TITLE: Expectation of max of Gaussian multiplied by a functional of Gaussian
QUESTION [5 upvotes]: Let $X \in \mathbb{R}^{d}$ follows the standard Gaussian distribution $N(0, I_d)$. Let $Y = \max_{j\in[d] } X_j$. It is not hard to see that 
\begin{align}
\mathbb{E}\left [ Y \cdot X\right] = \sum_{j=1}^d \mathbb{P}\left( j = \arg\max_{i \in [d]} X_i \right) \cdot e_j,
\end{align}
where $e_i$ is the standard basis in $\mathbb{R}^d$. Now I was wondering how to compute $$\mathbb{E} \left[ Y\cdot (X X^\top - I_d) \right ].$$
Is there a closed form solution?

REPLY [2 votes]: Let $n:=d$. Assume $n\ge3$. Let $F$ and $f$ denote the cdf and pdf of $N(0,1)$, and let $I\{\cdot\}$ denote the indicator function. 
Each of the diagonal entries of the matrix $EYXX^T$ equals
\begin{equation*}
 EX_1^2\max_iX_i=E_1+E_2,
\end{equation*}
where 
\begin{equation*}
 E_1:=EX_1^3\,I\{X_1>\max_2^n X_i\},\quad E_2:=(n-1)EX_1^2 X_2\,I\{X_2>X_1\vee\max_3^n X_i\}. 
\end{equation*}
Note that the cdf of the maximum of $k$ iid $N(0,1)$ random variables (r.v.'s) is $F^k$. So,
\begin{multline*}
 E_1=\int_{-\infty}^\infty d(F(c)^{n-1})\,EX_1^3\,I\{X_1>c\}
 =\int_{-\infty}^\infty d(F(c)^{n-1})\,\int_c^\infty dx\,x^3f(x) \\ 
 =\int_{-\infty}^\infty dc\, F(c)^{n-1}c^3f(c), 
\end{multline*}
by integration by parts. Next, using the identities 
$$EX_2\,I\{X_2>u\}=f(u)$$ 
and 
$$d_c f(u\vee c)=-cf(c)I\{c>u\}\,dc$$ 
for real $u$, and integrating by parts (twice), we have 
\begin{multline*}
 E_2:=\int_{-\infty}^\infty (n-1)d(F(c)^{n-2})\,EX_1^2 X_2\,I\{X_2>X_1\vee c\} \\
 =\int_{-\infty}^\infty (n-1)d(F(c)^{n-2})\,EX_1^2 f(X_1\vee c) \\ 
 =\int_{-\infty}^\infty (n-1)dc\, F(c)^{n-2}\,cf(c) EX_1^2\,I\{X_1<c\} \\
  =\int_{-\infty}^\infty  (n-1)dc\, F(c)^{n-2}\,cf(c) (F(c)-cf(c)) \\ 
  =\int_{-\infty}^\infty (n-1)dc\, F(c)^{n-1}\,cf(c) 
  -\int_{-\infty}^\infty (n-1)dc\, F(c)^{n-2}\,f(c)c^2f(c) \\ 
  =\int_{-\infty}^\infty (n-1)dc\, F(c)^{n-1}\,cf(c) 
  +\int_{-\infty}^\infty dc\, F(c)^{n-1}\,(2c-c^3)f(c) \\ 
  =\int_{-\infty}^\infty dc\, F(c)^{n-1}\,((n+1)c-c^3)f(c). 
\end{multline*}
Thus, each of the diagonal entries of the matrix $EYXX^T$ equals
\begin{equation*}
 EX_1^2\max_iX_i=E_1+E_2=(n+1)\int_{-\infty}^\infty dc\, F(c)^{n-1}\,f(c)c
\end{equation*}. 
Since 
\begin{equation*}
 EY=E\max_iX_i=n\int_{-\infty}^\infty dc\, F(c)^{n-1}\,f(c)c, \tag{1}
\end{equation*}
we conclude that each of the diagonal entries of the matrix $EY(XX^T-I)$ equals
\begin{equation*}
\text{diags}= EX_1^2\max_iX_i=\int_{-\infty}^\infty dc\, F(c)^{n-1}\,f(c)c=\tfrac1n\,EY. \tag{2}
\end{equation*}
Each of the off-diagonal entries of the matrix $EY(XX^T-I)$ equals
\begin{equation*}
 EX_1X_2\max_iX_i=\sum_1^n s_j, 
\end{equation*}
where
\begin{equation*}
 s_j:=EX_1 X_2 X_j\,I\{X_j\ge \max_1^n X_i\}.
\end{equation*}
If $j\notin\{1,2\}$, then, similarly to the previous multi-line display, 
\begin{multline*}
 s_j=EX_1 X_2 X_3\,I\{X_3>X_1\vee X_2\vee \max_4^n X_i\} \\ 
 =\int_{-\infty}^\infty d(F(c)^{n-3})\,EX_1 X_2 X_3\,I\{X_3>X_1\vee X_2\vee c\} \\ 
=\int_{-\infty}^\infty d(F(c)^{n-3})\,EX_1 X_2 f(X_1\vee X_2\vee c) \\ 
=\int_{-\infty}^\infty dc\,F(c)^{n-3}f(c)c\,EX_1 X_2 I\{X_1\vee X_2<c\} \\ 
=\int_{-\infty}^\infty dc\,F(c)^{n-3}f(c)c\,f(c)^2 \\ 
=\frac1{n-2}\,\int_{-\infty}^\infty dc\,F(c)^{n-2}(2c^2-1)\,f(c)^2 \\ 
=\frac2{(n-1)(n-2)}\,\int_{-\infty}^\infty dc\,F(c)^{n-1}(2c^3-5c)\,f(c). 
\end{multline*} 
If $j\in\{1,2\}$, then 
\begin{multline*}
 s_j=EX_1 X_2^2\,I\{X_2>X_1\vee \max_3^n X_i\} \\ 
 =\int_{-\infty}^\infty d(F(c)^{n-2})\,EX_1 I\{X_1<X_2\}X_2^2\,I\{X_2>c\} \\ 
 =-\int_{-\infty}^\infty d(F(c)^{n-2})\,Ef(X_2)X_2^2\,I\{X_2>c\} \\ 
 =-\int_{-\infty}^\infty d(F(c)^{n-2})\,\int_c^\infty dx\, f(x)^2x^2 \\ 
  =-\int_{-\infty}^\infty dc\,F(c)^{n-2}\, f(c)c^2 \\ 
 =\frac1{n-1}\,\int_{-\infty}^\infty dc\, F(c)^{n-1}\,(2c-c^3)f(c). 
\end{multline*}
So, each of the off-diagonal entries of the matrix $EY(XX^T-I)$ equals
\begin{multline*}
 \text{off-diags}=EX_1X_2\max_iX_i=2s_1+(n-2)s_3=
 \frac2{n-1}\int_{-\infty}^\infty dc\,F(c)^{n-1}(c^3-3c)\,f(c) \\
 =\frac2{(n-1)n}\,E(Y^3-3Y) \tag{3} 
\end{multline*}
-- cf. $(1)$, $(2)$. 
Mathematica claims that the value of the latter integral is $-\frac{1}{4 \sqrt{\pi }}\approx -0.141$ for $n=2,3$, 
but it cannot evaluate it in closed form for $n=4$ (an approximate value of that integral for $n=4$ is $-0.0969$). 
Yet, it is now easy to compute the diagonal and off-diagonal entries of the matrix $EY(XX^T-I)$ numerically, and it is also easy to find their asymptotics for large $n(=d)$, using standard methods of asymptotic analysis. 
Indeed, let us consider the integral 
\begin{equation*}
 J_n:=\int_{-\infty}^\infty dc\,F(c)^{n-1}(c^3-3c)\,f(c)
\end{equation*}
in $(3)$ for large $n$. Write
\begin{multline*}
 J_n=J_{1n}+J_{2n}+J_{3n},\\
  J_{1n}:=\int_{-\infty}^{c_n}\dots,\   J_{2n}:=\int_{c_n}^{d_n}\dots,\  J_{3n}:=\int_{d_n}^\infty\dots,  
\end{multline*}
where 
\begin{gather*}
 c_n:=F^{-1}(e^{-w_n/n}),\quad
 d_n:=F^{-1}(e^{-v_n/n}) \\ 
 v_n:=n^{-t_n},\quad w_n:=n^{t_n},\quad t_n:=1/\sqrt{\ln n}. 
\end{gather*}
Then 
\begin{gather*}
 0<v_n<w_n,\quad v_n\to0,\quad w_n=1/v_n\to\infty,\quad d_n>c_n\to\infty,\\
 1-F(c_n)\sim-\ln F(c_n)=w_n/n=n^{-1+o(1)},\quad 1-F(c_n)=\exp\{-c_n^2/(2+o(1))\}.   
\end{gather*}
Hence and from similar relations for $d_n$, we find that $c_n\sim\sqrt{2\ln n}\sim d_n$, and so, 
\begin{equation*}
c\sim\sqrt{2\ln n}\text{ uniformly in $c\in[c_n,d_n]$}. \tag{4}
\end{equation*}
Next, for real $c<c_n$ we have $F(c)^{n-1}<F(c_n)^{n-1}=e^{-(n-1)w_n/n}\sim e^{-w_n}
=\exp\{-e^{\sqrt{\ln n}}\}<\exp\{-\ln^2 n\}=o(1/n)$, whence 
\begin{equation*}
 |J_{1n}|\le F(c_n)^{n-1}\int_{-\infty}^{c_n} dc\,|c^3-3c|\,f(c)=o(1/n).
\end{equation*}
On the other hand, 
\begin{equation*}
 |J_{3n}|\le \int_{d_n}^\infty dc\,|c^3-3c|\,f(c)
 \sim d_n^3(1-F(d_n))\sim-d_n^3\ln F(d_n)\sim (\sqrt{2\ln n})^3 v_n/n
 =o(1/n).
\end{equation*}
Consider finally the integral $J_{2n}$, which is the main term of the asymptotics. In that integral, let us use the substitution $v=-n \ln F(c)$, so that the condition $c_n<c<d_n$ can be rewritten as $v_n<v<w_n$. 
Note also that $v=-n \ln F(c)\iff c=C_n(v):=F^{-1}(e^{-v/n})$. 
Also, by $(4)$, $C_n(v)\sim\sqrt{2\ln n}$ uniformly in $v\in[v_n,w_n]$. So, 
\begin{equation*}
 J_{2n}=\frac1n\,\int_{v_n}^{w_n}dv\,e^{-v}[C_n(v)^3-3C_n(v)]
 \sim\frac{(\sqrt{2\ln n})^3}n. 
\end{equation*}
Collecting all the pieces, we have 
\begin{equation*}
 \text{off-diags}=\frac2{(n-1)n}\,E(Y^3-3Y)
 \sim\frac{2^{5/2}\ln^{3/2} n}{n^2}. 
\end{equation*}
It is similar (and slightly simpler) to derive from $(2)$ the following:
\begin{equation*}
\text{diags}=\tfrac1n\,EY  
\sim\frac{2^{1/2}\ln^{1/2} n}{n}. 
\end{equation*}
So, one may also notice that 
\begin{equation*}
 \text{off-diags}\sim\tfrac2{n^2}\,(EY)^3. 
\end{equation*} 
Curiously, the off-diagonal entries of the matrix $EY(XX^T-I)$ seem to be negative for $n\in\{2,\dots,6\}$ but 
positive for $n\ge7$.