TITLE: Expectation involving maximum of Gaussian variables
QUESTION [7 upvotes]: Let $X\sim N(0, I_d)$ be a $d$-dimensional Gaussian random vector. Let $W_1, \ldots, W_k \in \mathbb{R}^d$ be $k$ fixed vectors in general positions. It is clear that $w_i^\top X, \ldots, w_k^\top X$ are jointly Gaussian random variables. Let $$ Y = \max _{i \in [k]} W_i^\top x ,$$ my question is how to compute $\mathbb{E} [ Y\cdot X]$. 
My idea is to use a smooth approximation of the max function and the Stein's identity. Let $$
f(y, \alpha ) = \alpha^{-1} \log \left [ \exp(\alpha y_1) + \ldots + \exp( \alpha y_k ) \right ],$$
it is known that $ \left |  \max_i y_i - f(y, \alpha) \right | \leq \log k / \alpha$. Then by Stein's identity, we consider 
\begin{align}
\mathbb{E} \left [ f(WX , \alpha) \cdot X \right ] = \mathbb{E} [ \nabla_{X} f(W X , \alpha) ] = \mathbb{E} \left [ \frac{ \sum_{i \in [k]} \exp ( \alpha \cdot W_i ^\top X ) \cdot W_i }{ \sum_{i \in [k]} \exp(\alpha \cdot W_i ^\top X )} \right ].
\end{align}
If I naively take $\alpha \rightarrow +\infty$, I would get 
\begin{align}
&\lim_{\alpha\rightarrow +\infty} \mathbb{E} \left [ f(WX , \alpha) \cdot X \right ]  = \mathbb{E} \left[ \sum_{j\in[k]} I\left\{   j =\arg\max _{i\in[k]} W_i^\top X \right \} \cdot W_j \right ] \\
&\quad = \sum_{j\in [k]} \mathbb{P}\left  (j =\arg\max _{i\in[k]} W_i^\top X   \right ) \cdot W_i. 
\end{align}
I wonder whether my naive derivation gets the correct answer and whether such derivation could be made rigorous.

REPLY [2 votes]: In the $d=2$ case the result is explicit and nice.
Let $x$ and $y$ be i.i.d standard normal, and let $(t,u)$ and $(v,w)$ be fixed vectors.  
Let $M = \begin{pmatrix}
t & u \\
v & w \\
\end{pmatrix},
\text{ and }
X =
\begin{pmatrix}
x \\
y \\
\end{pmatrix}
$.
We want $E\big[\max(MX)\,X\big]$.
The first coordinate is
\begin{align}
&E\big[\max(tx+uy,vx+wy)x\big] \\
={}&E\big[(tx+uy+vx+wy)x+\left|tx+uy-vx-wy\right|x\big]\ /\ 2 \\
={}&E\big[(tx+uy+vx+wy)x\big]\ /\ 2 \\
={}&(t+v)\ /\ 2
\end{align}
The first equation uses the identity $\max(a,b)=(a+b+|a-b|)/2$.  (This is the part specific to $d=2$:  are there any analogs with more than two variables being maximized?)
The second equation uses the symmetry of $(x,y)$ and $(-x,-y)$ in the distribution.
The third equation uses the standard normal variance and independence of $X$ and $Y$.
So using similar reasoning for the second coordinate,
$$
E\big[\max(MX)\,X\big]=
\frac{1}{2}
\begin{pmatrix}
t+v\\
u+w\\
\end{pmatrix}.
$$