I did not get the why there are N and N-1 while calculating population variance. When we use N and when we use N-1?

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It says that when population is very big there is no difference between N and N-1 but it does not tell why is there N-1 at the beginning.

Edit: Please don't confuse with n and n-1 which are used in estimating.

Edit2: I'm not talking about population estimation.

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    $\begingroup$ You can find an answer there: stats.stackexchange.com/questions/16008/…. Basically, you should use N-1 when you estimate a variance, and N when you compute it exactly. $\endgroup$
    – ocram
    Commented Nov 3, 2011 at 15:10
  • $\begingroup$ @ocram, as far as I know when we estimate a variance we use either n or n-1. $\endgroup$
    – ilhan
    Commented Nov 3, 2011 at 16:06
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    $\begingroup$ None of the answers below are written in terms of finite population inference. The word finite is absolutely crucial here; that's what Kish's book is about (and whoever was saying "The book is wrong" simply don't know enough about finite population surveys and samples). The quotient $N-1$ instead of $N$ just makes computations nicer and obviates the need to haul around factors like $1-1/N$. The full answer to this question would have to introduce the sampling inference where the sample indicators are random, and the values of observed characteristics $y$ are FIXED. Non-random. Set in stone. $\endgroup$
    – StasK
    Commented May 16, 2013 at 11:25
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    $\begingroup$ This doesn't really add to the other answers. That different divisors give different answers, or even that the difference diminishes with N, is not at issue. The question is when and why to use either divisor. $\endgroup$
    – Nick Cox
    Commented Aug 8, 2013 at 9:49
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    $\begingroup$ I would just like to mention to the other readers that this issue is called "Bassel's correction". You can check it out on Wikipedia en.wikipedia.org/wiki/… $\endgroup$ Commented Aug 11, 2018 at 9:07

5 Answers 5


Instead of going into maths I'll try to put it in plain words. If you have the whole population at your disposal then its variance (population variance) is computed with the denominator N. Likewise, if you have only sample and want to compute this sample's variance, you use denominator N (n of the sample, in this case). In both cases, note, you don't estimate anything: the mean that you measured is the true mean and the variance you computed from that mean is the true variance.

Now, you have only sample and want to infer about the unknown mean and variance in the population. In other words, you want estimates. You take your sample mean for the estimate of population mean (because your sample is representative), OK. To obtain estimate of population variance, you have to pretend that that mean is really population mean and therefore it is not dependent on your sample anymore since when you computed it. To "show" that you now take it as fixed you reserve one (any) observation from your sample to "support" the mean's value: whatever your sample might have happened, one reserved observation could always bring the mean to the value that you've got and which believe is insensitive to sampling contingencies. One reserved observation is "-1" and so you have N-1 in computing the variance estimate. The unbiased estimate is called sample variance (not to be confused with the sample's variance) which is an argot; it is better call what it is: sample unbiased estimate of population variance estimated with the sample's mean.

[Pasting here from my below comments: Imagine you are taking repeatedly samples of N=3 size. Of the 3 values in a sample, only 2 values express random deviatedness of observations from the population mean, but the left one expresses (takes on itself) the shift of the sample's mean from the population mean. Thus the "degree of free" observational variability is 2 of the 3, in each separate sample. When we estimate variability on a sample but want it to be an unbiased (unshifted) estimate of populational variability, we "believe" only those 2 free observations. We "pay" for the decision to measure variability off the sample mean as if it were the population mean, for we need to infer about the population variability. This "fee" (N-1 denominator, the Bessel correction) makes the variability wider, incorporating the oscillation of sample means within the variance, but it makes such variance an unbiased estimator.]

But imagine now that you somehow know the true population mean, but want to estimate variance from the sample. Then you will substitute that true mean into the formula for variance and apply denominator N: no "-1" is needed here since you know the true mean, you didn't estimate it from this same sample.

  • $\begingroup$ But my question has nothing to do with estimation. It is about computing population variance; with N and N-1. I'm not talking about n and n-1. $\endgroup$
    – ilhan
    Commented Nov 3, 2011 at 16:12
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    $\begingroup$ @ilhan, in my reply, I used N for both N and n. N is a size of a totality at hand, either population or sample. To compute population variance, you must have population at your disposal. If you have only sample you can either compute this sample's variance or compute population estimate variance. No other way round. $\endgroup$
    – ttnphns
    Commented Nov 3, 2011 at 16:21
  • $\begingroup$ I have a complete information about my population; all the values are know. I'm not interested in estimation. $\endgroup$
    – ilhan
    Commented Nov 3, 2011 at 16:35
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    $\begingroup$ If you do have your population then use N. N-1 would be illogical to use. $\endgroup$
    – ttnphns
    Commented Nov 3, 2011 at 17:00
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    $\begingroup$ @ilhan - Couldn't comment directly on your comment to ttnphns post, but here is an explanation of what you see in the book and how you should infer it. The symbol 'S' when used to imply variance always refers to sample variance. The Greek letter sigma is used to refer to the population variance. That is the reason why you see the book mention S = N * sigma / (N - 1) $\endgroup$
    – Arvind
    Commented Nov 13, 2011 at 9:23

$N$ is the population size and $n$ is the sample size. The question asks why the population variance is the mean squared deviation from the mean rather than $(N-1)/N = 1-(1/N)$ times it. For that matter, why stop there? Why not multiply the mean squared deviation by $1-2/N$, or $1-17/N$, or $\exp(-1/N)$, for instance?

There actually is a good reason not to. Any of these figures I just mentioned would serve just fine as a way to quantify a "typical spread" within the population. However, without prior knowledge of the population size, it would be impossible to use a random sample to find an unbiased estimator of such a figure. We know that the sample variance, which multiplies the mean squared deviation from the sample mean by $(n-1)/n$, is an unbiased estimator of the usual population variance when sampling with replacement. (There is no problem with making this correction, because we know $n$!) The sample variance would therefore be a biased estimator of any multiple of the population variance where that multiple, such as $1-1/N$, is not exactly known beforehand.

This problem of some unknown amount of bias would propagate to all statistical tests that use the sample variance, including t-tests and F-tests. In effect, dividing by anything other than $N$ in the population variance formula would require us to change all statistical tabulations of t-statistics and F-statistics (and many other tables as well), but the adjustment would depend on the population size. Nobody wants to have to make tables for every possible $N$! Especially when it's not necessary.

As a practical matter, when $N$ is small enough that using $N-1$ instead of $N$ in formulas makes a difference, you usually do know the population size (or can guess it accurately) and you would likely resort to much more substantial small-population corrections when working with random samples (without replacement) from the population. In all other cases, who cares? The difference doesn't matter. For these reasons, guided by pedagogical considerations (namely, of focusing on details that matter and glossing over details that don't), some excellent introductory statistics texts don't even bother to teach the difference: they simply provide a single variance formula (divide by $N$ or $n$ as the case may be).


There has, in the past been an argument that you should use N for a non-inferential variance but I wouldn't recommended that anymore. You should always use N-1. As sample size decreases N-1 is a pretty good correction for the fact that the sample variance gets lower (you're just more likely to sample near the peak of the distribution---see figure). If sample size is really big then it doesn't matter any meaningful amount.

An alternative explanation is that the population is a theoretical construct that's impossible to achieve. Therefore, always use N-1 because whatever you're doing you're, at best, estimating the population variance.

Also, you're going to be seeing N-1 for variance estimates from here on in. You'll likely not ever encounter this issue... except on a test when your teacher might ask you to make a distinction between an inferential and non-inferential variance measure. In that case don't use whuber's answer or mine, refer to ttnphns's answer.

Figure 1

Note, in this figure the variance should be close to 1. Look how much it varies with sample size when you use N to estimate the variance. (this is the "bias" referred to elswhere)

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    $\begingroup$ Please, tell me why N "not recommended anymore" with true population at hand? Population is not always a theoretical construct. Sometimes your sample is a bona fide population for you. $\endgroup$
    – ttnphns
    Commented Nov 3, 2011 at 17:10
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    $\begingroup$ ilhan, N can be used for your sample, or it can be used for the population size, if one exists. In most cases the distinction between big N and small n is dependent upon the topic. For example, n might be the number of cases in each condition in an experiment while N might be the number for the experiment. They're both samples. There is no global rule. $\endgroup$
    – John
    Commented Nov 3, 2011 at 17:39
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    $\begingroup$ ttnphns, it depends on what you mean by population. I would argue that if your whole population is so small that N-1 matters then it's questionable whether calculating a mean squared deviation is remotely useful at all. Show all the values, their shape and range. Furthermore, the whole old argument that you actually have N degrees of freedom if you're not making an inference is questionable. You lost one when you calculated the mean, that you needed to calculate the variance. $\endgroup$
    – John
    Commented Nov 3, 2011 at 17:42
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    $\begingroup$ @John, if you calculate mean within population you just state the fact about the parameter, so you spend no degrees of freedom. If you calculate it in sample and want to infer about the population, then you do spend one. Also, I can have population with N=1. With denominator N-1, it appeares that such parameter as variance does not exist for it. It is nonsense. $\endgroup$
    – ttnphns
    Commented Nov 3, 2011 at 17:55
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    $\begingroup$ @ilhan Please, consider updating your question (as you did) and point to the updated version rather than leaving such non-constructive comments. Everything is debatable, especially when the question itself lacks some context. Here it seems that the problem stands from defining what a population really is. $\endgroup$
    – chl
    Commented Nov 3, 2011 at 22:41

Generally, when one has only a fraction of the population, i.e. a sample, you should divide by n-1. There is a good reason to do so, we know that the sample variance, which multiplies the mean squared deviation from the sample mean by (n−1)/n, is an unbiased estimator of the population variance.

You can find a proof that the estimator of the sample variance is unbiased here: https://economictheoryblog.com/2012/06/28/latexlatexs2/

Further, if one were to apply the estimator of the population variance, that is the version of the variance estimator that divides by n, on a sample of instead of the population, the obtained estimate would biased.

  • $\begingroup$ This seems to answer a different question concerning estimating the population variance. It looks circular: isn't this answer predicated on assuming a specific convention for defining the population variance in the first place? $\endgroup$
    – whuber
    Commented Sep 1, 2016 at 16:16

The population variance is the sum of the squared deviations of all of the values in the population divided by the number of values in the population. When we are estimating the variance of a population from a sample, though, we encounter the problem that the deviations of the sample values from the mean of the sample are, on average, a little less than the deviations of those sample values from the (unknown) true population mean. That results in a variance calculated from the sample being a little less than the true population variance. Using an n-1 divisor instead of n corrects for that underestimation.

  • $\begingroup$ @ Bunnenburg, If you got answer to your question. Please clear to me now, what you got? It is a big confusion to me as well. $\endgroup$
    – Bilal Para
    Commented Jul 20, 2017 at 6:23
  • $\begingroup$ to compensate for that little less variance we get, why can't one use n-2, n-3, etc.? why n-1 in particular? why not a constant...??? $\endgroup$ Commented May 15, 2018 at 19:02
  • $\begingroup$ @SaravanabalagiRamachandran The discrepancy varies with the sample size and so a constant will not serve. The correction using n-1 is closer works better than the others you mention. $\endgroup$ Commented May 15, 2018 at 20:18

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