Hermite–Fourier expansion for the median

Question

Let $n$ be an odd positive integer. Let $M : \mathbb{R}^n \to \mathbb{R}$ be the median function: $M(x_1,\dots,x_n)$ is the median of $x_1,\dots,x_n$. What can be said about the Hermite–Fourier expansion of $M$?

Of course, an actual formula will be of interest but I will be happy to see also some numerical data. The case $n=3$ is already interesting to me. In particular, what is the contribution (sum of squares) of the coefficients for degree-$k$ terms?

Clarification: Note that $M$ is a function of $n$ variables so we need to find the inner product of $M$ with $H_{k_1}(x_1)H_{k_2}(x_2)\dotsm H_{k_n}(x_n)$. These functions form an orthonormal basis with respect to the Gaussian measure on $\mathbb R^n$.

Small Addition

A similar question can be asked for $MAX (x_1,x_2,\dots,x_n)$. Here $n$ can be even and even for $n=2$ is of interest. For the case $n=3$ it could be interesting to compare between the Hermite-Fourier expansion of $M(x_1,x_2,x_3)$ and of $MAX (x_1,x_2,x_3)$.

Answer 1

I presume you want the coefficients $$a_k=\int_{-\infty}^\infty p(M)\psi_k(M)\,dM,$$ with $$\psi_k(x)=\frac{1}{\pi^{1/4}\sqrt{ 2^k k!}} e^{-x^2/2} H_k(x)$$ the normalized Hermite function of order $k$ and $p(M)$ the probability density function of the median $M$ of the $n$ Gaussian random variables.

For $n\gg 1$ the median of $n$ i.i.d. normal random variables has a Gaussian distribution $p(M)=(2\pi\sigma^2)^{-1/2}e^{-M^2/2\sigma^2}$, with $\sigma^2=\tfrac{1}{2}\pi/n$, resulting in $a_{2m+1}=0$, $$a_{2m}=\frac{(-1)^m\sqrt{(2m)!}}{2^{m}\pi^{1/4}m!}\frac{\bigl(1-\tfrac{1}{2}\pi/n\bigr)^m}{\bigl(1+\tfrac{1}{2}\pi/n\bigr)^{m+1/2}}.$$

Answer 2

$\newcommand{\si}{\sigma}\newcommand{\R}{\mathbb R}$As I understand the problem, it is to compute $$e_{k_1,\dots,k_n}:=EM_nH_{k_1}(X_1)\cdots H_{k_n}(X_n),$$ where $n\ge1$ is an odd integer, the $H_k$'s are the (probabilist's) Hermite polynomials, $X_1,\dots,X_n$ are independent standard normal random variables, and $M_n$ is the sample median of $X_1,\dots,X_n$.

For $n=3$, it is possible to obtain explicit expressions for $e_{k_1,\dots,k_n}$, which are rather simple at least for small values of the $k_i$'s. In particular, $$e_{1,0,0}=1/3. $$ See details on this below.

However, this seems impossible to do already for $n=5$. Indeed, with (say) $(k_1,\dots,k_5)=(0,0,1,1,1)$ we will apparently have to compute $$\int_{\Bbb R^5}\prod_{i=1}^5(dx_i\,f(x_i))\,x_3^2 x_4 x_5\,1(x_1\vee x_2<x_3<x_4\wedge x_5)=\int_{\Bbb R}dx_3\,x_3^2 f(x_3)^3 F(x_3)^2,$$ where $a\vee b:=\max(a,b)$, $a\wedge b:=\min(a,b)$, and $f$ and $F$ are respectively the p.d.f. and the c.d.f. of the standard normal distribution. Mathematica cannot do anything with the latter integral. (However, Mathematica can find $\int_{\Bbb R}dt\,t f(t)^2 F(t)^2=\frac1{4\pi\sqrt3}$.) So, apparently, the latter displayed integral cannot be expressed in elementary or special functions.

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Details on case $n=3$: Note that $H_k=(-1)^k f^{(k)}/f$ for $\ge0$, where $f$ and $F$ still respectively denote the p.d.f. and the c.d.f. of the standard normal distribution. So, for (integers) $k_1,k_2,k_3\ge0$, \begin{equation*} \begin{aligned} e_{k_1,k_2,k_3}&= EM_3H_{k_1}(X_1) H_{k_2}(X_2) H_{k_3}(X_3) \\ &=(-1)^{k_1+k_2+k_3}\sum_{\si\in S_3}a(k_{\si(1)},k_{\si(2)},k_{\si(3)}), \end{aligned} \tag{10}\label{10} \end{equation*} where $S_3$ is the set of all permutations of the set $\{1,2,3\}$ and, for $k,l,m\ge0$, \begin{equation*} \begin{aligned} &a(k,l,m):=\int_{\R^3}dx\,dy\,dz\,1(x<y<z)y f^{(k)}(x) f^{(l)}(y) f^{(m)}(z) \\ &=\int_\R dy\, y f^{(l)}(y)\int_{-\infty}^y dx\,f^{(k)}(x)\, \int_y^\infty dz\,f^{(m)}(z) \\ &=\int_\R I f^{(l)}f^{(k-1)}\,\bar f^{(m-1)} \\ &\to\int_\R Iff^{(k-1)}\,f^{(m-1)},\ \int_\R ff^{(k-1)}\,f^{(m-1)},\ \int_\R I ff^{(k-1)}, \int_\R ff^{(k-1)} \\ &\to\int_\R ff^{(k-1)} f^{(m-1)},\ \int_\R ff^{(k)}, \int_\R fF, \end{aligned} \end{equation*} where $I$ is the identity function, so that $I(y):=y$ for real $y$, \begin{equation*} f^{(-1)}:=F, \quad \bar f^{(m-1)}:=1(m=0)-f^{(m-1)}, \end{equation*}
and $\to$ in this context means "reduces (by possibly repeated integration by parts) to integral(s) of the following form -- for nonnegative integers $k$ and $m$, which may be different in different expressions".

Next, \begin{equation*} \int_\R fF=\int_\R dF(y)\,F(y)=1/2 \end{equation*} and \begin{equation*} \int_\R ff^{(k)}=(-1)^k\int_\R f^2H_k, \end{equation*} which latter can be easily evaluated explicitly using the explicit expression for $H_k$.

It remains to evaluate \begin{equation*} b(k,m):=\int_\R ff^{(k-1)}f^{(m-1)}. \end{equation*} If $k,m\ge1$, then \begin{equation*} b(k,m)=(-1)^{k+m}\int_\R f^3H_{k-1}H_{m-1}, \end{equation*} which again can be easily evaluated explicitly using the explicit expression for Hermite polynomials. Finally, \begin{equation*} b(0,0)=\int_\R fF^2=\int_\R dF(y)\,F(y)^2=1/3 \end{equation*} and \begin{equation*} -b(1,0)=-b(0,1)=\int_\R f^2F =\frac1{2\sqrt\pi}\,P(Y_1/\sqrt2<Y) \\ =\frac1{2\sqrt\pi}\,P(\sqrt{\tfrac32}\,Y>0) =\frac1{4\sqrt\pi}, \end{equation*} where $Y$ and $Y_1$ are independent standard normal random variables; here we used the fact that $2\sqrt\pi\,f^2$ is the p.d.f. of $Y_1/\sqrt2$.

Thus, $e_{k_1,k_2,k_3}$ can be evaluated explicitly for all (integers) $k_1,k_2,k_3\ge0$. $\quad\Box$

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$E\max(X_1,X_2,X_3)H_{k_1}(X_1) H_{k_2}(X_2) H_{k_3}(X_3)$ and $E\min(X_1,X_2,X_3)H_{k_1}(X_1) H_{k_2}(X_2) H_{k_3}(X_3)$ can be evaluated similarly. In particular (cf. \eqref{10}), for (integers) $k_1,k_2,k_3\ge0$, \begin{equation*} \begin{aligned} &E\min(X_1,X_2,X_3)H_{k_1}(X_1) H_{k_2}(X_2) H_{k_3}(X_3) \\ &=(-1)^{k_1+k_2+k_3}\sum_{\si\in S_3}b(k_{\si(1)},k_{\si(2)},k_{\si(3)}), \end{aligned} \end{equation*} where, for $k,l,m\ge0$, \begin{equation*} \begin{aligned} &b(k,l,m):=\int_{\R^3}dx\,dy\,dz\,1(x<y<z)x f^{(k)}(x) f^{(l)}(y) f^{(m)}(z) . \end{aligned} \end{equation*}