Say we have a random variable. Say a dice. Say we roll it 10 times.
Say we got $x_1$ $x_2$ $x_3$ ... $x_10$
Estimate of the mean is $(x_1+x_2+x_3 + ... + x_10)/10$
Say we got that estimate. let's call it $\mu$.
Say we want to compute the variance estimate.
We then compute the value of $(x_1-\mu)^2 + (x_2-\mu)^2$ and so on and so on
After that we divide by 9. Not 10.
The reason is because $\mu$ is not the true mean. It's just an estimate of the mean. Estimating the variance by 10 will make the estimate too little. That's because m would tend to move toward the direction of the deviation.
Let's see the dice. The true "$\mu$" is 3.5. Say we roll the dice and we got estimate 4. That means through our rolling we happen to have high rolls on that dice. If we divide by 10 will not get the true estimate. We can divide by 10 only if we subtract by the true m. We subtract by 4 which is closer to all those numbers. So we divide by 9 instead of 10.
In general we divide n if we knows the true m. We divide by (n-1) if we uses estimate of "$\mu$".
Why $(n-1)$? Why not $(n-2)$?
I did some pencil pushing when I were young. I do indeed need to divide by $(n-1)$. I don't like my own derivation because it's too complex. I wonder if there is simpler derivation.
I sort of wonder if there is an elegant way to show those.