I was checking the definition of pooled variance, and although I think it makes sense intuitively, I was wondering how can one obtain that estimator. For the case of only one group, I understand the MLE of the variance assuming gaussian iid samples in which case we obtain a biased estimate. After computing the Expected value, we can see that is actually biased and we can correct the estimation by dividing by $n-1$. However, I have not been able to find a way to arrive to the pooled variance estimation equation:

$s_p^2=frac{sum_i (n_i-1)s_i^2}{sum_i (n_i-1)}$

Where $i$ is the index of the groups.

How could I obtain that equation?

Thanks!

**Contents**hide

#### Best Answer

I'm taking a stab at this, as I think it is just a weighted average: $$begin{align}E[s_p^2] & = Eleft[frac{Sigma(n_i-1)s_i^2}{Sigma(n_i-1)}right] \ & = frac{1}{Sigma(n_i-1)}Eleft[Sigma(n_i-1)s_i^2right] \ & = frac{1}{Sigma(n_i-1)}left(Sigma(n_i-1)E[s_i^2]right) \ & = frac{1}{Sigma(n_i-1)}left(Sigma(n_i-1)sigma^2right) \ & = frac{1}{Sigma(n_i-1)}left(sigma^2Sigma(n_i-1)right) \ & = sigma^2 end{align}$$ Sorry about the comment re: multiple regression…I think this is just using the rules for expectations.

### Similar Posts:

- Solved – Is a biased or unbiased estimator used for pooled SD in calculating Cohen’s d
- Solved – what is the log of the PDF for a Normal Distribution
- Solved – What does pooled variance “actually” mean
- Solved – Odds ratio and confidence interval in meta-analysis
- Solved – Normally distributed random variables with $N(0,sigma^2)$