英文:
Logic error of cumulative statement and lag function
问题
方法1是正确的，因为它按照给定的规则计算X(t)。方法2和方法3是错误的，因为它们在计算时没有正确地遵循规则。要修复它们，可以按照以下方式修改代码：
方法2：
data want;
set have;
*Method 2;
y = 0.5 * y - 0.5 * lag(y) + e;
run;
方法3：
data want;
set have;
*Method 3;
retain z;
if _n_ = 1 then z = e;
else if _n_ = 2 then z = 1.5 * z + e;
else z = 1.5 * z - 0.5 * lag2_z + e;
lag2_z = lag2(z);
run;
这些修改会确保方法2和方法3按照规则正确计算X(t)。
英文:
This task is going to derive a time-series variable, The rule is:
> X(t) = 1.5 * X(t-1) - 0.5 * X(t-2) + e;
> X(1) = e;
> X(2) = 1.5 * X(1) + e;
where e ~ N(0,1). Here is how I generate this sequence:
data have;
call streaminit(42);
do i=1 to 200;
e=rand("normal",0,1);
output;
end;
run;
data want;
set have;
*Method 1;
x+sum(0.5*x,-0.5*lag(x),e);
*Method 2;
y+0.5*y-0.5*lag(y)+e;
*Method 3;
retain z;
if _n_=1 then z=e;
if _n_=2 then z=1.5*z+e;
lag2_z=lag2(z);
if _n_>2 then z=1.5*z-0.5*lag2_z+e;
run;
Run this code, I just find only result of method 1 is right. Why Method 2 and Method 3 are wrong? How to fix them?
TIA.
答案1
得分: 1
Data:
我会使用稍微简单一些的数据，例如：
data have;
call streaminit(42);
do i=1 to 20;
e=1; /*rand("normal",0,1);*/
output;
end;
run;
Metchod1:
这个表达式 x+sum(0.5*x,-0.5*lag(x),e); 不同于 X(t) = 1.5 * X(t-1) - 0.5 * X(t-2) + e;，你在 0.5 中漏掉了 1。
Method2:
在第一次迭代中，这个表达式 0.5*y-0.5*lag(y)+e; 的值为缺失 (.)，因为 lag(y) 是缺失的，所以 sum 语句实际上是 y+.;，这将导致错误的结果。
Method3:
添加一些 put 语句以查看计算过程，例如：
*Method 3;
retain z;
put z= lag2_z= e=;
if _n_=1 then z=e;
if _n_=2 then z=1.5*z+e;
lag2_z=lag2(z);
put z= lag2_z= e=;
if _n_ > 2 then z=1.5*z-0.5*lag2_z+e;
put _N_ z= / /;
我会这样做：
%let n = 100;
%let e=1; /* 将 1 更改为 Rand() */
/*
%let e = rand("normal",0,1);
*/
data want;
call streaminit(42);
e=&e.;
a=e;
output;
b=1.5*a + e;
output;
do i=1 to &n.-2;
e=&e.;
c= 1.5 * b - 0.5 * a + e;
a=b;
b=c;
put _all_;
end;
run;
英文:
Data:
I would go with a bit simpler data, e.g.
data have;
call streaminit(42);
do i=1 to 20;
e=1; /*rand("normal",0,1);*/
output;
end;
run;
Metchod1:
This x+sum(0.5*x,-0.5*lag(x),e); is different from X(t) = 1.5 * X(t-1) - 0.5 * X(t-2) + e;, you are missing 1 in 0.5.
Method2:
In the first iteration this expression 0.5*y-0.5*lag(y)+e; evaluates to missing (.) since lag(y) is missing, so the sum statement is in fact y+.; which gives wrong result.
Method3:
Add some put statements to see how it calculates, e.g.
*Method 3;
retain z;
put z= lag2_z= e=;
if _n_=1 then z=e;
if _n_=2 then z=1.5*z+e;
lag2_z=lag2(z);
put z= lag2_z= e=;
if _n_>2 then z=1.5*z-0.5*lag2_z+e;
put _N_ z= / /;
I would do it like this:
%let n = 100;
%let e=1; /* change 1 to Rand() */
/*
%let e = rand("normal",0,1);
*/
data want;
call streaminit(42);
e=&e.;
a=e;
output;
b=1.5*a + e;
output;
do i=1 to &n.-2;
e=&e.;
c= 1.5 * b - 0.5 * a + e;
a=b;
b=c;
put _all_;
end;
run;
答案2
得分: 0
sum()函数在求和时会忽略缺失值。
sum语句也会忽略缺失值，然而在方法2中，summand是常规算术求和，其中包括lag(y)，而lag(y)是缺失的，因此在t=2时e从不起作用。换句话说，在t=1时，0.5*y-0.5*lag(y)+e是缺失的。
方法3.存在问题，因为对于_n_>2，lag2在重新计算z之前。
将这些添加到输出可能有助于你理解。
英文:
sum() function ignores missing values in it's summation.
The sum statement also ignores missing values, however the summand in method 2 is normal arithmetic sum with lag(y) which is missing and thus the e never gets into play at t=2. In other words, at t=1 0.5*y-0.5*lag(y)+e is missing.
Method 3. is hiccupping because the lag2 for n>2 is before the recalculated z
Adding these to output might help you suss it out in your head.
* method 3;
retain z;
if _n_=1 then z=e;
if _n_=2 then z=1.5*z+e;
lag2_z=lag2(z);
pre_z = z;
if _n_>2 then z=1.5*z-0.5*lag2_z+e;
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