📄 Extracted Text (383 words)
Moments of {x„}
The purpose of this subsection is to see if something can be said about the ratio
(Ek kx,)/(1k X0 when {xjn, is a non-zero solution to the system
0 = -q) n,1 Oarx - (TS, + di)x,
0 = q San., xn.I - (qta,, + d„) x„.
(1.28)
with a suitable constant. To this end, introduce by way of notation; = I ,,A)a„x„. The
equations in (1.28) can be used to derive two expressions for ;, these being
• Xn = oisani+ do (110.iska (goatir+k do )Citaj Xj for n 2.
• xy, — mani+ do (niska (qoacitak+kdo )(1 q)S.
(1.29)
Note that 4) must be such that
- ct vn Oak — 1.
q z—,M1 I k=1 (Oak +dk)
(1.30)
This last condition can be restated as saying that
ins2(qOan+dn)x• = cic-
(1.31)
and therefore
q; - (with, +di); + En.g, d„ x„ = q;
(1.32)
This tells us that
E„,d„x„ =(qtal +dax, = (1 -q); ,
(1.33)
where the left hand inequality comes via the n = 1 version of (1.29).
What is written in (1.33) is of at least two identities involving `moments' of @ca.
To elaborate, introduce a variable t and use (1.29) to see the equality between the
following two formal series:
((clOan+d,,)x.) = En, t° (94)a„x,,)
(1.35)
Let Q(t) denote the series En, ta(qta„x„) and let p(t) denote En, Then (1.35) says
that
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f l Q(t) + p(t) = (2(0+ (q. a +d 1)x1 .
(1.36)
This in turn can be rewritten using (1.33) as
p(t) = (t- 1) Q(t) + -q);
(1.37)
Taking t = I on both sides recovers (1.33): E,„ dnxn = (1- q);. Differentiating
once and setting t = 1 finds
E.Irldnx„= C2(1)+0 -(05.
(1.38)
To go further, use (1.31) to see that
Q(1)= - F.aIdn + q; + (qt al +di)x, =qS,
(1.39)
Granted this last equality, then (1.38) asserts that
En:Indnx„ = ; .
(1.40)
This with (1.33) says that
nd„x„ 1
Ea' d„x„ (1.q)
(1.41)
In the case d„ = d for all n, this asserts what is conjectured by Martin.
Identities for `moments' of the form Eno.?d„ x„ for p ≥ 2 require knowing
something of the (p-1)'st derivative of Q at t = 1. I don't know any good way to obtain
these.
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