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

Define:

$$g_G(x)=\mbox{med}_Y|x-Y|$$

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.

reference:

  • [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
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    $\begingroup$ 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
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    $\begingroup$ @StéphaneLaurent: I apologize for the confusion. Meanwhile I have corrected the oversight. $\endgroup$ – user603 Aug 29 '14 at 17:23
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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 ${\rm med}_{X}g_{\Phi}(X)$.

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  • $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ @user603 I updated my answer. Hope it is clear enough. $\endgroup$ – higgsss Aug 29 '14 at 12:00
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    $\begingroup$ Oh! The question should be edited then ! $\endgroup$ – Stéphane Laurent Aug 29 '14 at 12:29

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