Weighted Moving Average of Cubic

[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.

 

[Ray Vickson]

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?

 

[Scootertaj]

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

 

[SammyS]

...

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

What is I₄ ?

 

[Scootertaj]

Sorry, it just follows the same pattern as I2:

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

 

[Ray Vickson]

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]?

 

[Scootertaj]

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 B_i² are?

 

[Scootertaj]

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]

 

[Scootertaj]

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.