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Statistics & Probability Letters 3 (1985) 3537
ON A CHARACTERIZATION QUESTION FOR SYMMETRIC RANDOM VARIABLES
University of Toronto, Ontario, Canada
J. Michael STEELE
Princeton University, Prineeton, 1Q USA
Received July 1984
Revised September 1984
A bstract: If X,, X2 are independent with common density g symmetric about zero, then P ( X, + a X2 > 0) = '2 for all real a. We provide a counter example to show that the converse is false and thus settle a question posed by Burdick (1972).
AMS 1980 Subject Classifications: Primary 62E10; Secondary 60E10.
Keywords: symmetric random variables, medians, fractional moments, logarithmic moment, characteristic funtions.
A substantial part of the theory of characteriza exists for some e > 0 than (1) implies g(x)
tion problems is devoted to the deduction of prop g(x) almost everywhere. Burdick then asked if
erties of a summand from properties of the char the condition that a fractional moment exists could
acteristic funtion of the sum (see e.g. Lukacs (1970) be omitted. In this note we construct an asymmet
or Kagan, Linnik and Rao (1973) for numerous ric g for which (1) holds and so provide a negative
instances). Because it is so simple to characterize answer to Burdick's question. Our construction is
symmetric distributions as exactly those with real based on a lemma of Freedman and Diaconis
characteristic functions, it is intriguing to note that (1982).
as soon as summand inference is involved subtle We first state the following result.
ties can begin to arize. The problem studied in
Burdick (1972) and pursued here is an instance of Lemma. Let X, X2 be independent with a common
this phenomenon. continuous distribution function G. Then
Suppose that Xt and X2 are independent ran
dom variables with common density g(x). If g(x) P(Xl+aX2<0)=' foralla (3)
is symmetric about the origin then we may readily if and only if
P ( X1 + aX2 > 0) = 21 01(t)02(t)=0 forallt (4)
where shl, and 02 are the FourierStieltjes tran
for all real a. It is somewhat tempting to believe forms
that characterizes symmetric densities. Burdick .00 e'txd[G(ex) G(ex)l
(1972) proved that if a fractional moment 01(t)
f 1 x ieg(x) dx < oc (2) 00 Y't d[G(y) G(y)l (5)
01677152/85/$3.30 0 1985, Elsevier Science Publishers BY (NorthHolland) 35
and Remark. The condition in the lemma that G be
00 continuous may be omitted provided we replace
o,(t)=f e"xd[G(ex) + G(e)l (3) by
f00 P(X, + ax, < 0) + 1 P(X, + aX2 = 0)
Y d[G(y) + G(y)]. (6) 2 2
cc The same proof continues to hold except that
Proof. By the convolution formula (e.g. Feller instead of using rightcontinuous G we use the
(1970, p. 144)) we have that (3) is equivalent to symmetric form G (x) = '(G (x + ) G (x )) with
appropriate conventions for the Steiltjes integral.
Turning now to the counterexample let G have
G( x ) dG(x) =2 a * 0, (7)
a denstiy g so that
together with oi(t) =f 00 ei',xu(x) dx
G(O) = ' (8) 00
Further (7) is equivalent to the pair of conditions and
'P2W F eitxv(x) dx
[G(!) G( x dG(x) = 0, a > 0, (9) 00
[G(:~)+G( x dG (x) = 1, a> 0. (10)
a a u(x) = ex(g(ex) + g( ex))
But actually (8) is a consequence of (10): just and
make the change of variables xIa y and take v (x) = ex(g(ex) g( ex))
the limit a co. Furthermore, (9) and (10) actu
ally are equivalent to each other. To see this, first Observe that u and v here are in correspon
write (9) and (10), respectively, in the forms dence with the even and odd parts of g (but need
not in themselves be even or odd.) Now, according
f [G( x + :!) d[G(x) + G(x)l = 0, to the proof of Lemma 3 of Freedman and Di
a a aconis (1982), there exists a probability densityof
(11) the form
f [G( x + + X )l d[G(x) G( x)l = 1 f=C(fl + M)
(12) having characteristic function
and then apply intergration by parts to (12); make 1= C(I1 + 812) .
the change of variables xIa x; finally the change such that fl > 0, fl is real and vanishes off 1, 1],
of notation a 1/a. Hence, altogether, (3) and and 12 is purely imaginary and vanishes off
(11) are equivalent. [ 3, 21 U [2, 31.
Next in (11) put a = e' and make the variable The functions fl and f2 can be selected so that
change x eY to get 8 * 0, thus making sure that f is not symmetric. If
00 we now set
00 u(x)=cf,(x) and v(x)=c8f2(
.d[G(ey)+G(eY)1=0 forallt. (13) the nonzero segments of the Fourier transforms of
But this is the convolution of G (ex) G e x) u and v will not overlap. On solving for g we
and G(e')+G(e') whose FourierStieRjes obtain
transforms are 01 and 02. Consequently (13) is g(u) c 1 u 1 fl(Ini u 1) + (sgn u) 8f2(lni u 1)}.
equivalent to (4).
By our construction we see that g is a density density satisfying (1) which posses logarithmic mo
which satisfies (4) and therefore (3). Because 8 * 0 ments of all orders.
we have that g is not symmetric, and thus the
problem posed by Burdick (1972) is solved. References
As a final observation, we note that by Burdick's Burdick, D.L. (1972), A note on symmetric random variables,
theorem, g cannot possess a fractional moment. A nn. Math. Statis. 43(6), 20392040
We can however show that g can posses a teth Feller, W. (1970), An Introduction to Probabilit v Theoty and Its
logarithmic moment for any particular value of K. Applications, Vol. 2, 2nd ed. (Wiley, New York).
This follows from an immediate change of varia Freedman, D. and P. Diaconis (1982), De Finetti's Theorem for
bles and from the fact that the construction of f, symmetric location families, Ann. Statist 10(1), 184189.
and permits them to have moments of arbi Kagan' A.M., Yu.V. Linnik and C.R. Rao (1973), Characteriza
f2 tion Problems in Mathematical Statistics (Wiley, New York).
trarily large order. Lukacs, E. (1970), Characteristic Functions, 2nd ed. (Hufner,
We do not know if it is possible to construct a New York).