/*  Homework 3

1.  Work problem 24 page 106.

For the series Yt = 1.2 Y-.36 Ye , the autocovariances have the form
Gamma(h) = (A+Bh)*(.6)**h.   Also our notes on page 31 show weights
Y(t) = Sum j=0 to infinity of (1+j)(.6)e(t-j) for this series.
 Find the smallest M for which (1+X)*(.6)**X <M(.8)**X for X>0.
Make a quick check by running this program after changing M: */

data a;  M= 0.001;
x=0; fx=1; bound=M; output;
do x=0.2 to 5 by .2;
fx=(1+x)*0.6**x; bound = M*0.8**x;
output; end;
proc gplot; plot fx*x bound*x/overlay;
symbol1 v=dot i=join c=red;
symbol2 v=none i=join c=green;  run;


/*  2.  Work problem  4 page 102.  Also run this
SAS program and explain why it is of
interest in the context of this problem.  */

data a;
do a = -3 to 3 by .1; do B = -3 to 3 by .1;
  rho = (A+A*B)/(1+A*A+B*B); output; end; end;
proc g3d; plot A*B=rho;
plot a*b=rho/rotate=80 tilt=80;
run;


/* 3.  Run this program and explain why the results
are so different when the recursions are almost
the same.  Notice the handy lagging method (SAS also
has a lag function).  Will either series X or Y
converge?  If so, to what limit?  */

data a;
X2=0; X1=1;X=0;
Y2=0; Y1=1;Y=0; output;
do t=1 to 100;
Y = 1.5*Y1 - 0.499*Y2;
X = 1.5*X1 - 0.501*X2;
output;
** lagging step***;
Y2=Y1; Y1=Y; X2=X1; X1=X;
end;

proc print data=a(obs=10);
proc gplot; plot Y*t  X*t/overlay;
run;

do t=1 to 100;
Y = 1.5*Y1 - 0.499*Y2;
X = 1.5*X1 - 0.501*X2;
output; Y2=Y1; Y1=Y; X2=X1; X1=X;
end;
proc print data=a(obs=10);
proc gplot; plot Y*t  X*t/overlay;
run;