Calculating Variance (without Variance expression)

I am learning Mathematica on the fly, one of my tasks is to find the variance of white noise. I followed the tutorial for finding white noise by using the code:

WN = WhiteNoiseProcess[NormalDistribution[0, 10]];
data = RandomFunction[WN, {0, 10000}];

I know I can use the following code to find the variance:Variance[data]

However, I would like to find it by using the formula for variance. I checked the reference built into Mathematica and it says I can simply use:


I input data for the list:


When I do this, I don’t get the same output as when I use the Variance[data] code, but instead get:

So, I am curious what I am doing wrong? I’m sure it’s something simple I am not doing, but after spending a couple of hours wrestling with this, I am breaking down to ask. Sorry if this is a dumb question. Thank you in advance for your time.




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– bbgodfrey
May 15 ’15 at 22:24



You must extract the values (path) – see the doc for details under properties of temporaldata…
– ciao
May 15 ’15 at 22:31



I did some digging in there and used: Total[(data[“Values”] – Mean[data[“Values”]])^2]/(Length[ data[“Values”]] – 1) This resolved the same as the Variance expression.
– Sean Alto
May 15 ’15 at 22:39


2 Answers


The specific approach is as follows. Convert data to an ordinary list, eliminate an extra set of {}, and insert the list into your formula:

dta = First@Normal@data;
Last@Total[(dta – Last@Mean[dta])^2]/(Length[dta] – 1)

which gives the same result as


namely 102.245 for the particular set of random numbers used.



Outstanding, I was positive it was something small I was missing. Thank you for your help.
– Sean Alto
May 15 ’15 at 22:41

It is valuable to look at the properties of these complex objects, e.g. in your example:


To do your own variance:

val = First@data[“ValueList”];
Total[(val – Mean[val])^2]/(Length[val] – 1)

You can compare results of Variance and your mimic.