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Lucas Roberts
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I am looking for a derivation of the fact that $\frac{1}{\Phi^{-1}(3/4)}$ is the multiplier needed for the Median Absolute Deviation (MAD) to be an unbiased estimator of $\sigma$ when $x_i\sim N(0, \sigma^2)$. Recall the MAD is defined as:

$$ \lambda = b\times Median(| X-Median(X)|). $$

for some $b$ chosen to meet a given criteria, (e.g. often unbiased-ness) the claim is stated in many places (wikipedia and journal articles) but I cannot find a proof.

I am looking for a derivation of the fact that $\frac{1}{\Phi^{-1}(3/4)}$ is the multiplier needed for the Median Absolute Deviation (MAD) to be an unbiased estimator of $\sigma$ when $x_i\sim N(0, \sigma^2)$. Recall the MAD is defined as:

$$ \lambda = b\times Median(| X-Median(X)|). $$

the claim is stated in many places (wikipedia and journal articles) but I cannot find a proof.

I am looking for a derivation of the fact that $\frac{1}{\Phi^{-1}(3/4)}$ is the multiplier needed for the Median Absolute Deviation (MAD) to be an unbiased estimator of $\sigma$ when $x_i\sim N(0, \sigma^2)$. Recall the MAD is defined as:

$$ \lambda = b\times Median(| X-Median(X)|). $$

for some $b$ chosen to meet a given criteria, (e.g. often unbiased-ness) the claim is stated in many places (wikipedia and journal articles) but I cannot find a proof.

forgot square in sigma
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Lucas Roberts
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How to make a Median absolute deviation of $N(0, \sigma\sigma^2)$ an unbiased estimator of $\sigma$, asymptotically?

Source Link
Lucas Roberts
  • 4.4k
  • 1
  • 21
  • 49

How to make a Median absolute deviation of $N(0, \sigma)$ an unbiased estimator of $\sigma$, asymptotically?

I am looking for a derivation of the fact that $\frac{1}{\Phi^{-1}(3/4)}$ is the multiplier needed for the Median Absolute Deviation (MAD) to be an unbiased estimator of $\sigma$ when $x_i\sim N(0, \sigma^2)$. Recall the MAD is defined as:

$$ \lambda = b\times Median(| X-Median(X)|). $$

the claim is stated in many places (wikipedia and journal articles) but I cannot find a proof.