However, if one wants to add intuition to a formula the already mentioned suggestions appear reasonable. First, observations of a sample are on average closer to the sample mean than to the population mean.
The variance estimator makes use of the sample mean and as a consequence underestimates the true variance of the population. Dividing by n-1 instead of n corrects for that bias. Furthermore, dividing by n-1 make the variance of a one-element sample undefined rather than zero.
At the suggestion of whuber , this answer has been copied over from another similar question. Bessel's correction is adopted to correct for bias in using the sample variance as an estimator of the true variance.
The bias in the uncorrected statistic occurs because the sample mean is closer to the middle of the observations than the true mean, and so the squared deviations around the sample mean systematically underestimates the squared deviations around the true mean.
To see this phenomenon algebraically, just derive the expected value of a sample variance without Bessel's correction and see what it looks like. In regression analysis this is extended to the more general case where the estimated mean is a linear function of multiple predictors, and in this latter case, the denominator is reduced further, for the lower number of degrees-of-freedom. This also agrees with defining variance of a random variable as the expectation of the pairwise energy, i.
To go from the random variable defintion of variance to the defintion of sample variance is a matter of estimating a expectation by a mean which is can be justified by the philosophical principle of typicality: The sample is a typical representation the distribution. Note, this is related to, but not the same as estimation by moments.
To answer this question, we must go back to the definition of an unbiased estimator. An unbiased estimator is one whose expectation tends to the true expectation. The sample mean is an unbiased estimator. To see why:. Suppose that you have a random phenomenon. Oddly, the variance would be null with only one sample. This makes no sense. The illusion of a zero-squared-error can only be counterbalanced by dividing by the number of points minus the number of dofs.
This issue is particularly sensitive when dealing with very small experimental datasets. Generally using "n" in the denominator gives smaller values than the population variance which is what we want to estimate.
This especially happens if the small samples are taken. If you are looking for an intuitive explanation, you should let your students see the reason for themselves by actually taking samples! Watch this, it precisely answers your question. There is one constraint which is that the sum of the deviations is zero.
I think it's worth pointing out the connection to Bayesian estimation. You want to draw conclusions about the population. The Bayesian approach would be to evaluate the posterior predictive distribution over the sample, which is a generalized Student's T distribution the origin of the T-test.
The generalized Student's T distribution has three parameters and makes use of all three of your statistics. If you decide to throw out some information, you can further approximate your data using a two-parameter normal distribution as described in your question.
From a Bayesian standpoint, you can imagine that uncertainty in the hyperparameters of the model distributions over the mean and variance cause the variance of the posterior predictive to be greater than the population variance. I'm jumping VERY late into this, but would like to offer an answer that is possibly more intuitive than others, albeit incomplete. The non-bold numeric cells shows the squared difference.
My goodness it's getting complicated! I thought the simple answer was You just don't have enough data outside to ensure you get all the data points you need randomly. The n-1 helps expand toward the "real" standard deviation. Sign up to join this community. The best answers are voted up and rise to the top. Stack Overflow for Teams — Collaborate and share knowledge with a private group. Create a free Team What is Teams? Learn more. Ask Question. Asked 11 years ago. Active 10 months ago.
Viewed k times. Improve this question. Tal Galili Tal Galili You ask them "why this? Watch this, it precisely answers you question.
Add a comment. Active Oldest Votes. Improve this answer. Michael Lew Michael Lew In essence, the correction is n-1 rather than n-2 etc because the n-1 correction gives results that are very close to what we need. More exact corrections are shown here: en. What if it overestimates? Show 1 more comment.
Dror Atariah 2 2 silver badges 15 15 bronze badges. Why is it that the total variance of the population would be the sum of the variance of the sample from the sample mean and the variance of the sample mean itself? How come we sum the variances? See here for intuition and proof. Show 4 more comments. I have to teach the students with the n-1 correction, so dividing in n alone is not an option. As written before me, to mention the connection to the second moment is not an option.
Although to mention how the mean was already estimated thereby leaving us with less "data" for the sd - that's important. Regarding the bias of the sd - I remembered encountering it - thanks for driving that point home.
In other words, I interpreted "intuitive" in your question to mean intuitive to you. Thank you for the vote of confidence :. The loose of the degree of freedom for the estimation of the expectancy is one that I was thinking of using in class.
But combining it with some of the other answers given in this thread will be useful to me, and I hope others in the future. Show 3 more comments. Based on some of the responses I received, I will try to describe my favorite way of looking at the issue. If you want to follow along in R, you can copy the code from each code section; beginning with some setup code. The variance is a measure of the dispersion around the mean, and in that sense this formula makes sense.
We then divide this sum by the number of observations as a scaling factor. If we ignore this number, we could get a very high variance simply by observing a lot of data. So, to fix that problem, we divide by the total number of observations.
However, this is the formula for the population variance. The formula for calculating the variance of a sample is:. If you Google this question, you will get a variety of answers. It does not actually help me understand 1 the problem and 2 why the solution is the solution that it is. So, below I am going to try to figure it out in a way that actually makes conceptual and intuitive sense to me.
The problem with using the population variance formula to calculate the variance of a sample is that it is biased. It is biased in that it produces an underestimation of the true variance. We simulate a population of data points from a uniform distribution with a range from 1 to Below I show the histogram that represents our population.
The variance is 8. To start, we can draw a single sample of size 5. Say we do that and get the following values: 7, 6, 3, 5, 5.
In the former case, this will result in 1. Below I show the results of draws from our population. I simulated drawing samples of size 2 to 10, each different times. We see that the biased measure of variance is indeed biased. The average variance is lower than the true variance indicated by the dashed line , for each sample size. We also see that the unbiased variance is indeed unbiased. On average, the sample variance matches that of the population variance.
The results of using the biased measure of variance reveals several clues for understanding the solution to the bias. We see that the amount of bias is larger when the sample size of the samples is smaller.
So the solution should be a function of sample size, such that the required correction will be smaller as the sample size increases. Ideally we would estimate the variance of the sample by subtracting each value from the population mean. This is where the bias comes in.
In fact, the mean of a sample minimizes the sum of squared deviations from the mean. This means that the sum of deviations from the sample mean is always smaller than the sum of deviations from the population mean. The only exception to that is when the sample mean happens to be the population mean. Below are two graphs. In each graph I show 10 data points that represent our population. I also highlight two data points from this population, which represents our sample.
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