Jason,

Personally, I always find the lag functions slightly confusing. The important thing to remember is that lag<n>(x) doesn't return the value of x n observations ago, but the value of x when lag<n> was called n times ago. Therefore, conditionally executing lag<n> functions almost always returns results that you don't expect (in my experience at least!). You ca easily achieve what you want using thr retain statement:

data one; input x; cards; 1 2 3 4 5 6 ; run;

data two; retain t; set one; if _n_ eq 1 then t=1; else t=t*0.8; run;

proc print data=two; run;

HTH,

John "Jason Zhang" <qiyuzhang2@YAHOO.COM> wrote in message news:20041118044125.79499.qmail@web60602.mail.yahoo.com... > Hi, > > I want to generate a value for a variable, say t. But > tĄ¯s value depends on previous observationĄ¯s t value. > I used the code: > T=lag1(t)*0.8; > > It doesnĄ¯t work. The following is an example of what > I want to do: > data one; > input x; > cards; > 1 > 2 > 3 > 4 > 5 > 6 > ; > run; > > data two; > set one; > if _n_=1 then t=1; else > t=lag1(t)*0.8; > run; > > what I expected for tĄ¯s values are: > 1 > 0.8 > 0.64 > Ą­ > > Can anyone help me with this? Thanks a lot!!! > Jason > > > > > > __________________________________ > Do you Yahoo!? > The all-new My Yahoo! - Get yours free! > http://my.yahoo.com