# Show that $y_t=\sum_{j=1}^\infty(1+\theta)(-\theta)^{j-1}y_{t-j}+\epsilon_t$

Suppose that $y_t$ is a $ARIMA(0,1,1)$ model $y_t=y_{t-1}+\epsilon_t+\theta\epsilon_{t-1}$. Show that $$y_t=\sum_{j=1}^\infty(1+\theta)(-\theta)^{j-1}y_{t-j}+\epsilon_t$$ where $\epsilon_t$ is White Noise with $E[\epsilon_t]=0$ and $Var(\epsilon_t)=\sigma^2$

Let $x_t=\epsilon_t+\theta\epsilon_{t-1}$ a $MA(1)$ process, then $$y_t=y_{t-1}+x_t$$ where $x_t=(1+\theta B)\epsilon_t$ where $B$ is the lag operator. Then assuming that $|\theta|<1$ the process is invertible and we can write $x_t$ as $$x_t=\sum_{j=1}^\infty (-\theta)^jx_{t-j}+\epsilon_t$$ The problem is when I substitute $x_t=y_t-y_{t-1}$ I don't get the result.

• It is the same, you just need to rewrite the series collecting the $y_t$s , each one appearing twice. Commented Feb 17, 2018 at 8:06

You have

$$y_t=y_{t-1}+\epsilon_{t}+\theta\epsilon_{t-1}$$

As well as

$$\epsilon_{t-j}=y_{t-j}-y_{t-j-1}-\theta\epsilon_{t-j-1}, \ j=1,2,...$$

Set $j=1$ and substitute $\epsilon_{t-1}$ in first equation with $\epsilon_{t-1}=y_{t-1}-y_{t-2}-\theta\epsilon_{t-2}$. Then set $j=2$ and similarly substitute $\epsilon_{t-2}$. Keep iterating.

Now $y_t$ converges to the desired expression assuming that $|\theta| <1$.

• I don't get what you mean. Commented Dec 18, 2016 at 21:46