ST 730 -- Fall 2005
  Homework #4  -- Solutions 
1)
  a) Usual form is (y(t) - mu) = rho*(y(t-1)-mu) + a(t)
     gets reworked to   y(t) = mu*(1-rho) + rho*y(t) + a(t)
     so that rho = 0.6 and mu = 5 so that
                        y(t) = 2 + 0.6y(t) + a(t)

  b) For an AR(1) process, Var( Y(t) ) = sigma-sq-a / (1-rho**2)
     Here Var( a(t) ) = sigma-sq-a = 9, (1 - 0.6**2)= 0.64 so
      Var( y(t) ) = 9/0.64

  Forecast mean for AR(1) is mu + (rho**L)*(Y(N)-mu)
    and forecast variance is sigma-sq-a*(1 + rho**2 + ... + rho**(2L-2))

  c) Here y(N) = 4, so y(N)-mu = -1, so for 
                 forecast mean          forecast standard error
       y(N+1)     5 + 0.6*(-1) = 4.4    sqrt(9) = 3
       y(N+2)     5 + .36*(-1) = 4.64   sqrt(9*(1 + .36) ) = 3.4986
       y(N+1000)  5 + tiny*(-1) = 5     sqrt(Var(y(t))=sqrt(9/.64)=15/4
  d) Here y(N) = 7, so y(N)-mu = 2, so
                 forecast mean          forecast standard error
       y(N+1)     5 + 0.6*(2) = 6.2    sqrt(9) = 3
       y(N+2)     5 + .36*(2) = 5.72   sqrt(9*(1 + .36) ) = 3.4986
       y(N+1000)  5 + tiny*(2) = 5     sqrt(Var(y(t))=sqrt(9/.64)=15/4

2) See code in hwk054bs.sas
  a) see proc plot /gplot
  b) using one-sample model assumptions, confidence interval for 
     mean is ybar +/- t*se(mean), or .04563 +/- 1.97*0.0010264
     or (.04361, .04765 )
     se(mean) = sqrt( varest/N) ) = sqrt(.000225426/214)) = .0010264
 
  c) mu-hat = 0.4526  se(mu-hat) = .0023184  (twice as big!)
     rho-hat = 0.68193 se(rho-hat) = .05028

  *) Yes, AR(1) model looks adequate -- p-values on goodness-of-fit
      Ljung-Box statistic are large
  
  d) Using AR(1) model, CI for mu is mu-hat +/- z*se(mu-hat), or
      .04526 +/- 1.96*.0023184 or ( .04072, .04980 )
 
  e) read forecasts from output