# Bessel's correction for standard deviation

Why does Bessel's correction is biased downward as stated in wiki:

while the sample variance (using Bessel's correction) is an unbiased estimator of the population variance, its square root, the sample standard deviation, is a biased estimate of the population standard deviation; because the square root is a concave function, the bias is downward, by Jensen's inequality.

But by Jensen inequality for a concave function, we have

where f is the square root function, i.e. the sample standard deviation is greater than the expectation of the srandard deviation, i.e. its biased upward. Where am I wrong?

You have correctly identified that $$f$$ is the square root and the convex combination is the integral (expectation).

Therefore, your left hand side $$f\bigl( (1-\alpha) x + \alpha y \bigr)$$ corresponds to $$\sqrt{ E[ S^2 ] }$$, where the square root is outside of (taken after) the expectation. Note that $$\sqrt{ E[ S^2 ] } = \sqrt{ \sigma^2} = \sigma$$ since $$S^2$$ is unbiased for $$\sigma^2$$.

At the same time, the right hand side $$(1-\alpha)\,f(x) + \alpha\, f(y)$$ corresponds to $$E\bigl[ \sqrt{S^2} \bigr]$$.

Although I don't know where you got that image of Jensen's inequality from, indeed it is correct for the case of concave $$f$$. This gives us

$$\text{L.H.S.} > \text{R.H.S.} \quad \implies \quad \sigma > E[S]$$

So we see that the $$S \equiv \sqrt{S^2}$$ is underestimating $$\sigma$$, while $$S^2$$ is spot on for $$\sigma^2$$.

"Where am I wrong?"

What is being pushed upwards by the concavity is $$\sigma$$, not $$S$$. In relative terms, one might choose to say that it is $$S$$ being pushed downward.

The quantity that is bent upward (compared to a straight line) is the one that is "on the curve", which is literally $$~f(\cdot)$$ and that is $$~f\bigl( E[\text{whatever}] \bigr)$$ here.