B
Balvinder Singh
Member
Can someone please explain how does the maths work in the solution of part (iv) of Q8.
Thanks
Balvin
Thanks
Balvin
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We have:
\((1-0.4B)Y_t = (0.1+0.9e_{t-1}+e_t)\)
So we multiply both sides by the inverse of \((1-0.4B)\):
\(Y_t = (1-0.4B)^{-1}(0.1+0.9e_{t-1}+e_t)\)
We can then use a series expansion on the \((1-0.4B)^{-1}\) part using the formula given on page 2 of the Tables.
We have:
\((1-0.4B)Y_t = (0.1+0.9e_{t-1}+e_t)\)
So we multiply both sides by the inverse of \((1-0.4B)\):
\(Y_t = (1-0.4B)^{-1}(0.1+0.9e_{t-1}+e_t)\)
We can then use a series expansion on the \((1-0.4B)^{-1}\) part using the formula given on page 2 of the Tables.
B
Balvinder Singh
Member
Thank you.