Proof of multiplier formula

54 Views Asked by At

I don't understand two parts in the formula. In the first place how can we remove the sommation? Finally, in the last part how are the beta zero and the phi before the beta zero handled? Thank you in advance

$$\beta_0+\sum_{j=1}^\infty\phi^{j-1}(\phi\beta_0+\beta_1)=\beta_0+\frac{\phi\beta_0+\beta_1}{1-\phi}=\frac{\beta_0+\beta_1}{1-\phi}.$$

1

There are 1 best solutions below

3
On

Since\begin{align}\sum_{j=1}^N\phi^{j-1}(\phi\beta_0+\beta_1)&=\beta_0\sum_{j=1}^N\phi^j+\beta_1\sum_{j=1}^N\phi^{j-1}\\&=\beta_0\frac{\phi-\phi^{N+1}}{1-\phi}+\beta_1\frac{1-\phi^N}{1-\phi},\end{align}we have\begin{align}\sum_{j=1}^\infty\phi^{j-1}(\phi\beta_0+\beta_1)&=\beta_0\frac\phi{1-\phi}+\frac{\beta_1}{1-\phi}\\&=\frac{\beta_0\phi+\beta_1}{1-\phi}.\end{align}It should be clear that$$\beta_0+\frac{\beta_0\phi+\beta_1}{1-\phi}=\frac{\beta_0+\beta_1}{1-\phi}.$$