# How can we write the sample variance's formal definition of a continuous random variable considering Bessel's correction? [closed]

I am trying to find the formal way of writing the sample variance of a continuous random variable considering Bessel's correction.

I ask because the sample variance is usually written this way:

$$S^2 = \frac{\sum(X-\overline X)^2}{N-1}$$

But I don't see how can we consider the Bessel's correction of a sample of a continuous random variable, specially if the probability measure is not the same for each observation.

$$\operatorname{Var}(X) = \sigma^2 = \int (x-\mu)^2 f(x)\, dx$$

I searched in google and in this website to no avail.

Thanks for your help, as always.

• Bessels correction only makes sense for sample variances. It is used to ensure that when you use a sample to estimate the population variance, that the estimation is unbiased. If you know the population, you can just compute it’s variance, you don’t need to estimate it. Jun 20 '19 at 18:31
• The sample variance of a continuous random variable, I meant. How can I reflect the Bessel's correction? Jun 20 '19 at 19:14
• I don't quite follow. If you have a continuous random variable, say a normally distributed one, then have a sample from it, and want to unbiasedly estimate the population variance using the sample, then you may use Bessel's correction. Jun 20 '19 at 20:55
• I find your question confusing. 1. Samples are discrete, not continuous (if you think otherwise, can you give an example?) and each observation is $\frac{1}{n}$-th of the sample. 2. Bessel's correction applies when you replace $\mu$ by $\bar{X}$. Jun 20 '19 at 22:25
• Bessel's correction consists of dividing by $N-1$ rather than by $N$ when one writes $$\widehat\sigma^2 = \frac{\sum_{i=1}^N (\,X_i - \overline X\,)^2}{N-1}.$$ That is true regardess of whether the probability distribution of $X_i$ is discrete or continuous. If one somehow knew the value of $\mu$, one could write $$\widehat \sigma^2 = \frac{\sum_{i=1}^N (\,X_i - \mu\,)^2}{N}$$ and it would be an unbiased estimator of $\sigma^2$ with $N$ rather than $N-1$ in the denominator. So Bessel's correction is intended to correct the bias introduced by the use of $\overline X$ rather than $\mu.\,\,$ Jun 21 '19 at 2:47

Sample variance: The variance $$S^2$$ of a random sample $$X_1, X_2, \dots, X_n$$ from a population with variance $$\sigma^2$$ is usually defined as $$S^2 = \frac{1}{n-1}\sum_{i=1}^n (X_i - \bar X)^2,$$ where $$\bar X =\frac 1n \sum_{i=1}^n X_i.$$ The use of $$n-1$$ instead of $$n$$ in the denominator of $$S^2$$ makes $$S^2$$ an unbiased estimator or $$\sigma^2;$$ that is, $$E(S^2) = \sigma^2.$$

Furthermore, if the data are from a normal distribution we have $$\frac {(n-1)S^2} {\sigma^2} \sim \mathsf{Chisq}(\nu = n-1),$$

a relationship used to make confidence intervals for $$\sigma^2$$ and to do tests involving $$\sigma^2$$ based on $$S^2.$$

Sample standard deviation: The sample standard deviation is usually defined as $$S = \sqrt{S^2} = \sqrt{\frac{1}{n-1}\sum_{i=1}^n (X_i - \bar X)^2}.$$

Because expectation is a linear operator and taking the square root is not a lineat transformation, we do not generally have $$E(S) = \sigma,$$ so $$S$$ is not an unbiased estimate of $$\sigma.$$

For a normal sample of size $$n,$$ the exact relationship is $$E(S_n) = \sigma\sqrt{\frac{2}{n-1}}\Gamma\left(\frac n2\right)/ \Gamma\left(\frac{n-1}{2}\right),$$ where $$\Gamma(\cdot)$$ is the gamma function. Thus for a random sample of size $$n = 5$$ from a normal population with standard deviation $$\sigma,$$ we have $$E(S_5) \approx 0.940 \sigma.$$ For $$n = 100,\,$$ $$E(S_{100}) \approx 0.9975\sigma.$$ Computations in R:

sqrt(2/4)*gamma(5/2)/gamma(4/2)
[1] 0.9399856
sqrt(2/99)*gamma(100/2)/gamma(99/2)
[1] 0.997478


For small $$n,$$ the bias is not large enough to be a difficulty in many applications, and for large $$n,$$ the bias is often ignored.

Addendum on the estimation of $$\sigma^2:$$ It seems that there are compromises to be made all around in making inferences about normal population variances.

A popular criterion for judging the usefulness of an estimator is 'root mean square error' (RMSE). The RMSE of an estimator $$T$$ of a parameter $$\tau$$ is defined as $$\sqrt{E[(T-\tau))^2]}.$$ A small RMSE is considered desirable.

With $$Q = \sum_i (X_i - \bar X)^2,$$ denote the sample variance $$V_1 = S^2 = Q/(n-1),$$ the MLE as $$V_2 = Q/n.$$ Also, $$V_3 = Q/(n+1)$$ and $$V_4 = Q/(n+2).$$

According the RMSE criterion, the sample variance $$V_1 = S^2$$ has a slightly larger RMSE than the MLE $$V_1,$$ so one might argue in favor of using the MLE. However, $$V_3$$ has still smaller RMSE, but its use is resisted because it is even more biased than the MLE.

For the case $$n = 10, \sigma = 15, \sigma^2 = 225,$$ the following simulation illustrates some of the properties of these estimators. (Estimator $$V_4$$ in included just to show that $$Q/(n+2)$$ has a larger RMSE than does $$V_3.)$$

set.seed(620);  n = 10;  sg = 15;  m = 10^6
v1 = replicate(m, var(rnorm(n,0,sg)))
v2 = ((n-1)/n)*v1;  v3 = ((n-1)/(n+1))*v1
v4 = ((n-1)/(n+2))*v1

mean(v1); mean(v2); mean(v3); mean(v4)
[1] 225.0488   # aprx E(S) = 225
[1] 202.5439
[1] 184.1308
[1] 168.7866

sqrt(mean((v1-sg^2)^2))
[1] 106.05              # RMSE of MLE
sqrt(mean((v2-sg^2)^2))
[1] 98.05116            # RMSE of S
sqrt(mean((v3-sg^2)^2))
[1] 95.91148            # smallest of 4 RMSEs
sqrt(mean((v4-sg^2)^2))
[1] 97.39696


Histograms of the simulated distributions of the four variance estimators, with vertical bars at $$\sigma^2 = 225.$$