a>l, and even if it can be I do not know an explicit solution wAtM t"in your (10) 4not an in teger. I have sAen your paper about the latter but that is not an ex plicit formula. Best wishes, Sincerely, 1<^A LLJ^Uv PRINCETON INN w e40?,. kA archh 31, 1993 Lear Prof. Takacs: Thank you for your reply and reprints. I think there is a good chance that one can express the solution of my general problem in a satisfactory way, for example a finite T (fix ed.>aaSindependent of I. Here is the new lemma I proved Let Xk be iid with P(X) =(ba)/b, P(ba)=a/b, where a and b are relatively prime positive integers) a P (Sbnl = a) = PO(Sbn1 = ab) 6o llft cout,% .", where P0 denotes the taboo prob. with 0 taboo. Therefore either probability above is also equal to fbn = P(Sbn = 0) I discover this only a month ago from examining a diagram, and then proved it. Two or three "experts" have since been consulted and none knew it. You are next. Now when a = la the second [not the first] prob. above can be evaluated by the general ballot problem. What you show ed in your letter is equivalent to this. Hence I have now a combinatorial proof of my result of c. 1940. My old "proof" was by induction. Vhat you showed is the old LagrangeLaplace method, which led to identities such as Hagen's (cf. Gould). Who knows but there is a similar (of course harder) identity for any a? You try. it in your retirement for the pleasure of it,. By the ways here is a other easier problem: can one deduce Hagen's apparent ly moae formula from the case above with a = i?/ I have not seen you since the ejaly fifties and am glad tb have this correspon dence. Kind regards, [I shall lecture on the above in La Jolla tomorrow.] AMiJ Apr. ; .' P. to.I KkfZ3s6,6XA +M me" /4 i( April 18, 1993 Dear Prof. Takacs: lvy proof is simpler than yours, and can be at once extended to the general p and q as you have. Renewal equation (or rather: first entrance decomDostion): n n n k kPnk where f =P(To=k), pn is either P(Sbn=a) or P(S bn=ab), is the desired taboo probability in either case. Now compute the ratio axyixixaxaxyxikxw xixxnpx xEheBmxXxE of the two different p in the 9Rxfk two S tig: cases: it is (ba)p:aq for all n. Therefore by the equation the same ratio holds between the two cases of p'o If you have any trouble understanding the above you will under stand why I have trouble following your indications for Hagen. First of all your (3) cannot be true when k>n. If indeed you can show that my old formula (for a=l, as announced in the Monthly in 1946) does imply aagen's identity (the converse is true according to Bould), I(d be Aftj to see a somewhat more detailed exposition. Please begin by proving the correct part of your (3) by the ballot problem, and print out on your wordprocessor (which I do not have and do not want) the derivation of the exact form of Hagen's identity. That would SMXQ k be simpler than the proof of the latter by BlackwellDubins, which I never read. Did you know the latter? How about HagenRothe? Can that also be derived by ballot? Finally, one must try again to solve my unsolved problem for general a and b. There must be a neat formula for f n albeit not as neat as when a=l. Best wishes FA ETEL STANFORD UNIVERSITY STANFORD, CALIFORNIA 94305I2z5 DEPARTMENT OF MATHEMATICS (415) 7256284 April k 1 7 Dear Takacs: In my last letter, fk anould be P(T =bk), and my remark about your ( ) should be deleted because O=c.0 for any c. But please be so kind as to detail your arguments from (5) to A (5); and 9 9f@! Gould's book p. 41 for the identities by agen ,lothe and me. Cf. 0lso Go4id'tts X . In case you have a copy of Aepili's 192/4) original paper, could you send one to me? I am also curious about another rroof of the generall ballot problem by Gruenbaum, cited by another author. I am too lazy to look it up, but maybe you can tell me about it? Thanks and best wishes (for the solution of my problem). /' ' S,\/es iA II 9/234, 1993 Dear Takacs: It occurred to me that there is still an unsolved problem even in the case b>a=l except when b=2 (Andre's), as follows: what is the ex plicit expression for (A) PX{Sbn=0 Sk/V for 0 for all x and all n. My old result proved by new mebt a few months ago is the case x=0. 'Jith outrmn maJThe case x>0 can be dealt with by the general ballot problem' but not the case x<0. I have an idea which might works but I am asking your expert opinion lest I missed something simpler. You gave a proof that my old problem is equivalent to Hagen's identity which might just also imply an explicit formula for (A). Pxmeans of course the random walk starts at x. A much shorter nota tion for (A) is f(n) xo in my notation for the first entrance time from x to o as a Markov chain. It is this extension that brought to mind the new problem. When b=2 the problem is solved by symmetry. Sincerely. 71 4a/r (61 ) xl 'h(bQJ~ (hi)k1X~u CAAA'4 X {t xV f, wLs L04 A C/)n x 0 S1 4 (f)+ 1 > '1al) n oc ( l) 0X tiS 7n J c 1, xs  ~9v1tX ) fvv' ~~2( >1 g jA^;t JfiOf K o) I^ 9 ^ ^^^^^2, 4^ c 4 / 1 1 S,I1 i, 2  L2L 2.t 2*j~ X >0 = 0 'C z T TO VA (I (b hty + x .,,r /k > ^e 1XI krj a^so+n+(bi n 4G6 P,) ",4 c^ rA 2 Let (Xk} be an iid sequence of random variables taking va i an mardg1 thx 1t et" S cement let Sn / / /W / l/d"Va fil and TO = min (nl: S =O). =E Xk Then for ^^adT/ /mm n>l, and l Proof. (Shao) For n=2 this is trivial. Suppose the result is true for l The left member above is equal to the sum over m from 1 to N/ of the follww ing (3) E(Xk; T0=m; SN=0}= E{Xk; T0=m; SNSm=0} Oj//d For m (4) P(TO=m).E(Xki SNW = 0 because the second factor in (4) is zero by symmetry. For k (5) E(ik; T0=m). P{SNS =0) which is zero because the first factor in (5) is equal to zero by the induction hypothesis applied to m fx~m for m from 1 to N is zero. Hence the quantity inl(3) for the last value m=N must also be zero. That quantity is nothing but the left mem ber of (1) with n replaced by N. Hence the induction is complete and the result is proved. Commentary. This is really another renewal decomposition like my proof of the backward lemmum (see previous communication). But the induc tion is new here. I am sure your proof is on the way. Whether it is the same as the above or not* please send me a reference if it has appeared in print before. E(45 T~ze'o) May 26, 1994 Dear Takacs: Since I encouraged you to write that article (scheduled for October* 1993) I feel obligated to tell you my opinion now that I have read part of it. I had originally hoped that you would "expose"'thegeneral ballot problem in a way that most unacquainted readers could follow the argument easily [as in the case mu.= 1, done already in Borel's book, e. g.]. But heing spent at least twenty minutes over two. of your proofs I must tell you that I did not find your exposition satisfactory. In both the induc tion proof on pp. 78, and your version of Aeppli's proof on pp89, you seem to have glossed over exactly the part of the argument which should have been made clearer. In the induction case you did not explain how the two bounda conditions determine the solution. In the Aeppli proof you omitted (saying as Laplace used to say and was severely criticized by his readers: 'est file a voir", essentially) the crucial onetoone correspondence asserted oh line 2, p. 9. Of course the result is 71 years old and proofs are known, but if your article is meant for a "young reader" to whom it is new, and he is not so smart that he can figure out all the omitted argument, then I must say you have not done it.1 On the other hand, your forma, hore advanced proof that I read some months ago [with that bad misprint] is OK, though certain readers would be frightenened away by the modern'notation. I think youreproduced that also as the third or fourth proof, but of course I did not read it. Now some mistakes: p. 2, line 18: can't be right p. 4, the line after (14): English is incorrect. Yours sincerely, \ ( 2 d P.S. As I told you, my interest in this is disinterested. y (i ^"J /UL c~j '^ A^^~i^ j ou~ru^ cC ^ /wieo, ^ {^ u}^ 40,^^) OALAA,'^^ OL^~~~ 4e*m~,c^t fltc7 ~kY'~7 4~t(frte9 (I) A.12 f~e~f~. ?, ^ 2S ., )A44 e / SMW_ I; ,t 4~CeLk / I~ %%~ 'lCc.J}t .. / (2? w2 OztciCIA ^/ /... pA 2  W 2 Ae4ce 1 fA? 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