0
$\begingroup$

Let's say I have a data set of $10,20,30$. My mean and variance here are mean= $20$ and variance = $66.667$. Is there a formula that lets me calculate the new variance value if I was to remove $10$ and add $50$ to the data set turning it into $20,30,50$?

$\endgroup$
  • $\begingroup$ you may find this relevant :math.stackexchange.com/questions/102978/… $\endgroup$ – Sean Lee Feb 14 '19 at 13:30
  • $\begingroup$ @SeanLee That link focuses on if we were just adding data to the dataset, but what about removing? $\endgroup$ – dude8998 Feb 14 '19 at 13:42
  • $\begingroup$ I haven't worked it out in detail, but I believe if you know how to calculate the incremental SD by adding data, this formulation should also allow you to calculate the incremental SD by removing data. $\endgroup$ – Sean Lee Feb 14 '19 at 13:44
  • $\begingroup$ Yeah, I tried to derive but I just can't wrap my head around it, would help if someone else worked it out $\endgroup$ – dude8998 Feb 14 '19 at 13:52
1
$\begingroup$

Suppose there are $n$ values in the data set and we replace a value $x$ with a new value $x'$.

First calculate the new mean $M'$:

$M' = M + \frac{x'-x}{n}$

where $M$ is the old mean. Then calculate the new variance:

$V' = V + (M'-M)^2 + \frac{(x'-M')^2-(x-M')^2}{n}$

where $V$ is the old variance. $(M'-M)^2$ is the change due to the movement of the mean and $\frac{(x'-M')^2-(x-M')^2}{n}$ is the change due to the replacement of $x$ by $x'$.

In your example, $n=3$, $x=10$, $x'=50$ so:

$M' = 20 +\frac{50-10}{3}=\frac{100}{3}$

$V' = \frac{200}{3} + \frac{40^2}{9} + \frac{50^2-70^2}{27} = \frac{1400}{9}$

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Denote the running SD (of window length $n$) at the $k$-th time step as $s_{k:n+k-1}$, and the corresponding running mean as $\bar{X}_{k:n+k-1}$ (The subscript specifies the datapoints that we are taking in our calculations, which will be relevant for later).

What you're asking, is essentially, for every time step, that given $s_{k:n+k-1}$ to:

  1. Calculate a temporary SD $s_{k+1:n+k-1}$ first by removing the "old" data point
  2. Use $s_{k+1:n+k-1}$ to calculate the new SD $s_{k+1:n+k}$

The rest follows directly from incremental computation of standard deviation:

and it is easy to show that the summation term above is equal to $0$ which gives $$ s^2_n = \frac{(n - 2)s^2_{n - 1} + (n - 1)(\bar X_{n - 1} - \bar X_n)^2 + (X_n - \bar X_{n})^2}{n - 1}. $$

Or if I were to write it in the notation that I have introduced, where I treat $X_k$ as the "new" datapoint (although it's the datapoint we want to remove):

$$ s^2_{k:n+k-1} = \frac{(n - 2)s^2_{k+1:n+k-1} + (n - 1)(\bar X_{k+1:n+k-1} - \bar X_{k:n+k-1})^2 + (X_k - \bar X_{k:n+k-1})^2}{n - 1}. $$

The following step would just be simple algebra:

$$ s^2_{k+1:n+k-1} = \frac{(n-1) s^2_{k:n+k-1} - (n - 1)(\bar X_{k+1:n+k-1} - \bar X_{k:n+k-1})^2 - (X_k - \bar X_{k:n+k-1})^2}{n-2} $$

Now, since we have $s^2_{k+1:n+k-1}$, we can calculate $s^2_{k+1:n+k}$, which is what we want. Of course, we just apply the formula that we were given again:

$$ s^2_{k+1:n+k} = \frac{(n - 2)s^2_{k+1:n+k-1} + (n - 1)(\bar X_{k+1:n+k-1} - \bar X_{k+1:n+k})^2 + (X_{k+n} - \bar X_{k+1:n+k})^2}{n - 1}. $$

And we have obtained the running SD (or Variance) which you want. I believe you've already figured out how to calculate the running means, so I won't go through that.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.