10-26-2012 06:11 PM
Can someone please help me with the following coding?
I need to create a data set of 1000 values for the following model:
X(t) = (1/((1-x)**0.4))e ,where e is a standard normal random variable and L is the lag operator. I don't know how to deal with the lag operator. finally I need to plot X vs. t.
But I need use the 1st 100 terms of the INFINITE BINOMIAL expansion of the term (1/((1-x)**0.4)).
10-29-2012 08:56 AM
Since I can't observe the lag operator in your expression, I'll try to explain you the usage of lag function in general.
x = lagN(varname); - this will produce the value of 'varname' on a step t - N.
So y = x(t-1) will be
y = lag1(x);
y = x(t-3) will be
y = lag3(x);
Notice: consider reading help about lag function and using it in conditional statements.
X(t) = (1/((1-X(t-1))**0.4))* e;
var t X e;
do t = 1 to 1000;
e = rand('NORMAL', 0, 1);
X =( 1/( (1 - coalesce(lag1(X),0))**0.4 ) ) * e ;