Why we use (n -1) in the denominator of standard deviation formula ?




In short, if  x1, x2 ,…, xN  be independent and identically distributed random variables. The standard deviation of the entire population is given by the formula :



However, if  x1, x2 ,…, xn  ( n < N )  is not a set variables of the entire population, but only a sample of the population, then we use the standard deviation formula :







Why we do like this ?


Not so technical, we use  (n – 1)  because we just like to make the "spread" (or deviation) a little larger to reflect the fact that, since we are using a sample, not the entire population, we have more uncertainty.





Degree of freedom


           Degrees of freedom is a measure of how much precision an estimate of variation has. A general rule is that the degrees of freedom decrease when we have to estimate more parameters.


           Before you can compute the standard deviation, we have to first estimate a mean. This causes you to lose one degree of freedom and you should divide by (n – 1)  rather than n.  In more complex situations, like Analysis of Variance and Multiple Linear Regression, we usually have to estimate more than one parameter. Measures of variation from these procedures have even smaller degrees of freedom.





Unbiased estimator


           A more formal way to clarify the situation is to say that  s  (or the sample standard deviation) is an unbiased estimator of  s  , the population standard deviation if the denominator of  s  is  (n – 1). 


           Suppose we are trying to estimate the parameter  Q  using an estimator  θ  (that is, some function of the observed data).  Then the bias of  θ  is defined to be    E(θ) – Q  , in short, "the expected value of the estimator  Q  minus the true value  θ . "


If  E(θ) – Q = 0 ,  then the estimator is unbiased.





Variance formula using  n  instead of  (n – 1)  is biased


Suppose we use the variance formula (square of standard deviation)




We are going to show that 




Since  E(s2) ¹ s2 ,  s2  is a biased estimator of  s2  if we use (3) instead of  (2).







We would like to clarify some points before we like to prove  (4) .

Firstly, if  X  and  Y  are independent and identically distributed, we have:


    Var (kX) = k2 Var (X)                                                ....   (5)

    Var (X + Y) = Var (X) + Var (Y)                               ....   (6)


The proofs of  (5)  and  (6)  are left to the reader.


Secondly, if    is the sample mean  and  m  is the population mean, then






since  x1 , x2 , …, xn  are independent and identically distributed  and having the same variance  s2 .





Proof of  (2)



Taking the summation, we have






Taking the expection of  (8),