We talked about this general version of CLT.

Why did we use biased variance here? Can it not be S_(n-1)_ ? Which is the unbiased variance

We talked about this general version of CLT.

Why did we use biased variance here? Can it not be S_(n-1)_ ? Which is the unbiased variance

Hi @pushpakruhil,

Requesting you to mention Video title and timeframe along with the doubts, so that easy for everyone to keep track

Week 19 : normal approximation of binomial theorem, we study this general version of CLT where each random sample has it’s own mean and variance.

But later on in the course, in week 20 we study “Sn²" is the biased variance, and S(n-1)² is the unbiased variance. Shouldn’t this Sn² be S(n-1)² because Sn²= (1 - 1/n)sigma² and S(n-1)² = sigma²

@ pushpakruhil

In general version of CLT, each sample is drawn from different population which have their own *population mean* and *population variance*. for e.g. drawing sample from 3 different cities to observe mean salary. Each of these 3 cities are most likely to have different mean salary and variance.

(In context)Right, S_n^2 is biased as its expected value is not equal to the respective population variance, but when we used the modified formula and use S^2_{n-1} instead, its expectation is equal to the respective population variance.

But, *I’m not able to understand how you related biased and unbiased variance here with the screenshot in post at the top.*

In your original post (first one), In the formula s^2_n=\sum_1^n \sigma_i^2, ~~~s^2_n is not same as biased variance. Its a different quantity with a different formula (also notice its small s not capital S)

If I missed your point, please add a bit more detail if possible.