Maximum Likelihood --->Hessian matrix H(the final one)--->covariance matrix for parameters =- inverse of H--->std = square root of diagonal of COV matrix
Poisson has fixed scale =1, no std at all. Of note: sometimes forcing scale =1 is not realistic at all, due to overdispersion in the real data. seen here:
http://blog.sina.com.cn/s/blog_a3a92636010128im.html
you can look at the covariance matrix by:



jingju

Thank you for your help!
I notice that the Hessian matrix is the expectation of the d2(L)/d(b1)d(b2), whether could I use the information matrix J , and covariance matrix cov(b)=J^-1 , where J * J^-1 = I ? Then I could just find the variance of the i th parameter bi from the main diagonal element.  

agree. same for using observed information matrix.

有一个关于芯片瑕疵数量的模型：
说A、B工艺分别生产出10个芯片，其瑕疵数量为：
A: 8, 7, 6, 6, 3, 4, 7, 2, 3, 4
B: 9, 9, 8, 14, 8, 13, 11, 5, 7, 6
假设瑕疵数服从独立的泊松分布muA和muB。
模型是log（mu）=a+bx，A对应x=0，B对应x=1。由题意，muA、muB的估计值分别为5和9。
通过log（5）=a ；log（9）=a+b 解得a=1.6094，b=0.5878
我用信息矩阵的方法：J=x'*w*x
x=[1 0; 1 1], w=[5 0; 0 9],
得到J=[14 9; 9 9]
J^-1=  [0.2000  -0.2000; -0.2000 0.3111]
那么var（a）=0.2 ， var（b）=0.311？ 可是与sas的数据不一样啊？

having trouble to understand what you have done. you might be correct anyway.