would someone please explain why the degrees of freedom for a random sample is n-1 instead of n ?

I'm looking for an explanation that is intuitive and easily understood by a high school student.



4 Answers 4


The short answer is that dividing by n returns a biased approximation of the population standard deviation (which is usually what we are trying to estimate from our sample.) Such a calculation for sample standard deviation will be biased low (i.e. an underestimate) relative to the population standard deviation.

Dividing by n-1 makes the sample variance an unbiased estimator, and the sample standard deviation a less biased estimator (this bias is still an issue while n is small.)

Wikipedia provides some details here.

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    $\begingroup$ See also stats.stackexchange.com/q/11707/16974 for more technical explanations (which is not what the OP wanted, but for those interested in theoretical underpinnings) $\endgroup$ May 6, 2013 at 1:14
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    $\begingroup$ I'd suggest avoiding talking about bias in the standard deviation as a motivation; the $n-1$ denominator is related to unbiasing the variance. Consequently - by the Jensen inequality - that guarantees bias in the standard deviation. $\endgroup$
    – Glen_b
    May 6, 2013 at 3:31
  • $\begingroup$ Thank you both for the help! I'll definitely take a look at the information linked above later this week. $\endgroup$ May 6, 2013 at 4:15
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    $\begingroup$ @Glen_b Fair point -- most of the (applied biostats) textbooks I've seen simply say that using n-1 gives a better estimate of variance, and leave it at that. $\endgroup$ May 6, 2013 at 4:22
  • $\begingroup$ Interesting though, dividing by $n $ gives a smaller mean square error $E [(s^2 - \sigma^2) ^2]$, and dividing by $n+1$ smaller still! Also, you don't get an unbiased estimate for $\sigma $ by dividing by $n-1$ (which is what $s^2$ is used for more likely) $\endgroup$ May 24, 2016 at 10:30

Version 1 ("they complain that the real answer is too hard, so here's some handwaving for non-mathematically-inclined students"):

If you knew the real distribution average, you could construct a sum of squares of independent identically distributed standard normal variables $\xi_i$, similar to $S^2$, and it would've been $\chi_n^2$-distributed: $\frac{(n-1)\hat{S}^2}{\sigma^2} = \sum \limits_{i=1}^{n} (\frac{\xi_i-\mu}{\sigma})^2 \sim \chi^2_n$ with $n$ degrees of freedom.

However, you need to know the real distribution mean $\mu$ and distribution variance $\sigma^2$ to achieve the desired sum, and they are not available in real-life small sample situation.

As you don't know the real distribution mean $\mu$, you have to approximate it with sample average $\bar{\xi}=\frac{\sum \limits_{i=1}^n \xi_i}{n}$.

But this takes away one degree of freedom (if you know the sample mean, then only $\xi_i$ from $1$ to $n-1$ can take arbitrary values, but the $n$th has to be $\xi_n = \hat{\xi} - \sum \limits_{i=1}^{n-1}\xi_i$).

So your real $S^2$ loses one degree of freedom: $\frac{(n-1)S^2}{\sigma^2} = \sum \limits_{i=1}^{n} (\frac{\xi_i-\bar{\xi}}{\sigma})^2 \sim \chi^2_{n-1}$

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Oh, and here's a cute kitten for you.

Version 2 ("the real answer, if you do care, as short as I can make it and with as little background required, as possible"):

  1. If you had sum of squares of i.i.d. standard normal variables $\xi_i$, it would've been $\chi_n^2$-distributed: $\sum \limits_{i=1}^{n} (\frac{\xi_i-\mu}{\sigma})^2 \sim \chi^2_n$ with $n$ degrees of freedom.

  2. However, you need to know the real distribution mean $\mu$ and distribution variance $\sigma^2$ to achieve the desired sum that are not available in real-life small sample situation. Your best bet is to use sample average $\bar{\xi}=\frac{\sum \limits_{i=1}^n \xi_i}{n}$ and sample variance $S^2 = \frac{1}{n-1}\sum \limits_{i=1}^n (\xi_i-\bar{\xi})^2$ instead. If you substituted sample mean for the distribution mean, notice that the squares of normal random variables in the sum $\sum \limits_{i=1}^n (\frac{\xi_i-\bar{\xi}}{\sigma})^2$ stop being independent (e.g. if $n=2$ two normal variables $\frac{\xi_1-\hat{\xi}}{\sigma}$ and $\frac{\xi_2-\bar{\xi}}{\sigma}$ are exact opposites of each other, if $\frac{\xi_1-\bar{\xi}}{\sigma} = k$, you know that $\frac{\xi_2-\bar{\xi}}{\sigma} = -k$). Thus you can no longer assume that $\sum \limits_{i=1}^n (\frac{\xi_i-\bar{\xi}}{\sigma})^2$ is chi-square distributed.

  3. Let us find another way to achieve chi-square-distributed random variables. You can express your true sum of independent random variables through the approximate one (add and subtract the sample mean to the numerator of each term of your sum and rearrange):

$\sum \limits_{i=1}^n \frac{(\xi_i - \mu)^2}{\sigma^2} = \sum \limits_{i=1}^n \frac{(\xi_i - \bar{\xi} + \bar{\xi} - \mu)^2}{\sigma^2} = \sum \limits_{i=1}^n (\frac{(\xi_i-\bar{\xi})^2}{\sigma^2} + \underbrace{2 \frac{(\xi_i - \bar{\xi})(\bar{\xi} - \mu)}{\sigma^2}}_{0 \text{ due to }\sum \limits_{i=1}^n (\xi_i - \bar{\xi}) = 0} + \frac{(\bar{\xi} - \mu)^2}{\sigma^2})$

$\sum \limits_{i=1}^n \frac{(\xi_i - \mu)^2}{\sigma^2} = (n-1)\frac{S^2}{\sigma^2} + n\frac{(\bar{\xi} - \mu)^2}{\sigma^2}$, where $\sum \limits_{i=1}^n \frac{(\xi_i - \mu)^2}{\sigma^2} \sim \chi^2_n$, $n\frac{(\bar{\xi} - \mu)^2}{\sigma^2} \sim \chi^2_1$.

  1. By Cochran's theorem $S^2$ is independent of $\bar{\xi}$, thus, $(n-1)\frac{S^2}{\sigma^2}$ is independent of $n\frac{(\bar{\xi} - \mu)^2}{\sigma^2}$. For independent random variables we can apply convolution formula to their probability density functions: $f_{\eta+\psi}(t) = \int \limits_{s=-\infty}^{\infty}f_{\eta}(t-s)f_{\psi}(s)ds$.

$f_{\sum \limits_{i=1}^n \frac{(\xi_i - \mu)^2}{\sigma^2}}(t) = \int \limits_{s=-\infty}^{\infty}f_{(n-1)\frac{S^2}{\sigma^2}}(t-s) \cdot f_{n\frac{(\bar{\xi} - \mu)^2}{\sigma^2}}(s)ds$

  1. Substitute pdf of chi-square distributions with n and 1 degrees of freedom into the last formula:

$\frac{t^{n/2-1}e^{-t/2}}{\Gamma(n/2)2^{n/2}} = \int \limits_{s=-\infty}^{\infty} f_{(n-1)\frac{S^2}{\sigma^2}}(t-s) \cdot \frac{s^{1/2-1}e^{-s/2}}{\Gamma(1/2)2^{1/2}} ds$.

Suppose that $f_{(n-1)\frac{S^2}{\sigma^2}}(t-s) \sim \chi_{n-1}^2$, then $f_{(n-1)\frac{S^2}{\sigma^2}}(t-s) = \frac{(t-s)^{(n-1)/2-1}e^{-(t-s)/2}}{\Gamma((n-1)/2)2^{(n-1)/2}}$. Substitute it into the integral:

$\frac{t^{n/2-1}e^{-t/2}}{\Gamma(n/2)2^{n/2}} = \int \limits_{s=-\infty}^{\infty} \frac{(t-s)^{(n-1)/2-1}e^{-(t-s)/2}}{\Gamma((n-1)/2)2^{(n-1)/2}} \cdot \frac{s^{1/2-1}e^{-s/2}}{\Gamma(1/2)2^{1/2}} ds = \frac{e^{-t/2}}{2^{n/2} \Gamma((n-1)/2) \Gamma(1/2)} \int \limits_{s=-\infty}^{\infty} (t-s)^{(n-1)/2-1}s^{1/2-1}ds = $

$= \frac{e^{-t/2}}{2^{n/2}\Gamma(n/2)} \int \limits_{s=-\infty}^{\infty} \frac{\Gamma(n/2)}{\Gamma((n-1)/2) \Gamma(1/2)} (t-s)^{(n-1)/2-1}s^{1/2-1}ds$.

Do a variable substitution $s = tu$, $ds = tdu$:

$\frac{e^{-t/2}}{2^{n/2}\Gamma(n/2)} \int \limits_{s=-\infty}^{\infty} \frac{\Gamma(n/2)}{\Gamma((n-1)/2) \Gamma(1/2)} (t-tu)^{(n-1)/2-1}(tu)^{1/2-1}tdu = \frac{t^{n/2-1}e^{-t/2}}{\Gamma(n/2)2^{n/2}} \underbrace{\int \limits_{s=-\infty}^{\infty} (1-u)^{(n-1)/2-1}u^{(1/2-1)}du}_{\beta-\text{distribution integral } =1}$.

Thus, we've shown that $\chi^2_{n-1}(x)$ fits as the distribution of $f_{(n-1)\frac{S^2}{\sigma^2}}(x)$.

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    $\begingroup$ Unfortunately I have to downvote this answer, as the mathematical complexity involved is unlikely to be accessible and easily understood by a high school student who perhaps just had their first classes in statistics. $\endgroup$
    – B.Liu
    Jun 30, 2021 at 9:51
  • $\begingroup$ @B.Liu Thanks for your suggestion. I added a "simplified" version for students, who don't really care. I don't know a way to prove this statement without getting one's hands dirty. Note that I did my best to avoid using complicated mathematical tools, such as moment-generating functions/cumulants or characteristic functions/Fourier transform. To the best of my knowledge this proof is as simple as it gets. $\endgroup$ Jun 30, 2021 at 10:29
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    $\begingroup$ I have retracted my downvote, but my concern remains. I don't believe every answer requires a proof, but feel there is no problem whatsoever for opinions to diverge on that. $\endgroup$
    – B.Liu
    Jul 7, 2021 at 11:30

Overall, degrees of freedom appear when we evaluate some objetive within a mathematical system that may or may not have constraints. That is why the most simple answer usually is: "variables - independent parameters". However, I would argue that many students studying statistics, data science, machine learning, phycology (etc) would probably not be satisfied with this answer because there is a lot of things happening behind the "n-1" idea, although it does come simply from a mathematical restriction (for example: we know the mean and want to estimate an statistic like the sample variance).

I believe a good way to illustrate all this and empower the audience would be to first explain Dimension, Basis, and Subspaces in an easy way. After all, many things in data science and statistics can be viewed within that domain.

Then, move to the idea of restriction within a space. Then, the degrees of freedom idea would arise.

Maybe illustrate the concept with the geometry of a Linear Regression, the vector of residuals, and the estimated variance. This would probably be a good way to start. I don't have much time now, but I'll get back to it when I can.

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    $\begingroup$ How does this answer the question? $\endgroup$ May 16, 2023 at 17:04
  • $\begingroup$ Although some questions about df (including this one) can instructively be answered using all this linear algebra machinery, it's not necessary. Moreover, to "fully grasp" the issues it is helpful to see the limitations of the linear algebraic approach and to become aware of some of the subtleties of df calculations. See stats.stackexchange.com/questions/16921 for alternatives. And, once all the smoke clears, it should be possible to see that the answer is ... the Pythagorean Theorem. That brings us full circle to appreciating what the linear algebra does for us. $\endgroup$
    – whuber
    May 16, 2023 at 19:47
  • $\begingroup$ @ShawnHemelstrand I edited to be more helpful. I don't have much time, so at least I thought I could provide some ideas. $\endgroup$ May 17, 2023 at 22:02

The first step in computing the SD is to compute the difference between each value and the mean of those values. You don't know the true mean of the population; all you know is the mean of your sample. Except for the rare cases where the sample mean happens to equal the population mean, the values will tend to be closer to the sample mean than to the true population mean. So the sum of the square of those differences will be smaller (and can't be larger) than what it would have been had you used the true population mean in the first step.

To make up for the underestimation of the sum-of-squares, when calculating the average squared difference (the variance), you need to divide by a value smaller than n. Why is the correct denominator n-1? If you knew the sample mean, and all but one of the values, you could calculate what that last value must be. Statisticians say there are n-1 *degrees of freedom.


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