I am aware of Jensen's inequality where, given the concave square root function, the mean of the square root is lesser than the square root of the mean.

However, I cannot figure out why the square root of the unbiased estimator for population variance will be less than the true population standard deviation. A lot of text on the web state Jensen's inequality to explain that it is indeed smaller without actually explaining why.

Am I right to think that since the square root of the unbiased estimator is lesser so it must somehow relate to the "mean of the square root" while the population standard deviation somehow relates to "square root of the mean" (to use the terms from the Jensen's inequality example)?

**Responding to Timothy Chow:**

I understand, in the description of your procedure, how Jensen's inequality comes into play in both cases. It is apparent in your procedure because you are taking multiple samplings from the population.

I suppose all descriptions of estimating population variance (or standard deviation) assume that one can take multiple samplings from the population for analysis. I am interested in estimating population variables using one single sampling from the population. I am aware that one single sampling from the population is extremely unreliable but in certain situations, once is all you get. The background of my question is in financial applications.

**I re-frame my question as follows:**

In estimating population standard deviation from a single sampling of data, will the square root of the unbiased estimator for population variance underestimate population standard deviation? If it does, why is this so, since there is no averaging of unbiased estimators for population variance across multiple samplings?

cansay that the value you get willtendto be less than 4.5 because theaveragevalue of a die roll is 3.5. That is why averages come into play even if you are taking just one sample and trying to predict what will happen. – Timothy Chow Feb 9 '12 at 3:17