2.1-1
15 dPlxt
l 1+2+3+4+5+=21
2 {n+et+¥+si+¥+sh2=am
4 {n+2h+?+sh+?+su#=am
is um+sm+sm+sm+sm+smrhzam
In {11th} + 211: + 31th) + 4100 + 51th] + l')1f1' =
ee mast{1,2,3,4,} =
Let I = [2:11 2:2, . . . ,arn]. and let are be the largest element ufat fur same
3.!1-15 {a} The (i, kjth element efR = E[:e[t]x[t] is
M = E[{A1e-i1i+ nale +e[e]}]
= E{|A1E2]ej'_m + STEELE
With
5(w1]=[11Eiwll*i21jlu j-lm-lhwi ]!
the {i1 kjth element ef 5(w1)s[w1}H is eJwH. Cemhimng |lheee ehservatiens tegether. we ehtain
a = E[x{i]rH[
ECE-Sl Analytie Metheds in Systems
Selntien #2
3.1r1 (a) We ran write
tTil1
Jr.
Lr._2
E1:=[P1.Ps:.u.11s] 2 dis-3
a
l
Then
eyes : elllhe.
Identifying
He 1 he
AHA- =
l: l [ hf rte
we see that
Had = ALLIAtH the Grammian matrix
and
he = All" 113's
and
H
Tia;
ll.2~1
H.233
{e} It 1:" 3: 131 then when 11 «a: y at: 1 we always; cheese He, en :1 = ll. 1When 1 a: y «c: 2", we weuld cheese H1;
hmvewn', thie eccnre with prehebility le, en '5' = 1f2.
If tr {1: 1f?~ we elwnjfe ellenee H1, ecu a = 1 and '13 = 1.
11'