E[M^k]=0,1,0,3,0,15,0,105 for k=1,2,3,4,5,6,7,8
which implies
p x^2+q y^2=1/2
p x^4+q x^4=3/2
p x^6+q y^6=15/2
p x^8+q y^8=105/2
set s=x^2 and t=y^2
as 15/2=(-15)*1/2+10*3/2
and 105=(-15)*3/2+10*15/2
we have s and t are the roots of the equation a^2=-15+10 a, which are distinct and positive.
s=5-sqrt(10); t=5+sqrt(10)
or x=sqrt(5-sqrt(10)); y=sqrt(5+sqrt(10))
insert them to the first two equations, we have
p=(7+2sqrt(10))/60,q=(7-2sqrt(10)/60. p and q satisfies 0<p,q<p+q<1/2.
thus Z has the distribution for those x,y,q,p. any change of order is allowed.
E[M^9]=E[Z^9]=0 is evident.
E[M^10]=945
E[Z^10]=-15 E[Z^6]+10 E[Z^8]=825
[ 本帖最后由 reynolds_wwy 于 2007-6-19 19:53 编辑 ]
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