# Why do my weights have to equal one?

I'm currently learning the very basics of exponential smoothing.

As follows:

The book first presents the following model:

$$\sum{\theta^tY_{T-1}}$$

It then claims that the sum of all weights add up to

$$\sum \theta^t=\frac{1-\theta^T}{1-\theta}$$

...Which doesn't necessarily add up to 1. I don't contest any of this.

If we do want the sum of our weights to add up to 1, we have to multiply every term with $\frac {1-\theta}{1-\theta^T}$ For large sums (since $\theta^T$ goes towards zero) this can be written as:

$$y_T=(1-\theta)\sum{\theta^tY_{T-1}}$$

.....And that's all very fine and dandy, but why would I want the sum of my weights to add up to one in the first place?

Why is this last step necessary?

## 1 Answer

Could it be that the formula is $y_T = \sum_t \theta^t Y_{T-t}$ ?

I think that you will want your ''smoothed series'' to have values that are somewhere in between the values of your original series' values, at least that makes sense to me.

If you use weights $w_i$ that are non-negative, at most one and sum up to one, i.e. $0 \le w_t \le 1, \sum_t w_t=1$, then the weighted average $\sum_t w_t Y_{T-t}$ will be a value that is somewhere in between all the values $Y_{T-t}$. ($\sum_t w_i Y_{T-t}$ is called a convex combination of the $Y_{T-t}$).

You can easily see this when you do it for only two terms $a$ and $b$; in that case the weighted average will be like $w a + (1-w) b$. If $0 \le w \le 1$ then $w a + (1-w) b$ will always be between $a$ and $b$.

You can easily try it with the R-code below:

a<-2
b<-5

w1<-seq(0,1,0.01)

x<- w1 * a + (1-w1) * b

cat(x)


EDIT: The R-code takes two values $a=2$ and $b=5$, and it computes $x=w_1 a + w_2 b$ where $w_1+w_2=1$ or $w_2=1-w_1$ so it computes $x=w_1a + (1-w_1)b$ for $w_1=0, 0.01, 0.02, 0.03, \dots , 1$. Whatever values for $w_1$ between $0$ and $1$ you take, this $x$ will always be between a(=2) and b(=5).

• You probably meant $y_T = \sum_t \theta^t Y_{T-t}$ in your first paragraph..? – Tim Dec 31 '15 at 12:14
• @Tim: You are rirght, thanks ! I edited the answer (+1) – user83346 Dec 31 '15 at 12:20
• That's correct! About the formula I mean. Still trying to work out why thee weights need to add up to one for the smoother to end up between the original values – Magnus Dec 31 '15 at 12:26
• Read the wiki article: en.wikipedia.org/wiki/Convex_combination. Will file this under "things I need a math degree in order to understand". – Magnus Dec 31 '15 at 12:59
• @Magnus: if the weights add up to 1 then the smoothed value will be in between. This will always be. If the weights do not add up to one then you can never guarantee this. In the R-code this is shown for two values a and b: $w_1 a + w_2 b$ will always be between a and be when $w_1+w_2=1$ (or when $w_2 = 1 - w_1$). When you take weights that do not sum up to one, you may get a values larger than b or smaller than a. Note that the arithmetic average is a special case of ''convex combination'' namey one where $w_i=\frac{1}{n}, i=1,2,\dots n$. Note that $\sum_{i=1}^{n} \frac{1}{n} = 1$... – user83346 Dec 31 '15 at 13:24