# Weighted Moving Average of Cubic

• Scootertaj
In summary, applying a second-order weighted moving average will not change the result of a cubic polynomial.

#### Scootertaj

1. Show that applying a second-order weighted moving average to a cubic polynomial will not change anything.
$$X_t = a_0 + a_1t + a_2t^2 + a_3t^3$$ is our polynomial
Second-order weighted moving average: $$\sum_{i=-L}^{i=L} B_iX_{t+i}$$
where $$B_i=(1-i^2I_2/I_4)/(2L+1-I_2^{2}/I_4)$$
where $$I_2=\sum_{i=-L}^{i=L} i^2$$

2. I did a similar problem where applying a linear moving average to a linear equation returns the same thing, but I'm not sure how to proceed here.

Scootertaj said:
1. Show that applying a second-order weighted moving average to a cubic polynomial will not change anything.
$$X_t = a_0 + a_1t + a_2t^2 + a_3t^3$$ is our polynomial
Second-order weighted moving average: $$\sum_{i=-L}^{i=L} B_iX_{t+i}$$
where $$B_i=(1-i^2I_2/I_4)/(2L+1-I_2^{2}/I_4)$$
where $$I_2=\sum_{i=-L}^{i=L} i^2$$

2. I did a similar problem where applying a linear moving average to a linear equation returns the same thing, but I'm not sure how to proceed here.

What is L supposed to be in your case?

L is any arbitrary number.
For any L, this should be true.

Scootertaj said:
...

where $$I_2=\sum_{i=-L}^{i=L} i^2$$
...

What is I4 ?

Sorry, it just follows the same pattern as I2:

$$I_4=\sum_{i=-L}^{i=L} i^{4}$$

Scootertaj said:
1. Show that applying a second-order weighted moving average to a cubic polynomial will not change anything.
$$X_t = a_0 + a_1t + a_2t^2 + a_3t^3$$ is our polynomial
Second-order weighted moving average: $$\sum_{i=-L}^{i=L} B_iX_{t+i}$$
where $$B_i=(1-i^2I_2/I_4)/(2L+1-I_2^{2}/I_4)$$
where $$I_2=\sum_{i=-L}^{i=L} i^2$$

2. I did a similar problem where applying a linear moving average to a linear equation returns the same thing, but I'm not sure how to proceed here.

So you have weighting factors of the form
$$B_i = A - B i^2, i=-L, \ldots, L.$$ What happens when you compute
$$\sum_{i=-L}^{L} (A - B i^2) [a_0 + a_1 (t+i) + a_2 (t+i)^2 + a_3 (t+i)^3]?$$

Ray Vickson said:
So you have weighting factors of the form
$$B_i = A - B i^2, i=-L, \ldots, L.$$ What happens when you compute
$$\sum_{i=-L}^{L} (A - B i^2) [a_0 + a_1 (t+i) + a_2 (t+i)^2 + a_3 (t+i)^3]?$$

I'm unsure as to how you rearranged the weights to get $$B_i = A - B i^2$$, would you mind clarifying what A and Bi2 are?

Last edited:
I mean, one way to write what we have is
$$\frac{1}{2L+1-I_2^{2}/I_4} \sum_{i=-L}^{i=L} (1-i^2\frac{I_2}{I_4}) [a_0 + a_1 (t+i) + a_2 (t+i)^2 + a_3 (t+i)^3]$$

I THINK I may have got it. I basically looked at what we had, and what we need.

In order for us to get back a0, for example, we need:

$$\frac{1}{2L+1-\frac{I_2^{2}}{I_4}}\sum_{i=-L}^{i=L}a_0(1-i^2\frac{I_2}{I_4}) = a_0$$
Well, let's multiply through:

$$\frac{a_0}{2L+1-\frac{I_2^{2}}{I_4}}\sum_{i=-L}^{i=L} 1 - a_0\frac{\sum_{i=-L}^{i=L} i^2\frac{I_2}{I_4}}{2L+1-\frac{I_2^{2}}{I_4}}$$
$$\frac{a_0(2L+1) - a_0\frac{I_2^{2}}{I_4}}{2L+1-\frac{I_2^{2}}{I_4}} = a_0$$

Then, do the same for the rest.

## What is a Weighted Moving Average of Cubic?

A Weighted Moving Average of Cubic is a data analysis technique used to smooth out a time series by calculating the average of a specified number of data points. It gives more weight to recent data points, making it more responsive to changes in the data over time.

## How is a Weighted Moving Average of Cubic calculated?

To calculate the Weighted Moving Average of Cubic, you would first assign weights to each data point, typically starting with the most recent data point and decreasing in weight with each preceding data point. Then, you would multiply each data point by its assigned weight and take the average of the resulting values.

## What are the advantages of using a Weighted Moving Average of Cubic?

One advantage of using a Weighted Moving Average of Cubic is that it gives more weight to recent data points, making it more responsive to changes in the data over time. This can help to identify trends and patterns in the data more accurately. Additionally, the use of weights allows for more flexibility in adjusting the smoothing process to fit the specific data set being analyzed.

## What are the limitations of a Weighted Moving Average of Cubic?

One limitation of using a Weighted Moving Average of Cubic is that it can be more complex and time-consuming to calculate compared to other moving average methods. Additionally, if the data contains a lot of noise or outliers, the weighted average may not accurately reflect the underlying trend in the data.

## How is a Weighted Moving Average of Cubic used in real-world applications?

A Weighted Moving Average of Cubic is commonly used in financial analysis, forecasting, and trend analysis. It can also be used to smooth out noisy data in fields such as medicine, economics, and engineering. Additionally, it is often used in combination with other statistical techniques to gain deeper insights into the data.