--For further background into the question, one can refer to equations 2.3 and 2.6 (page 1275 of [0])--



where $X$ and $Y$ are independent stochastic variables with distribution function $G$;

In the above cited paper, the authors derive a closed form expression for $g_G'(x)$ when $G=\Phi$.

my questions is: How does it come about that when $G=\Phi$ and $x=q$: $$g_G'(x)=\frac{\phi(x-c^{-1})-\phi(x+c^{-1})}{\phi(x+c^{-1})+\phi(x-c^{-1})}$$ where $c^{-1}=\mbox{med}_X g_{\Phi}(X)$. That part is not clear to me.


  • [0]:Rousseeuw, Peter J.; Croux, Christophe (December 1993), "Alternatives to the Median Absolute Deviation", Journal of the American Statistical Association (American Statistical Association) 88 (424): 1273–1283, doi:10.2307/2291267
  • 1
    $\begingroup$ I have spent ~20 minutes on your question (before someone answered). I would never have found because you have not mentionned that the equality holds for $x=q$ and not for all $x$. $\endgroup$ – Stéphane Laurent Aug 29 '14 at 15:05
  • 1
    $\begingroup$ @StéphaneLaurent: I apologize for the confusion. Meanwhile I have corrected the oversight. $\endgroup$ – user603 Aug 29 '14 at 17:23

It seems to me that $\phi$ and $\Phi$ are the PDF and CDF of the standard Gaussian; that is, \begin{equation} \phi(x) = \Phi^\prime (x). \end{equation} To obtain $g_{\Phi}(x) \equiv {\rm med}_{Y} |x-Y|$ (the CDF of $Y$ is $\Phi$), one should find a distance $a$ such that for exactly half of the members of the Gaussian distribution, the distance from $x$ is greater than $a$ (and of course smaller than $a$ for the other half). This condition amounts to \begin{equation} F(x,a) \equiv \Phi(x+a) - \Phi(x-a) = \frac{1}{2}. \end{equation} Solving this for $a$ gives $g_{\Phi}(x)$. Therefore obtaining $g_{\Phi}^{\prime}(x)$ boils down to differentiating the implicit function given above. That is, \begin{equation} \begin{split} g_{\Phi}^{\prime}(x) &= - \frac{\partial F/\partial x}{\partial F/\partial a} = \frac{\phi(x-a)-\phi(x+a)}{\phi(x+a)+\phi(x-a)}\\ &=\frac{\phi(x-g_{\Phi}(x))-\phi(x+g_{\Phi}(x))}{\phi(x+g_{\Phi}(x))+\phi(x-g_{\Phi}(x))}. \end{split} \end{equation} Then, using the fact that $g_{\Phi}(q) = 1/c = {\rm med}_{X}g_{\Phi}(X)$, where $q$ is defined by $\Phi(q) = 3/4$, gives \begin{equation} g_{\Phi}^{\prime}(q) = \frac{\phi(q - c^{-1})-\phi(q + c^{-1})}{\phi(q + c^{-1})+\phi(q - c^{-1})}. \end{equation}

Update: How do we know that ${\rm med}_{X}g_{\Phi}(X) = g_{\Phi}(q)$?

Claim: $g_{\Phi}(x)$ is an even function that monotonically decreases as a function of $|x|$.

That this function is even follows from its definition (as an implicit function) given above and the fact that $\Phi(-x) = 1 - \Phi(x)$. It monotonically decreases in $|x|$ because \begin{equation} g_{\Phi}^{\prime}(x) \ \Bigg\{\begin{array}{ccc} >0 \ (x>0)\\=0\ (x=0)\\<0\ (x<0)\end{array}. \end{equation} One can see this from the expression for $g_{\Phi}^{\prime}(x)$ and the shape of the Gaussian function $\phi$. More rigorously, one should be able to deduce this fact from the following considerations:

  1. $\phi(x) = \phi(-x)$.

  2. $\phi(x)$ monotonically decreases in $|x|$.

  3. $a = g_{\Phi}(x)>0$.

Then, the value of $x$ giving the median of $g_{\Phi}(x)$, which we denote as $q$, is where exactly half of the members of the Gaussian distribution falls within $|q|$. Hence $\Phi(q) = 1/4$ or $3/4$. Evaluating $g_{\Phi}$ at either of the two values (they only differ by their signs, and $g_{\Phi}$ is even) would give ${\rm med}_{X}g_{\Phi}(X)$.

  • $\begingroup$ Thanks for the clear answer! I was wondering, could you also recommend a graduate level book with more background on derivatives of order statistics? $\endgroup$ – user603 Aug 29 '14 at 11:23
  • 1
    $\begingroup$ @user603 Sorry. I don't really come from statistics background and don't know about books in this field. I found that you accepted my answer and then decided to revoke your decision. Did you want more explanation on why $g_{\Phi}(q) = 1/c$? In that case, I can update my answer. $\endgroup$ – higgsss Aug 29 '14 at 11:27
  • $\begingroup$ :yes that would be great! – $\endgroup$ – user603 Aug 29 '14 at 11:45
  • 1
    $\begingroup$ @user603 I updated my answer. Hope it is clear enough. $\endgroup$ – higgsss Aug 29 '14 at 12:00
  • 1
    $\begingroup$ Oh! The question should be edited then ! $\endgroup$ – Stéphane Laurent Aug 29 '14 at 12:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.