# simplifying a summation for a time series

I was curious how to simplify this summation that is giving me a bit of trouble

$Y_t$=$-\sum_{j=1}^\infty (\frac{1}{3})^j*e_{t+j}$

So i get that this is a geometric sum that converges and I would was thinking I would do the following

Sum = -$\frac{\frac{1}{3}*e_{t+1}}{1-\frac{1}{3}}=-\frac{1}{2}*e_{t+1}$

but the index on $e_{t+j}$ is causing me some problems. Do I need to take the $e_{t+j}$ into account in the denominator when I subtract the ratio?

Any help would be appreciated.

• $\sum_{n=1}^\infty n^2\times\left(\frac12\right)^n$ is not a geometric series, for example. And the reason they are called geometric series, as a fun fact, is because you can make really cool geometric representations of the series. Anyways, what is $e_{t+j}$? – Simply Beautiful Art Oct 1 '16 at 22:10
• its an error term for a time series....but the $e_{t+j}$ is a subscript not a power so its still not a geometric series? – statsGuy Oct 1 '16 at 22:20
• If $e_{t+j}$ changes as $j$ changes, then it is not a geometric sequence. If it didn't change much, though, you could say it is approximately the geometric series multiplied by the average value of $e_{t+j}$. Do you have a closed form for $e_{t+j}$? – Simply Beautiful Art Oct 1 '16 at 22:23
• only that they are iid distributed as being normal (0,1) ....so I were to assume they did not change much what form would the solution take? – statsGuy Oct 1 '16 at 22:27
• I can't tell you exactly, as I don't know what it is exactly, so you can basically use your evaluation above, or you can use the average value of $e_{t+j}$. – Simply Beautiful Art Oct 1 '16 at 22:29

You can approximate the summation as you have tried:

$$\text{Sum}\approx-\frac12e_{t+1}$$

If you wanted to other values of $e_{t+j}$ into consideration,

$$\text{Sum}\approx-\frac12\operatorname{avg}(\{e_{t+j}\}_{j=1}^\infty)$$

Or, you could use the old fashioned method:

$$\text{Sum}=-\frac13e_{t+1}-\frac19e_{t+2}-\frac1{27}e_{t+3}-\dots$$

And just cut the sum short once you feel accurate enough.