# (S_n)² vs (S_n-1)²

E[Sn²] = (n-1/n)sigma²

This is biased variance. So to get unbiased variance, we take E[(Sn-1)²] which according to videos is = sigma²

how is this equal to sigma²? How did we claim that this is indeed equal to sigma²?

Instead of giving the answer directly, can you just tell me how to proceed with equating E[(Sn-1)²] = sigma²

Because if I’m not wrong, if we were to replace n with n-1 in the E[Sn²] formula, we should have ended up on just sigma², but we don’t. So I’m stuck. Help me please

Recall S_n^2 and S^2_{n-1} definition:

S_n^2=\frac{1}{n}\sum_1^n (X_i - \bar{X})^2, and, S_{n-1}^2=\frac{1}{n-1}\sum_1^n (X_i - \bar{X})^2

Further, taking expectation:

E[S_{n-1}^2] =E[\frac{1}{n-1}\sum_1^n (X_i - \bar{X})^2]

Rewriting:

E[S_{n-1}^2] =E[\frac{n}{n}*\frac{1}{n-1}\sum_1^n (X_i - \bar{X})^2]=E[\frac{n}{n-1}*\frac{1}{n}\sum_1^n (X_i - \bar{X})^2]

Now, to complete the proof and deduce E[S_{n-1}^2]=\sigma^2, try using the following 2 results in above:

• E[aX]=aE[X], and
• E[S_n^2]=\frac{n-1}{n}\sigma^2 (This is the same biased variance formula you mentioned at the top)

Leaving the rest of the steps for you to fill in.

I hope, I understood your question properly.

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