William Feller, 1957-1964


Material Information

William Feller, 1957-1964
Physical Description:
Feller, William
Chung, Kai Lai
Physical Location:
Box: 1
Folder: William Feller, 1957-1964


Subjects / Keywords:
Mathematics -- History -- 20th century

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Source Institution:
University of Florida
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All applicable rights reserved by the source institution and holding location.
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Full Text

Professor K. L. Chung




William Feller

e4. 1C~r U94
L <^ r7 ? ci~t- <^rry o < o )7 54


October 9, 1957

The m in (8.17) should be m + Thus (8.17) does not contain the
m + La, 1-xj
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 l-x].
p.549. (Footnote) From (9.7) it follows, as is well-known, 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,



Department of October 18, 1957 Address reply to
BOX 708

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 semi-group 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
With kindest regards.
Sincerely r,

WF:vn William Feller


Department of October 29, 1957 Address reply to
BOX 708
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
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+[,)l-x] 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
of P

(i,) (P(i,j) 6(i,j)}x(j)
= x(i) E in (i,j)x(j) x(i)m ZE a(k)Pin(k,J)x(j)
= x(i) (x(i) x,(i)) x((i)m[(a, x-x ]

= x,(i) x7(i)m,[a, x-x ]

= x(i)m~ [a, x-X])

= x,(i)mX(Xp + m + [a, 1-x,] [(, x-x ])

= xx(i)m({p + m + [a, 1-x])

m + [(, l-x]
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-
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-(An-Anl-) 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
Choquet-Deny) that f is a constant. The theorem then. follows from
Kolmogorov's 0-1 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 z-transience 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.

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