Hi, I need to solve the maximum likelihood estimation problem. Here is the equation all the matrix and scalar are given except the highlighted gamma*_hat (1x1 matrix), Beta*_hat (6x1 matrix) and Sigma*_hat (6x6 matrix).   Here's my part of proc iml codes for the above equations. (N = 6, T =60) I already have the initial guess number for beta and sigma. I need to plug my initial guess beta and sigma to gamma equation and get a new gamma, and then use this new gamma to calculate a new beta and sigma, and repeat this process until the results convergence.  /*gamma*/ gamma = ((I - beta)`*inv(sigma)*(mu - beta*mu_m))/ ((I - beta)`*inv(sigma)*(I - beta)); /*beta*/ /**Numerator*/ Nu = j(N,1,0); D = 0; do i = 1 to T by 1; Nu = Nu + (R[i, ]` - gamma*ID)*(Rm[i,1] - gamma); end; /*Denominator*/ do i = 1 to T by 1; D = D + (Rm[i,1] - gamma)##2; end; beta = Nu/D; /*sigma*/ sigma = j(6,6,0); do i = 1 to T by 1; sigma = sigma + (Rt[i, ]`-gamma*(I-beta) - beta*Rm_b[i,1])*(Rt[i, ]`-gamma*(I-beta) - beta*Rm[i,1])`; end; sigma_b = sigma_b/T; Does anyone know how can I repeat these codes? should I use do loop? is there a repeat code function in iml?  I really appreciate any help!  ... View more

Hi I have a stock quote data set with exchange information. I want to know how many minutes was a given exchange holding the quote for a given stock. Here's part of my raw data ( the column name is DATE Sym_root Time_M and EX)  Here's my current code proc sort data= rawdata; by sym_root descending time_m; run; data test; set rawdata; by sym_root descending time_m; inforce = abs(dif(time_m)); if first.sym_root then inforce = max(("16:00:00.000000000"t-time_m),0); run; data test; set test; if EX = 'A' then A = inforce/60; else A =0; if EX = 'B' then B = inforce/60; else B =0; if EX = 'C' then C = inforce/60; else C =0; if EX = 'J' then J = inforce/60; else J =0; if EX = 'K' then K = inforce/60; else K =0; if EX = 'M' then M = inforce/60; else M =0; if EX = 'N' then N = inforce/60; else N =0; if EX = 'P' then P = inforce/60; else P =0; if EX = 'Q' then Q = inforce/60; else Q =0; if EX = 'T' then T = inforce/60; else T =0; if EX = 'V' then V = inforce/60; else V =0; if EX = 'X' then X = inforce/60; else X =0; if EX = 'Y' then Y = inforce/60; else Y =0; if EX = 'Z' then Z = inforce/60; else Z =0; run; proc report data=test nowd out= timeshare (drop=_BREAK_); column date sym_root A B C J K M N P Q T V X Y Z; define date / group; define sym_root / group; define A/sum 'A'; define B/sum 'B'; define C/sum 'C'; define J/sum 'J'; define K/sum 'K'; define M/sum 'M'; define N/sum 'N'; define P/sum 'P'; define Q/sum 'Q'; define T/sum 'T'; define V/sum 'V'; define X/sum 'X'; define Y/sum 'Y'; define Z/sum 'Z'; run; I use 'Inforce' as variable name to calculate the time difference. because the normal stock time is from 9:30 to 16:30. so as you can see in my code, I use inforce = max(("16:00:00.000000000"t-time_m),0) to get the last inforce value.    However, one of my code problem is sometimes more than 1 exchange will hold the quote. I tried use lag(inforce) and code ' if inforce=0 then inforce=lag_inforce'. but it doesn't work when there are more than two exchange hold quote at the same time.      I really appreciate if someone can help me solve this issue. please see the attachment of part of rawdata. ... View more