Computing Sample Variance: Why Divide by N - 1?

Variance Estimation¶

In statistics we know that the mean and variance of a population $Y$ are defined to be:

\begin{equation} \left\{ \begin{aligned} \text{Mean}(Y) &= \mu = \frac{1}{N} \sum_{i=1}^{N} Y_i \\ \text{Var}(Y) &= \sigma^2 = \frac{1}{N} \sum_{i=1}^{N} (Y_i - \mu)^2 \\ \end{aligned} \right. \end{equation}

where $N$ is the size of the population.

Given the population $Y$, we can draw a sample $X$ and compute statistics for $X$:

\begin{equation} \left\{ \begin{aligned} \text{Mean}(X) &= \bar{X} = \frac{1}{n} \sum_{j=1}^{n} X_j \\ \text{Var}(X) &= s^2 = \frac{1}{n - 1} \sum_{j=1}^{n} (X_j - \bar{X})^2 \\ \end{aligned} \right. \end{equation}

where lowercase $n$ is the size of the sample, typically a much smaller number than $N$. One detail that is often not clearly explained in introductory statistics is why we should divide by $n - 1$ instead of $n$ in the calculation for the sample variance.

Why divide by n - 1?¶

It turns out that we should divide by $n - 1$ because dividing by $n$ would give us a biased estimator of the population variance. Let's look at a concrete example before diving into the math for why. Let's say we have a population of 100,000 data points. These can represent, for instance, a movie rating for each of 100,000 people.

In :
%matplotlib inline
import matplotlib.pyplot as plt
from IPython.core.pylabtools import figsize
figsize(15, 5)

In :
import pandas as pd
import numpy as np

In :
np.random.seed(42)

In :
N = 100000  # size of population

In :
population = pd.Series(np.random.randint(1, 11, N))


We can easily calculate the population mean and population variance:

In :
population.mean()

Out:
5.5155599999999998
In :
((population - population.mean()) ** 2).sum() / N

Out:
8.2592178864000001

Note that we are dividing by $N$ in the variance calculation, also that in numpy or pandas this is the same as simply using the method var with ddof=0

In :
population.var(ddof=0)

Out:
8.2592178864000001

where ddof=0 means to divide by $N$, and ddof=1 means to divide by $N - 1$.

Simulation¶

As usual in statistics, the population parameters are often unknown. But we can estimate them by drawing samples from the population. Here we are drawing a random sample of size $30$. As of version 0.16.1, pandas has a convenient Series.sample() function for this:

In :
samples = {}
n = 30  # size of each sample
num_samples = 500  # we are drawing 500 samples, each with size n
for i in range(num_samples):
samples[i] = population.sample(n).reset_index(drop=True)

In :
samples = pd.DataFrame(samples)
samples.T.tail()

Out:
0 1 2 3 4 5 6 7 8 9 ... 20 21 22 23 24 25 26 27 28 29
495 4 6 3 7 9 1 3 6 8 7 ... 2 9 4 5 3 1 3 5 9 6
496 8 3 5 4 7 6 10 4 1 5 ... 5 8 1 5 3 10 2 5 9 2
497 8 4 7 5 3 10 5 7 9 2 ... 6 9 10 3 10 3 2 6 5 8
498 3 3 6 6 1 4 6 8 10 8 ... 8 1 3 8 3 6 4 9 7 3
499 10 5 1 3 10 7 3 6 8 9 ... 7 7 7 4 9 6 7 8 9 2

5 rows × 30 columns

As we expect, if we average all the sample means we can see that the it is a good estimate for the true population mean:

In :
df = pd.DataFrame({'estimated mean': pd.expanding_mean(samples.mean()),
'actual population mean': pd.Series(population.mean(), index=samples.columns)})
df.plot(ylim=(4.5, 6.5))

Out:
<matplotlib.axes._subplots.AxesSubplot at 0x107a29198> Now let's compare the results we would get by using the biased estimator (dividing by $n$) and the unbiased estimator (dividing by $n-1$)

In :
df = pd.DataFrame({'biased var estimate (divide by n)': pd.expanding_mean(samples.var(ddof=0)),
'unbiased var estimate (divide by n - 1)': pd.expanding_mean(samples.var(ddof=1)),
'actual population var': pd.Series(population.var(ddof=0), index=samples.columns)})
df.plot(ylim=(6.5, 10.5))

Out:
<matplotlib.axes._subplots.AxesSubplot at 0x107a292b0> We can see that the biased estimator (dividing by $n$) is clearly not estimating the true population variance as accurately as the unbiased estimator (dividing by $n-1$).

Mathematical Proof¶

To prove that dividing by $n - 1$ is an unbiased estimator, we need to show that expected value of the estimaor is indeed $\sigma^2$: \begin{equation} E(s^2) = E\left(\frac{1}{n - 1} \sum_{j=1}^{n} (X_j - \bar{X})^2\right) = \sigma^2 \end{equation}

First we'll need to recall a few basic properties of expectation and variance:

\begin{equation} \left\{ \begin{aligned} & E(Z_1 + Z_2) = E(Z_1) + E(Z_2), \text{ for any } Z_1, Z_2 \\ & \text{Var}(a Z) = a^2 \text{Var}(Z), \text{ for any } Z \\ & \text{Var}(Z_1 + Z_2) = \text{Var}(Z_1) + \text{Var}(Z_2), \text{ if } Z_1 \text{ and } Z_2 \text{ are independent} \\ \end{aligned} \right. \end{equation}

Also, the following is a useful form for variance: \begin{equation} \text{Var}(Z) = E((Z - E(Z))^2) = E(Z^2 - 2ZE(Z) + E(Z)^2) = E(Z^2) - E(Z)^2 \end{equation}

This is equivalent to \begin{equation} E(Z^2) = \text{Var}(Z) + E(Z)^2 \end{equation}

Using the above properties we can now simplify the expression for $E(s^2)$:

\begin{aligned} E(s^2) = E\left(\frac{1}{n - 1} \sum_{j=1}^{n} (X_j - \bar{X})^2\right) = & \frac{1}{n - 1} E \left( \sum_{j=1}^{n} (X_j^2 - 2X_j\bar{X} + \bar{X}^2) \right) \\ = & \ \frac{1}{n - 1} E \left( \sum_{j=1}^{n} X_j^2 - 2n\bar{X}^2 + n\bar{X}^2 \right) \\ = & \ \frac{1}{n - 1} E \left( \sum_{j=1}^{n} X_j^2 - n\bar{X}^2 \right) \\ = & \ \frac{1}{n - 1} \left[ E \left( \sum_{j=1}^{n} X_j^2 \right) - E \left( n\bar{X}^2 \right) \right] \\ = & \ \frac{1}{n - 1} \left[ \sum_{j=1}^{n} E \left( X_j^2 \right) - n E \left( \bar{X}^2 \right) \right] \\ \end{aligned}

Now notice that the first term can be simplied as:

\begin{aligned} \sum_{j=1}^{n} E \left( X_j^2 \right) = & \sum_{j=1}^{n} \left( Var(X_j) + E(X_j)^2 \right) \\ = & \sum_{j=1}^{n} \left( \sigma^2 + \mu ^2 \right) \\ = & \ n \sigma^2 + n \mu ^2 \\ \end{aligned}

Using the same trick, the second term becomes:

\begin{aligned} E(\bar{X}^2) = & \ Var(\bar{X}) + E(\bar{X})^2 \\ = & Var(\frac{1}{n} \sum_{j=1}^{n} X_j) + \mu ^2 \\ = & \frac{1}{n^2} Var(\sum_{j=1}^{n} X_j) + \mu ^2 \\ = & \frac{1}{n^2} \sum_{j=1}^{n} Var(X_j) + \mu ^2, \text{ because all } X_j\text{'s are independent} \\ = & \frac{1}{n^2} n\sigma^2 + \mu ^2 \\ = & \frac{1}{n} \sigma^2 + \mu ^2 \\ \end{aligned}

Plugging the two terms back we finally get:

\begin{aligned} E(s^2) = & \ \frac{1}{n-1} \left[ \sum_{j=1}^{n} E \left( X_j^2 \right) - n E \left(\bar{X}^2 \right) \right] \\ = & \ \frac{1}{n-1} \left[n \sigma^2 + n \mu ^2 - n \left( \frac{1}{n} \sigma^2 + \mu ^2 \right) \right] \\ = & \ \frac{1}{n-1} \left[n \sigma^2 + n \mu ^2 - \sigma^2 - n \mu ^2 \right] \\ = & \ \sigma^2 \\ \end{aligned}

Dividing by $n-1$ gives us an unbiased estimate for the population variance indeed!

Source of Bias¶

One intuitive way to think about why the bias exists is to notice that we generally don't actually know the true population mean $\mu$, and therefore the sample variance is being computed using the estimated mean $\bar{X}$. However the quadratic form $\sum_{j=1}^{n} (X_j - a)^2$ is actually minimized by $a = \bar{X}$, which means that whatever the true population mean $\mu$ is, we will always have

\begin{equation} \sum_{j=1}^{n} (X_j - \mu)^2 \geq \sum_{j=1}^{n} (X_j - \bar{X})^2 \end{equation}

Therefore we are underestimating the true variance because we don't know the true mean.

In fact, we can see that we are underestimating by exactly $\sigma^2$ on average:

\begin{aligned} E\left(\sum_{j=1}^{n} (X_j - \mu)^2\right) = & \ E \left(\sum_{j=1}^{n} (X_j - \bar{X} + \bar{X} - \mu)^2\right) \\ = & \ E \left(\sum_{j=1}^{n} (X_j - \bar{X})^2 + \sum_{j=1}^{n} 2(X_j - \bar{X})(\bar{X} - \mu) + \sum_{j=1}^{n} (\bar{X} - \mu)^2 \right) \\ = & \ E \left(\sum_{j=1}^{n} (X_j - \bar{X})^2 + \sum_{j=1}^{n} (\bar{X} - \mu)^2 \right) \\ = & \ E \left(\sum_{j=1}^{n} (X_j - \bar{X})^2 \right) + E \left(\sum_{j=1}^{n} (\bar{X} - \mu)^2 \right) \\ = & \ E \left(\sum_{j=1}^{n} (X_j - \bar{X})^2 \right) + \sum_{j=1}^{n} E \left((\bar{X} - \mu)^2 \right) \\ = & \ E \left(\sum_{j=1}^{n} (X_j - \bar{X})^2 \right) + \sum_{j=1}^{n} \left( \text{Var} (\bar{X} - \mu) + E (\bar{X} - \mu)^2 \right) \\ = & \ E \left(\sum_{j=1}^{n} (X_j - \bar{X})^2 \right) + \sum_{j=1}^{n} \left( \text{Var} (\bar{X}) + E (\bar{X} - \mu)^2 \right) \\ = & \ E \left(\sum_{j=1}^{n} (X_j - \bar{X})^2 \right) + \sum_{j=1}^{n} \text{Var} (\bar{X}) \\ = & \ E \left(\sum_{j=1}^{n} (X_j - \bar{X})^2 \right) + n \text{Var} (\bar{X}) \\ = & \ E \left(\sum_{j=1}^{n} (X_j - \bar{X})^2 \right) + n \text{Var} (\frac{1}{n} \sum_{j=1}^{n} X_j) \\ = & \ E \left(\sum_{j=1}^{n} (X_j - \bar{X})^2 \right) + n \frac{1}{n^2} \sum_{j=1}^{n} \text{Var} (X_j) \\ = & \ E \left(\sum_{j=1}^{n} (X_j - \bar{X})^2 \right) + \sigma^2 \\ \end{aligned}

Combined with the result we have from the proof in the previous section, we can see that if we somehow magically knew the true mean $\mu$, dividing by $n$ would be unbiased:

\begin{aligned} E\left(\frac{1}{n} \sum_{j=1}^{n} (X_j - \mu)^2\right) = & \ \frac{1}{n} E \left(\sum_{j=1}^{n} (X_j - \mu)^2\right) \\ = & \ \frac{1}{n} \left[ E \left(\sum_{j=1}^{n} (X_j - \bar{X})^2\right) + \sigma^2 \right] \\ = & \ \frac{1}{n} \left[ (n - 1) \sigma^2 + \sigma^2 \right] \\ = & \ \sigma^2 \\ \end{aligned}

However since we don't know the true mean and are using the estimated mean $\bar{X}$ instead, we'd need to divide by $n - 1$ to correct for the bias. This is also known as Bessel's correction.