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Can someone help me wrap my mind around this:

When calculating the variance of a sample, we divide the squared distance of each data point from the mean by the sample size - 1. But how come we don't need to divide the sum by anything when we are calculating the variance of a discrete random variable?

Hannah Tang
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  • There is an essential difference between the variance of a sample (it is a random variable) and the variance of a random variable (it is a real number, if it exists). In the second case there is no (comparable) sum. – drhab Jul 25 '19 at 18:10
  • If I understand what you're saying, the first case you mention is actually the way you estimate the variance of a set of $n$ indipendent, identically distributed random variables. Division by $n-1$ is then necessary so that the random variable you obtain in this way, has expected value equal to the variance you want to estimate. – dfnu Jul 25 '19 at 19:28

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