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Given a time series $V_t_t=1^n$, where $t mapsto V_t in mathbbR$, I want to have some smoothed notion of the "derivative" of this time series. It was recommended that I look at

$$tildeV_t = fracV_t - text$k$-step moving average of V_ttext$k$-step std dev of $V_t$.$$

The $k$-step moving average of $V_t$ is $frac1k sum_i=0^k-1V_t-i$.

This quantity, empirically, seems to act something like derivative. (For a line it is constant, for a sine curve it is almost a cosine curve, etc.) Can someone explain why, mathematically, this would be a heuristic for a derivative?

Thanks.

real-analysis statistics discrete-mathematics time-series

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edited Apr 4 at 19:36

Théophile

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asked Apr 4 at 19:02

statsdadstatsdad

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statsdad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

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  $begingroup$
  Derivatives don't really make sense with discrete time, so it's better to compare it to the differences $V_t - V_t-1$. In both cases you're looking at measures of how quickly $V_t$ is changing relative to the recent past, so you would expect the metrics to have similar properties.
  $endgroup$
  – Jair Taylor
  Apr 4 at 20:05
  

  
  
  
  


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  $begingroup$
  @JairTaylor Derivatives do have a discrete analogue, finite differences.
  $endgroup$
  – Théophile
  Apr 4 at 20:11
  

  
  
  
  


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  $begingroup$
  @Théophile Sure, you could also compare to finite difference quotients over longer intervals.
  $endgroup$
  – Jair Taylor
  Apr 4 at 20:14
  
  

  
  
  
  

add a comment | 

2

1

$begingroup$

Given a time series $V_t_t=1^n$, where $t mapsto V_t in mathbbR$, I want to have some smoothed notion of the "derivative" of this time series. It was recommended that I look at

$$tildeV_t = fracV_t - text$k$-step moving average of V_ttext$k$-step std dev of $V_t$.$$

The $k$-step moving average of $V_t$ is $frac1k sum_i=0^k-1V_t-i$.

This quantity, empirically, seems to act something like derivative. (For a line it is constant, for a sine curve it is almost a cosine curve, etc.) Can someone explain why, mathematically, this would be a heuristic for a derivative?

Thanks.

real-analysis statistics discrete-mathematics time-series

share|cite|improve this question

edited Apr 4 at 19:36

Théophile

20.4k13047

asked Apr 4 at 19:02

statsdadstatsdad

111

New contributor

statsdad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Derivatives don't really make sense with discrete time, so it's better to compare it to the differences $V_t - V_t-1$. In both cases you're looking at measures of how quickly $V_t$ is changing relative to the recent past, so you would expect the metrics to have similar properties.
  $endgroup$
  – Jair Taylor
  Apr 4 at 20:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JairTaylor Derivatives do have a discrete analogue, finite differences.
  $endgroup$
  – Théophile
  Apr 4 at 20:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Théophile Sure, you could also compare to finite difference quotients over longer intervals.
  $endgroup$
  – Jair Taylor
  Apr 4 at 20:14
  
  

  
  
  
  

add a comment | 

$begingroup$

Given a time series $V_t_t=1^n$, where $t mapsto V_t in mathbbR$, I want to have some smoothed notion of the "derivative" of this time series. It was recommended that I look at

$$tildeV_t = fracV_t - text$k$-step moving average of V_ttext$k$-step std dev of $V_t$.$$

The $k$-step moving average of $V_t$ is $frac1k sum_i=0^k-1V_t-i$.

This quantity, empirically, seems to act something like derivative. (For a line it is constant, for a sine curve it is almost a cosine curve, etc.) Can someone explain why, mathematically, this would be a heuristic for a derivative?

Thanks.

real-analysis statistics discrete-mathematics time-series

share|cite|improve this question

edited Apr 4 at 19:36

Théophile

20.4k13047

asked Apr 4 at 19:02

statsdadstatsdad

111

New contributor

statsdad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Apr 4 at 19:36

Théophile

20.4k13047

asked Apr 4 at 19:02

statsdadstatsdad

111

New contributor

statsdad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Apr 4 at 19:36

Théophile

20.4k13047

asked Apr 4 at 19:02

statsdadstatsdad

111

New contributor

statsdad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Apr 4 at 19:36

Théophile

20.4k13047

edited Apr 4 at 19:36

Théophile

20.4k13047

asked Apr 4 at 19:02

statsdadstatsdad

111

New contributor

statsdad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Apr 4 at 19:02

statsdadstatsdad

111