Takacs, 1993-1994

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Takacs, 1993-1994
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Mathematics -- History -- 20th century

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und ve rTdte Gebiete
Springer-Ve 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=gr-gr-. Then your formula (10) becomes Z6r
g -g =n-k, 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 51-year 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-) =(b-a)/b, P(b-a)=a/b, where
a and b are relatively prime positive integers) a for any n we have Y9 orove a for of +.'s ser+s
P (Sbnl = a) = PO(Sbn-1 = a-b) 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 Lagrange-Laplace 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 to-morrow.]
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 kPn-k
where f =P(To=k), pn is either P(Sbn=a) or P(S bn=a-b), 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 (b-a)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 word-processor (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 Blackwell-Dubins, which
I never read. Did you know the latter?
How about Hagen-Rothe? 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 94305-I2z5

DEPARTMENT OF MATHEMATICS
(415) 725-6284 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








9/23-4, 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 n I o im
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.






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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 (1) E(Xk; T0=n) = 0
Proof. (Shao) For n=2 this is trivial. Suppose the result is true
for l (2) E{Xk; SN=0) = 0
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; SN-Sm=0}
Oj//d For m hence hyx the right member of (3) is equal to
(4) P(TO=m).E(Xki SN-W = 0
because the second factor in (4) is zero by symmetry. For k second famtburof (3) is equal to A
(5) E(ik; T0=m). P{SN-S =0)
which is zero because the first factor in (5) is equal to zero by the
induction hypothesis applied to m zero for m from 1 to N-1. But by (2) the sum of these quantiti!Ies
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. 7-8, and your version of Aeppli's proof on pp8-9, 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 one-to-one
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 you-reproduced 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.





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