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. 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 It is a parameter. And when we calculate, when we attempt to calculate something for a sample we would call that a statistic-- statistic. So how do we think about the mean for a population? Well, first of all, we denote it with the Greek letter mu.
And we essentially take every data point in our population. So we take the sum of every data point. So we start at the first data point and we go all the way to the capital Nth data point. So every data point we add up. So this is the i-th data point, so x sub 1 plus x sub 2 all the way to x sub capital N. And then we divide by the total number of data points we have. Well, how do we calculate the sample mean? Well, the sample mean-- we do a very similar thing with the sample.
And we denote it with a x with a bar over it. And that's going to be taking every data point in the sample, so going up to a lower case n, adding them up --so these are the sum of all the data points in our sample-- and then dividing by the number of data points that we actually had.
Now, the other thing that we're trying to calculate for the population, which was a parameter, and then we'll also try to calculate it for the sample and estimate it for the population, was the variance, which was a measure of how dispersed or how much of the data points vary from the mean.
So let's write variance right over here. And how do we denote any calculate variance for a population? Well, for population, we'd say that the variance --we use a Greek letter sigma squared-- is equal to-- and you can view it as the mean of the squared distances from the population mean. But what we do is we take, for each data point, so i equal 1 all the way to n, we take that data point, subtract from it the population mean.
So if you want to calculate this, you'd want to figure this out. Well, that's one way to do it. We'll see there's other ways to do it, where you can calculate them at the same time. But the easiest or the most intuitive is to calculate this first, then for each of the data points take the data point and subtract it from that, subtract the mean from that, square it, and then divide by the total number of data points you have. Now, we get to the interesting part-- sample variance. There's are several ways-- where when people talk about sample variance, there's several tools in their toolkits or there's several ways to calculate it.
One way is the biased sample variance, the non unbiased estimator of the population variance. And that's denoted, usually denoted, by s with a subscript n. And what is the biased estimator, how we calculate it? Well, we would calculate it very similar to how we calculated the variance right over here. But what we would do it for our sample, not our population. So for every data point in our sample --so we have n of them-- we take that data point.
And from it, we subtract our sample mean. We subtract our sample mean, square it, and then divide by the number of data points that we have. But we already talked about it in the last video. How would we find-- what is our best unbiased estimate of the population variance? This is usually what we're trying to get at. We're trying to find an unbiased estimate of the population variance.
Well, in the last video, we talked about that, if we want to have an unbiased estimate --and here, in this video, I want to give you a sense of the intuition why. We would take the sum.
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