Or, you can do it inside a datastep. data simualation;   df=1;   call streaminit(1234567); /*initialize the seed is optional*/;   do i=1 to 100000;   chisq=rand('chisq',df);   output;   end; run; ... View more

It is not so complicated, I dont find it neccessary to refer a paper. There exist a theorem that shows that any symetric matrix, V,  can be diagonalized into ADA T , where A consist of eigenvectors and D is a diagonal-matrix with eigenvaules. Then we use that if U is N(0,I) distributed, then BU is N(0,BB T )-distributed. With B=A sqrt(D) we get the result that BU has the wanted covariance. sqrt(D) should just be some diagonal-matrix  which multiplied with itself gives D. I agree by the way with Rick that the IML solution is more easy than the one I showed. My method is preferable if IML is not available or if you for some other reason want to do the calculation inside a datastep. ... View more

I dont know if there is any very easy way to do this. I do it manually by simulate vectors with independent entries with a variance equal to the eigenvalues. Then the linear transformation obtained by multiply the matrix with eigenvectors on that result in a vector with the wanted covariance. *I will need to do some matrix operations, so I use option cmplib to point on the place where I have put these. See my post about that here https://communities.sas.com/ideas/1570 option cmplib=(work.func); data corr;   input c1-c8; cards; 1    0.00885217    -0.088563853    0.170315918    -0.14616131    -0.019573222    -0.092562819    -0.160620889 0.00885217    1    -0.059626568    0.035719892    0.01249655    0.011717493    0.073782848    0.018419863 -0.088563853    -0.059626568    1    0.04983739    0.050539605    -0.030848973    0.081274999    -0.016919634 0.170315918    0.035719892    0.04983739    1    -0.155494382    -0.286070226    -0.171346052    -0.036586031 -0.14616131    0.01249655    0.050539605    -0.155494382    1    0.115005076    0.148470073    -0.0011815 -0.019573222    0.011717493    -0.030848973    -0.286070226    0.115005076    1    -0.10766422    0.054409451 -0.092562819    0.073782848    0.081274999    -0.171346052    0.148470073    -0.10766422    1    -0.158523541 -0.160620889    0.018419863    -0.016919634    -0.036586031    -0.0011815    0.054409451    -0.158523541    1 ; run; data simulate;   set corr end=end;   array c{8};   array corr{8,8} _temporary_;   array eigenvec{8,8} _temporary_;   array eigenval{8} _temporary_;   do i=1 to 8;     corr[_n_,i]=c;   end;   if end; *now the covariance matrix is loaded into an array; *then I find the eigenvectors and eigenvalues;   call jacobi(corr,eigenval,eigenvec,nrot);   array u{8,1} _temporary_;   array xtemp{8,1} _temporary_;   array x{8}; *Then I start simulating;   do j=1 to 100000;   do i=1 to 8;     u[i,1]=rand('normal',0,sqrt(eigenval));   end;   call multiplicer(eigenvec,u,xtemp);   do i=1 to 8;     x=xtemp[i,1];   end;   output;   end;   keep x1-x8; run; proc corr data=simulate pearson; var x1-x8; run; ... View more

I would like to emphasize that the offset only should be used in Poisson regression when the log is used as link function - which by the way is the default. As I mentioned above, when log is used as linkfunction, then one has a multiplicative structure: λ =time * exp(β X)=exp(log(time)+β X), and it turns out that the log(time) should be used as offset. But, if one use an additive mean structure, eg have identity as linkfunction, then the expected count is λ =time * β X = β (time*X). It turns out that one should not have log time as offset, but instead make the regression on the covariate vector multiplied with the time. ... View more

An offset term should be used when the model includes a term which should not be multiplied with any parameter. Often in Poisson regression you will have an offset because meanvalue will be proportional to the time the observation is observed. That is also the case in your question. Here I call the observation time PY (Person Years). Then the expected count is λ =PY * exp(β X)=exp(log(PY)+β X) Therefore, log(PY) is an offset in the model equation. This is only true when you model the mean with a multiplicative meanstructure (log as linkfunction) ... View more